29786323-Probablity - Probability Theory and Random...

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Unformatted text preview: Probability Theory and Random Processes Title : Probability Theory and Random Processes Course Code : 07B41MA106 (3­1­0), 4 Credits Pre­requisite : Nil Objectives : To study ● Probability: its applications in studying the outcomes of random experiments ● Random variables: types, characteristics, modeling random data ● Stochastic systems: their reliability ● Random Processes: types, properties and characteristics with special reference to signal processing and trunking theory. Learning Outcomes :students will be able to (i) model real life random processes using appropriate statistical distributions; (ii) compute the reliability of different stochastic systems; (iii) apply the knowledge of random processes in signal processing and trunking theory. Evaluation Scheme Evaluation Components Weightage ( in percent) Teacher Assessment 25 (based on assignments, quizzes,attendence etc.) T1 (1 hour) Reference Material : 1. T. Veerarajan. Probability, Statistics and Random processes. Tata McGraw-Hill. 2. J. I. Aunon & V. Chandrasekhar. Introduction to Probability and Random Processes. McGraw-Hill International Ed. 3. A. Papoulis & S. U. Pillai. Probability, Random Variables and Stochastic Processes. Tata WcGraw-Hill. 4. Stark, H. and Woods, J.M. Probability and Random Processes with Applications to Signal Processing. . Origins of Probability The study of probabilities originally came from gambling! Why are Probabilities Important? • They help you to make good decisions, e.g., – Decision theory • They help you to minimize risk, e.g., – Insurance • They are used in average-case time complexity analyses of - Computer algorithms. • They are used to model processes in - Engineering. Random Experiments • An experiment whose outcome or result can be predicted with certainty is called a deterministic experiment. •Although all possible outcomes of an experiment may be known in advance, the outcome of a particular performance of the experiment cannot be predicted owing to a number of unknown causes. Such an experiment is called a random experiment. •A random experiment is an experiment that can be repeated over and over, giving different results. •e.g A fair 6-faced cubic die, the no. of telephone calls received in a board in a 5-min. interval. Probability theory is a study of random or unpredictable experiments and is helpful in investigating the important features of these random experiments. Probability Definitions • For discrete math, we focus on the discrete version of probabilities. • For each random experiment, there is assumed to be a finite set of discrete possible results, called outcomes. Each time the experiment is run, one outcome occurs. The set of all possible outcomes is called the sample space. Example. If the experiment consists of flipping two coins, then the sample space is: S = {(H, H), (H, T), (T, H), (T, T)} Example. If the experiment consists of tossing two dice, then the sample space is: S = {(i, j) | i, j = 1, 2, 3, 4, 5, 6} More Probability Definitions • A subset (say E) of the sample space is called an event. In other words, events are sets of outcomes. ( If the outcome of the experiment is contained in E, then we say E has occurred.) For each event, we assign a number between 0 and 1, which is the probability that the event occurs. Example. If the experiment consists of flipping two coins, and E is the event that a head appears on the first coin, then E is: E = {(H, H), (H, T)} Example. If the experiment consists of tossing two dice, and E is the event that the sum of the two dice equals 7, then E is: E = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} Union: E∪ F Intersection:E ∩ F also denoted as EF (If E ∩ F = φ , then E and F are said to be mutual exclusive.) NOTE It may or may not be true that all outcomes are equally likely. If they are, then we assume their probabilities are the same. Definition of Probability in Text The probability of an event E is the sum of the probabilities of the outcomes in E p(E) = n(E)/n(S) number of cases favourable to E = exhasustive number of cases in S Example • Rolling a die is a random experiment. • The outcomes are: 1, 2, 3, 4, 5, and 6, presumably each having an equal probability of occurrence (1/6). • One event is “odd numbers”, which consists of outcomes 1, 3, and 5. The probability of this event is 1/6 + 1/6 + 1/6 = 3/6 = 0.5. Example • An urn contains 4 green balls and 6 red balls. What is the probability that a ball chosen from the urn will be green? • There are 10 possible outcomes, and all are assumed to be equally likely. Of these, 4 of them yield a green ball. So probability is 4/10 = 0.4. Example • What is the probability that a person wins the lottery by picking the correct 6 lucky numbers out of 40? It is assumed that every number has the same probability of being picked (equally likely). Using combinatorics, recall that the total number of ways we can choose 6 numbers out of 40 is: C(40,6) = 40! / (34! 6!) = 3,838,380. Therefore, the probability is 1/3,838,380. Examples • Consider an experiment in which a coin is tossed twice. • Sample space: { HH, HT, TH, TT } • Let E be the event that at least one head shows up on the two tosses. Then E = { HH, HT, TH } • Let F be the event that heads occurs on the first toss. Then F = { HH, HT } • A natural assumption is that all four possible events in the sample space are equally likely, i.e., each has probability ¼. Then the P(E) = ¾ and P(F) = ½. Probability as a Frequency Let a random experiment be repeated n times and let an event A occur n A times out of the n trials. nA The ratio is called the relative frequency of n the event A nA As n increases, shows a tendency to stabilise and to n approach a constant value. This value, denoted by P(A), is called the probability of the event A, i.e., P(A) = lim n →∞ nA . n Frequency Definition of Probability • Consider probability as a measure of the frequency of occurrence. – For example, the probability of “heads” in a coin flip is essentially equal to the number of heads observed in T trials, divided by T, as T approaches infinity. Pr( heads) ≈ lim T →∞ number of heads T Probability as a Frequency • Consider a random experiment with possible outcomes w1, w2, …,wn. For example, we roll a die and the possible outcomes are 1,2,3,4,5,6 corresponding to the side that turns up. Or we toss a coin with possible outcomes H (heads) or T (tails). • We assign a probability p(wj) to each possible outcome wj in such a way that: p(w1) + p(w2) +… + p(wn) = 1 • For the dice, each outcome has probability 1/6. For the coin, each outcome has probability ½. Example To find the probability that a spare part produced by a machine is defective. If , out of 10,000 items produced , 500 are defective, it is assumed that the probability of a defective item is 0.05 Axioms of Probability Axioms (where A and B are events): ∪ • 0 <= P(A) <= 1 • P(S) = 1; P({}) = 0 • P(A ∪ B) = P(A) + P(B) – P(A ∩ B) • If A and B are disjoint then P(A ∪ B) = P(A) + P(B) (mutually exclusive events) B A A ∩B B A A∪ B A • A Example • Rolling a die is a random experiment. • The outcomes are: 1, 2, 3, 4, 5, and 6. Suppose the die is “loaded” so that 3 appears twice as often as every other number. All other numbers are equally likely. Then to figure out the probabilities, we need to solve: p(1) + p(2) + p(3) + p(4) + p(5) + p(6) = 1 and p(3) = 2*p(1) and p(1) = p(2) = p(4) = p(5) = p(6). Solving, we get p(1) = p(2) = p(4) = p(5) = p(6) = 1/7 and p(3) = 2/7. • One event is “odd numbers”, which consists of outcomes 1, 3, and 5. The probability of this event is: p(odd) = p(1) + p(3) + p(5) = 1/7 + 2/7 + 1/7 = 4/7. Theorem 1 The probability of the impossible event is zero, i.e. , if Φ is the subset (event) containing no sample point, P( Φ ) = 0. Pr oof The certain event S and the impossible event Φ are mutually exclusive. Hence P(S ∪ Φ ) = P(S) + P( Φ ) But S ∪ Φ = S. ∴ P(S) = P(S) + P( Φ ) ∴ P( Φ ) = 0 Theorem If A and B are any 2 events, P(A ∪ B) = P(A) + P(B) - P(A ∩ B) ≤ P(A) + P(B) Pr oof : A is the union of the mutually exclusive events AB and AB and B is the union of the mutually exclusive events AB and AB. ∴ P( A ) = P( AB ) + P( AB ) & P(B ) = P( AB ) + P( AB ) S AB A AB AB B ∴ P( A ) + P( B ) [ = P( AB ) + P( AB ) + P( AB ) + P( AB ) = P( A ∪ B ) + P( A ∩ B ) Hence proved. S AB A AB AB B Conditional Probability If event F occurs, what is the probability that event E also occurs? This probability is called conditional probability and denoted as p(E|F). Definition of Conditional Probability If p(F) > 0, then p( E ∩ F ) p( E | F ) = p( F ) Example. An urn contains 8 red balls and 4 white balls. We draw 2 balls from the urn without replacement. What is the probability that both balls are red? Solution: Let E be the event that both balls drawn are red. Then p(E) = C(8, 2)/C(12, 2) or, we can solve the problem using conditional probability approach, Let E1 and E2 denote, respectively, the events that the first and second balls drawn are red. Then p(E1E2) = p(E1) p(E2 | E1 ) = (8/12) (7/11) Multiplication Rule p( E1 E2 E3 ...En ) = p( E1 ) p( E2 | E1 ) p( E3 | E1 E2 ) ... p( En | E1 E2 ...En−1 ) Example. A deck of 52 playing cards is randomly divided into 4 piles of 13 cards each. Compute the probability that each pile has exactly one ace. Solution: Define events E1 = the first pile has exactly one ace E2 = the second pile has exactly one ace E3 = the third pile has exactly one ace E4 = the fourth pile has exactly one ace p(E1E2E3E4) = p(E1)p(E2| E1)p(E3|E1E2)p(E4|E1E2E3) p(E1) = C(4,1)C(48,12)/C(52,13) p(E2|E1) = C(3,1)C(36,12)/C(39,13) p(E3|E1E2) = C(2,1)C(24,12)/C(26,13) p(E4|E1E2 E3) = C(1,1)C(12,12)/C(13,13) p(E1E2E3E4) ≈ 0.1055 Let E and F be events. We can express E as E = EF ∪ EFc Where Fc is the complementary event of F. Therefore, we have p(E) = p(EF) +p(EFc) = p(E|F)p(F) +p(E|Fc)p(Fc) Independent Events Two events E and F are independent if p(EF)=p(E)p(F) Two events are not independent are said to be dependent. p(EF) = p(E)p(F) if and only if p(E|F) = p(E). If E and F are independent, then so are E and Fc. Theorem If the events A and B are independent , the events A & B are also independents. Pr oof The events A ∩ B & A ∩ B are mutually exclusive such that P(A ∩ B) + P(A ∩ B) = P(B)(by addition theorem) P(A ∩ B) = P(B) − P(A ∩ B) = P(B) − P(A)P(B)(by product theorem) = P(B)[1 − P(A)] = P(A)P(B) Three events E, F, and G are independent if p(EFG) = p(E)p(F)p(G) p(EF) = p(E)p(F) p(EG) = p(E)p(G) p(FG) = p(F)p(G) Example. Two fair dice are thrown. Let E denote the event that the sum of the dice is 7. Let F denote the event that the first die is 4 and let G be the event that the second die is 3. Is E independent of F? Is E independent of G? Is E independent of FG? p(E) = 6/36 = 1/6 p(F) = 1/6 p(G) = 1/6 p(EF) = 1/36 ⇒ E and F are independent. p(EG) = 1/36 ⇒ E and G are independent. p(FG) = 1/36 ⇒ F and G are independent. p(EFG) = 1/36 ⇒ but, E and FG are NOT independent. Baye's Theorem Let B1 , B2 ,.......Bk be a partition of the sample space S. Let P(A/Bi ) & P(Bi / A) be the conditional probabilities for i = 1 to k, then P(Bi / A) = P ( A / Bi ) P ( Bi ) k ∑ P(A / B j ) P( B j ) i = 1,2,....., k j=1 Pr oof Given B1 , B2 ,....Bk be a partition of the sample space S. i.e., (i) Bi ∩ B j = φ∀i ≠ j. S B1 B2 B4 B3 A (ii) ∪ Bi = S(iii )P(Bi ) > 0∀i. Let A be the event associated with S. Then (A ∩ B1 ) ∪ (A ∩ B2 ) ∪ ....... ∪ (A ∩ Bk ) k i =1 = A ∩ (B1 ∪ B2 ) ∪ (A ∩ B3 ) ∪ ....... ∪ (A ∩ Bk ) (by distributive law) = A ∩ (B1 ∪ B 2 ∪ .... ∪ Bk ) = A ∩ S = A Also all the events (A ∩ B1 ), (A ∩ B2 ),......., (A ∩ Bk ) are pairwise mutually exclusive. For, (A ∩ B1 ) ∩ (A ∩ B2 ) ∩ ....... ∩ (A ∩ B k ) = A ∩ (Bi ∩ B j ), i ≠ j = A∩φ = φ Then P(A) = P(A ∩ B1 ) + P(A ∩ B2 ) + ... + P(A ∩ Bk ) However each term P(A B j ) may be expressed as P(A/B j )P(B j ) & hence we obtain the theorem on total probability. P(A) = P(A / B1 )P(B1 ) + P(A / B2 )P(B2 ) + ... + P(A / B k )P(B k ) k = ∑ P(A / B j )P(B j ) j=1 P ( Bi ∩ A ) = P ( Bi ) × P ( A / B i ) = P ( A ) × P ( Bi / A ) P ( Bi ) × P ( A / Bi ) ∴ P ( Bi / A ) = P(A) = P ( Bi ) × P ( A / Bi ) k ∑ P( B j ) × P(A / B j ) j=1 Example. In answering a question on a multiple-choice test, a student either knows the answer or guesses. Let p be the probability that the student knows the answer and 1−p the probability that the student guesses. Assume that a student who guesses at the answer will be correct with probability 1/m, where m is the number of multiple-choice alternatives. What is the (conditional) probability that a student knew the answer to a question, given that his answer is correct? Solution: C = the student answers the question correctly, K = the student actually knows the answer p( C | K ) p( K ) p( K | C ) = p( C | K ) p( K ) + p ( C |~ K ) p( ~ K ) p = p + (1 − p ) ( 1 / m ) mp = 1 + ( m − 1) p Example. When coin A is flipped it comes up heads with probability ¼, whereas when coin B is flipped it comes up heads with probability ¾. Suppose that one coin is randomly chosen and is flipped twice. If both flips land heads, what is the probability that coin B was the one chosen? Solution: C = coin B is chosen H = both flips show head p( H | C ) p( C ) p( C | H ) = p ( H | C ) p ( C ) + p ( H |~ C ) p( ~ C ) 3 3 1 × × 4 4 2 = 3 3 1 1 1 1 × × + × × 4 4 2 4 4 2 = 9 10 ≈ 0.9 Example. A laboratory test is 95 percent correct in detecting a certain disease when the disease is actually present. However, the test also yields a “false” result for 1 percent of the healthy people tested. If 0.5 percent of the population has the disease, what is the probability a person has the disease given that his test result is positive? Example. A suspect is believed 60 percent guilty. Suppose now a new piece of evidence shows the criminal is left-handed. If 20 percent of the population is left-handed,and it turns out that the suspect is also left-handed, then does this change the guilty probability of the suspect? By how much? Solution: D = the person has the disease E = the test result is positive p( E | D ) p( D ) p( D | E ) = p( E | D ) p( D ) + p( E |~ D ) p ( ~ D ) ( 0.95) ( 0.005) = ( 0.95) ( 0.005) + ( 0.01) ( 0.995) 95 = 294 ≈ 0.323 Example. A suspect is believed 60 percent guilty. Suppose now a new piece of evidence shows the criminal is left-handed. If 20 percent of the population is left-handed,and it turns out that the suspect is also left-handed, then does this change the guilty probability of the suspect? By how much? Solution: G = the suspect is guilty LH = the suspect is left-handed p ( LH | G ) p ( G ) p ( G | LH ) = p( LH | G ) p( G ) + p ( LH |~ G ) p( ~ G ) (1.0) ( 0.6) = (1.0) ( 0.6) + ( 0.2) ( 0.4) 60 = 68 ≈ 0.88 Random Variable Definition: A random variable (RV) is a function X : S → R that assigns a real number X(s) to every element s ∈ S,where S is the sample space corresponding to a random experiment E. Discrete Random Variable If X is a random variable (RV) which can take a finite number or countably infinite number of values, X is called a discrete RV. Eg. 1. The number shown when a die is thrown 2. The number of alpha particles emitted by a radioactive source are discrete RVs. Example Suppose that we toss two coins and consider the sample Space associated with this experiment. Then S={HH,HT,TH,TT}. Define the random variable X as follows: X is the number of heads obtained in the two tosses. Hence X(HH) = 2, X(HT) = 1 = X(TH) & X(TT) = 0. Note that to every s in S there corresponds exactly one value X(s). Different values of x may lead to the same value of S. Eg. X(HT) = X(TH) Probability Function If X is a discrete RV which can take the values x1 , x 2 , x 3 ,.... such that P(X = x i ) = p i , then p i is called the probability function or probability mass function or point probability function, provided p i (i = 1,2,3,...) satisfy the following conditions: (i)p i ≥ 0, ∀i,& (ii)∑ p i = 1 i Example of a Discrete PDF • Suppose that 10% of all households have no children, 30% have one child, 40% have two children, and 20% have three children. • Select a household at random and let X = number of children. • What is the pmf of X? Example of a Discrete PDF • We may list each value. – P(X = 0) = 0.10 – P(X = 1) = 0.30 – P(X = 2) = 0.40 – P(X = 3) = 0.20 Example of a Discrete PDF • Or we may present it as a chart. x 0 1 2 3 P(X = x) 0.10 0.30 0.40 0.20 Example of a Discrete PDF • Or we may present it as a stick graph. P(X = x) 0.40 0.30 0.20 0.10 0 1 2 3 x Example of a Discrete PDF • Or we may present it as a histogram. P(X = x) 0.40 0.30 0.20 0.10 0 1 2 3 x Example The pmf. of a random variable X is given by cλ p(i) = i = 0,1,2,....., where λ is some positive i! value. Find (i) P(X = 0)(ii)P(X > 2) i Solution. λ Since ∑ p(i) = 1, we have c ∑ = 1 i =0 i = 0 i! ∞ ∞ i λ λ s e = ∑ , we have ce = 1. i = 0 i! λ ∞ i -λ 0 eλ −λ Hence P(X = 0) = =e 0! P ( X > 2) = 1 − P ( X ≤ 2) −λ λe −λ = 1 − e + λe + 2 2 −λ -λ eλ P (X = x) = x! x If X represents the total number of heads obtained, when a fair coin is tossed 5 times, find the probability distribution of X. X :0 1 2 3 4 5 1 5 10 10 5 1 P: 32 32 32 32 32 32 Continuous Random Variable If X is an RV which can take all values (i.e., infinite number of values ) in an interval, then X is called a continuous RV. 0, x < 0, X( x ) = x ,0 ≤ x < 1, 1, x ≥ 1 Probability Density Function If X is a continuous RV,then f is said to be the probability density function (pdf) of X , if it satisfies the following conditions: ∞ (i)f ( x ) ≥ 0, ∀x ∈ R x , and (ii) ∫ f ( x )dx = 1 −∞ (iii) For any a, b with - ∞ < a < b < ∞, b P(a ≤ X ≤ b) = ∫ f ( x )dx a When X is a continuous RV a P(X = a ) = P(a ≤ X ≤ a ) = ∫ f ( x )dx = 0 a This means that it is almost impossible that a continuous RV assumes a specific value. Hence P(a ≤ X ≤ b) = P(a ≤ X < b) = P(a < X < b) Probability Density Function 0 if x < 0 f X ( x) = 1 − x /α if 0 ≤ x ⋅e α 0.6 0.5 0.4 0.3 0.2 0.1 0.0 f X (x ) -3 0 3 6 x 9 12 15 Example Check whether the function f(x) = 4x ,0 ≤ x ≤ 1 is a probability density function or not. 3 Solution. Clearly f(x) ≥ 0∀x ∈ [0,1]. 1 1 0 0 Also ∫ f ( x )dx = ∫ 4 x dx = 1. 3 A random variable X has the density function 1 / 4 - 2 < x < 2 f(x) = 0 elsewhere Obtain (i) P(X < 1)(ii)P( X > 1)(iii )P(2X + 3 > 5) (¾,1/2,1/4) Find the formula for the probability distribution of the number of heads when a fair coin is tossed 4 times. Cumulative Distribution Function (cdf) If X is an RV, discrete or continuous , then P(X<=x) is called the cumulative distribution function of X or distribution function of X and denoted as F(x). If X is discrete , F(x) = ∑ p j j X j≤x X If X is continuous, F(x) = P(-∞ < X ≤ x) = ∫ f(x)dx -∞ Probability Density Function 0.6 0.5 0.4 0.3 0.2 0.1 0.0 f X (x ) -3 0 if x < 0 f X ( x) = 1 − x / α if 0 ≤ x ⋅e α 0 3 6 x 9 12 15 Cumulative Distribution Function 1.2 1.0 0.8 0.6 0.4 0.2 0.0 F X (x ) -3 0 0 if x < 0 FX ( x) = − x /α 1 − e if 0 ≤ x 3 6 x 9 12 15 Probbility Density Function 0.0016 0.0012 f X (x ) 0.0008 0.0004 0.0000 -100 0 100 200 300 400 500 600 700 800 0 if x < 0 1 f X ( x) = if 0 ≤ x ≤ u u 0 if u < x x F X (x ) Cumulative Distribution Function 1.0 0.8 0.6 0.4 0.2 0.0 -100 0 100 200 300 400 500 600 700 800 0 if x < 0 x x FX ( x) = if 0 ≤ x ≤ u u 1 if u < x If the probability density of a random variable is K(1 - x 2 ),0 < x < 1 given by f(x) = 0 otherwise Find (i) K, (ii) the cumulative distribution function of the random variable. 3 / 2[ x − x 3 / 3] 0 < x < 1 = otherwise 0 A random variable X has the density fuction 1 if − ∞ < x < ∞ K. K = 1/ π f(x) = 1 + x 2 1 / π(tan x ),−∞ < x < ∞ F( x ) = 0 otherwise 0, otherwise Deter min e K and the distribution function. −1 Properties of the cdf F(x) 1.F(x) is a non - decreasing function of x, i.e. , if x1 < x 2 , then F(x1 ) ≤ F(x 2 ) . 2.F(−∞ ) = 0 & F(∞) = 1. 3. If X is a discrete RV taking values x1 , x 2 ,...., where x1 < x 2 < x 3 < .... < x i −1 < x i < ....., then P(X = x i ) = F( x i ) − F( x i −1 ). d 4.If X is a continuous RV, then F( x ) = f ( x ), at all dx points where F(x) is differentiable. Definitions If X is a discrete RV, then the expected value or the mean value of g(X) is defined as E{g(X)} = ∑ g(x i )p i , where p i = P(X = x i ) i the probability mass function of X. If X is a continuous RV with pdf f(x), then E{g(X)} = ∫ g(x)f(x)dx Rx Two expected values which are most commonly used for characterising a RV X are its mean µ x & var ianceσ 2 x µ x = E (X) = ∑ x i p i , if X is discrete i = ∫ xf(x)dx, if X is continuous Rx { Var ( x ) = σ = E (X − µ x ) 2 x 2 } = ∑ ( x i − µ x ) p i , if X is discrete 2 i = ∫ (x - µ x ) f ( x )dx , if X is continuous 2 Rx The square root of variance is called the standard deviation. Example Find the expected value of the number on a die when thrown.(7/2) Var (X) = E(X ) − {E(X)} 2 2 2 Var (X) = E{( X − µ x ) } = E{X − 2µ x X + µ x } 2 2 2 = E (X ) − 2µ x E (X) + µ x (sin ce µ x is a constant) 2 2 = E (X ) − µ x (sin ceµ x = E (X)) 2 Var (X) = E (X ) − {E (X)} 2 2 Example A random variable X has the following probability distribution x : - 2 -1 0 123 p(x) : 0.1 K 0.2 2K 0.3 3K (a) Find K, (b) Evaluate P(X < 2) and P(-2 < X < 2), (c) find the cdf of X and (d) evaluate the mean of X. The probability function of an infinite discrete distribution is given by P(X = j) = 1/2 ( j − 1,2,3,..). j Verify that the total probability is 1and find the mean and variance of the distribution. Find also P(X is even), P(X ≥ 5) & P(X is divisible by 3). xe x≥0 If p(x) = 0 x<0 (a )show that p(x) is a pdf(of a continuous RV X.) -x 2 / 2 (b) find its distribution function P(x). Example F( x ) = 1 − e − x / 2 , x ≥ 0 A continuous random variable has the probability 2 kxe -λx x ≥ 0, λ > 0 density function defined by f(x) = 0otherwise Deter min e the cons tan t k and find mean and variance. (k = λ ,2 / λ,6 / λ ) 2 2 Moments E(X r )is called rth If X is the discrete or continuous RV, order raw moment of S about the origin and denoted by µ'r . ∑ x r f ( x ) if X is discrete x r µ r ' = E(X ) = ∞ r ∫ x f ( x )dx if X is continuous - ∞ Since the first and second moments about the origin are given by µ1 = E (X) & µ1 = E (X 2 ), 1 2 mean = first moment about the origin () Var(X) = µ − ( E (X) ) = µ − µ . 1 2 2 1 2 12 1 E{(X − µ x ) n is called the nth order central moment of X and denoted by µ n . n n E{ X } & E{ X − µ x } are called absolute moments of X. n E{(X − a ) } & E{ X − a } are called generalised moments of X. n µ1 = E(X − E(X)) = E(X) − E(X) = 0 µ 2 = E (X − E(X)) = Var (X) 2 Two-Dimensional Random Variables Definitions : Let S be the sample space associated with a random experiment E. Let X = X(s) and Y = Y(s) be two functions each assigning a real number to each outcomes s ∈ S. Then (X, Y) is called a two - dimensional random variable. Two-dimensional continuous RV. Two-dimensional discrete RV. (X, Y) − ( x i , y i ), i = 1,2,3,...., m,......; j = 1,2,3,..., n,..... Probability Function of (X,Y) If (X, Y) is a two - dimensional discrete RV such that P(x = x i , y = y i ) = p ij then p ij is called the probability mass function of (X, Y) provided (i)p ij ≥ 0, ∀i & j (ii)∑ ∑ p ij = 1 ji The set of triplets{x i , y i , p ij}, i = 1,2,..., j = 1,2,3... is called the joint probability distribution of (X, Y) • If (X,Y) is a two-dimensional continuous RV. The joint probability density function (pdf) f is a function satisfying the following conditions f ( x, y) ≥ 0, ∞∞ ∫− ∞ ∫− ∞ −∞ < x < ∞ −∞ < y< ∞ f ( x, y)dxdy = 1 Pr[ x1 < X ≤ x 2 , y1 < Y ≤ y 2 ] = x 2 y2 ∫x1 ∫y1 f ( x, y)dydx Cumulative Distribution Function If (X, Y) is a two - dimensional RV(discrete or continuous), then F(x, y) = P{X ≤ x & Y ≤ y} is called the cdf of (X, Y) F( x , y) = ∑ ∑ p ij ji = xy ∫−∞ ∫−∞ f (u , v)dvdu •Properties of joint PDF 0 ≤ F ( x, y ) ≤ 1, −∞ < x < ∞ −∞ < y < ∞ F (−∞ , y ) = F ( x,−∞ ) = F (−∞ ,−∞ ) = 0 F ( ∞, ∞ ) = 1 F ( x, y ) is a nondecreasing function as either x or y, or both, increase. F (∞, y ) = FY ( y ) F ( ∞, x ) = F X ( x ) F ( x, y ) = Pr( X ≤ x, Y ≤ y ) ∂ 2 F ( x, y ) f ( x, y ) = ∂x∂y Examples tossing two coins X ... random variable associated with the first coin Y ... random variable associated with the second coin sample space {(0,0), (0,1), (1,0), (1,1)} 0 for tail, 1 for head F( y) x, 1 4 2 F(0,1) = 4 3 F(1,0) = 4 F(1,1) = 1 F(0,0) = 1/2 1/4 1 y x Example Three balls are drawn at random without replacement from a box containing 2 white,3red and 4 black balls. If X denotes the number of white balls drawn and Y denotes the no of red balls drawn, find the joint probability distribution of (X,Y). Solutions As there are only 2 white balls in the box, X can take the values 0,1,2and Y can take the values 0,1,2,3. P(X=0,Y=0)=P(drawing 3 balls none of which is white or red)=P(all the 3 balls drawn are black) = C3 / C3 = 1 / 21 4 9 P(X=0,Y=1)=3/14,P(X=0,Y=2)=1/7……… X Y 0 0 1 2 1 2 3 X Y 0 1 2 3 0 1/21 3/14 1/7 1/84 1 1/7 2/7 1/14 0 2 1/21 1/28 0 0 For the bivariate probability distribution of (X, Y) given below, find P(X ≤ 1), P(X ≤ 1, Y ≤ 3), P(X ≤ 1/Y ≤ 3) P(Y ≤ 3/X ≤ 1) & P(X + Y ≤ 4). 1 Y 2 3 X 0 0 0 1/32 2/32 2/32 3/32 1 1/16 1/16 1/8 2 1/32 1/32 1/64 1/64 0 (ans.7/8,9/32,18/32,9/28,13/32) 4 1/8 5 1/8 6 1/8 2/64 Example The joint density function of (X, Y) is given by 2e e 0 < x < ∞,0 < y < ∞ f(x, y) = 0otherwise Compute (i) P(X > 1, Y < 1), (ii)P(X < Y), (iii)P(X < a) -x −2 y ans. e -1 (1 − e − 2 ),1 / 3,1 − e −a Marginal probability density function For every fixed j p(xj, y1) + p(xj, y2) + p(xj, y3) + … = p{X= xj} = f(xj) and for every fixed k p(x1, yk) + p(x2, yk) + p(x3, yk) + … = p{Y= yk} = f(yk) The probability functions f(xj) and g(yk) are also called marginal probability density functions. f X (x) = ∞ ∫−∞ f ( x , y)dy, f Y ( y) = ∞ ∫−∞ f ( x , y)dx As an illustrative example, consider a joint pdf of the form 6 f ( x, y ) = (1 − x 2 y ) for 0 ≤ x ≤ 1, 5 =0 elsewhere 0 ≤ y ≤1 -Integrating this wrt y alone and wrt x alone gives the two marginal pdf - 6 x2 f X ( x ) = (1 − ) 5 2 6 y f Y ( y) = (1 − ) 5 3 0 ≤ x ≤1 0 ≤ y ≤1 Independent Random Variables Two random variables X and Y with joint probability density function f(x,y) and marginal probability functions f X ( x ) and f Y ( y) If F(x,y) = F(x) G(y) Or p(x,y) = f X ( x ) f ( y) Y for all x, y, then X and Y are independent. Example A machine is used for a particular job in the forenoon and for a different job in the afternoon. The joint probability of (X,Y), where X and Y represent the number of times the machine breaks down in the forenoon and in the afternoon respectively, is given in the following table. Examine if X and Y are independent RVs. XY 0 1 2 0 0.1 0.04 0.06 1 0.2 0.08 0.12 2 0.2 0.08 0.12 XY 0 1 2 0 0.1 0.2 0.2 1 0.04 0.08 0.08 2 0.06 0.12 0.12 X & Y are independent , if Pi* × P* j = Pij∀i, j P0* = f (0) = 0.1 + 0.04 + 0.06 = 0.2; P1* = 0.4; P2* = 0.4 P*0 = 0.5; P*1 = 0.2; P*2 = 0.3 P*0 × P0* = 0.2 × 0.5 = 0.1 = P00 P0* × P*1 = 0.2 × 0.2 = 0.04 = P01 Hence the RVs X and Y are independent The cumulative distribution function of the continuous random variable (X, Y) is given by 1 - e − e + e , x, y > 0 F(x, y) = 0otherwise Pr ove that X and Y are independent. -x -y -(x + y) x, y ≥ 0 ∂ F( x , y) e f ( x , y) = = ∂x∂y 0otherwise 2 −( x + y) e − x x ≥ 0 e − y y ≥ 0 f1 ( x ) = f 2 ( y) = 0otherwise 0otherwise 0x < 0 0 y < 0 F1 ( x ) = F2 ( y) = −x −x 1 − e x ≥ 0 1 − e y ≥ 0 −x −y F1 ( x )F2 ( y) = (1 − e )(1 − e ) = F( x , y) Example Let X and Y be the lifetimes of two electronic devices. Suppose that their joint pdf is given by f(x, y) = e -(x + y) , x ≥ 0, y ≥ 0 then X and Y are independent Example Suppose that f(x, y) = 8xy,0 ≤ x ≤ y ≤ 1. Check the independence of X and Y. Expectation of Product of random variables If X and Y are mutually independent random variables, then the expectation of their product exists and is E(XY) = E(X) E(Y) Example Assuming that the lifetime X and the brightness Y of a lightbulb are being modeled as continuous random variables Let the pdf be given by f(x, y) = λ1λ 2 e −( λ1x + λ 2 y ) ,0 < x < ∞,0 < y < ∞ Find the joint distribution function A line of length a units is divided into two parts. If the first part is of length X, find E(X), Var(X) and E{X(a-X)}. Expectation of Sum of random variables If X1, X2, …, Xn are random variables, then the expectation of their sum exists and is E(X1+ X2+…+ Xn) = E(X1) + E(X2) +… + E(Xn) E ( X ) + E ( Y ) = ∑ x j p ( x j , yk ) + ∑ yk p ( x j , yk ) j ,k j ,k = ∑ ( x j + yk ) p ( x j , yk ) j ,k = E( X + Y ) Example What is the mathematical expectation of the sum of points on n dice? Ans. (7/2)n n n A box contains 2 tickets among which C r tickets bear the number r (r = 0,1,2,…,n). A group of m tickets is drawn . Let S denote the sum of their numbers. Find E(S) and Var S. Ans. (n/2)m E ( XY ) = ∑ x j yk p ( x j , yk ) j ,k = ∑ x j yk f ( x j ) g ( yk ) j ,k ∑ x f ( x ) ∑ y g( y ) = j j k k j k = E( X ) E(Y ) Example If the joint pdf of (X, Y) is given by f(x, y) = 24y(1- x),0 ≤ y ≤ x ≤ 1 Find E(XY) y 11 E(XY) = ∫ ∫ xyf ( x , y)dxdy 0y y x= 0 Ans. 4/15 x Binomial Distribution (re-visit) Suppose that n Bernoulli trials, each of which results in a success with probability p and results in a failure with 1–p, are performed. If Sn represents the number of successes that occur in the n Bernoulli trials, then Sn is said to be a binomial random variable with parameter n and p. Let Xk be the number successes scored at the kth trial. Since Xk assumes only the values 0 and 1 with corresponding probabilities q and p, we have E(Xk) = 0 • q + 1 • p = p Since Sn = X1 + X2+…+ Xn We have E(Sn) = E(X1+ X2+…+ Xn) = E(X1) + E(X2) +… + E(Xn) = np Conditional Probability F ( x M ) = Pr[ X ≤ x M ] Pr{ x ≤ x , M } = Pr( M ) Pr( M ) > 0 Now, consider the case where the event M depends on some other random variable Y. Conditional Probability Density Function f ( x , y) or f ( y | x) = fX ( x ) f ( x , y) f ( x | y) = f Y ( y) - the continuous version of Bayes’ theorem f ( x | y ) fY ( y ) f ( y | x) = f X ( x) - another expression of the marginal pdf ∞ ∞ −∞ −∞ ∞ ∞ −∞ −∞ f X ( x) = ∫ f ( x, y )dy = ∫ f ( x | y ) fY ( y )dy fY ( y ) = ∫ f ( x, y )dx = ∫ f ( y | x) f X ( x)dx For the bivariate probability distribution of (X, Y) given below, find P(X ≤ 1), P(X ≤ 1, Y ≤ 3), P(X ≤ 1/Y ≤ 3) P(Y ≤ 3/X ≤ 1) & P(X + Y ≤ 4). 1 2 3 X 0 0 0 1/32 2/32 2/32 3/32 1 1/16 1/16 1/8 2 1/32 1/32 1/64 1/64 0 Y (ans.7/8,9/32,18/32,9/28,13/32) 4 1/8 5 1/8 6 1/8 2/64 Suppose that p(x,y) the joint probability mass function of X and Y , is given by p(0,0) =.4 p(0,1)=.2,p(1,0)=.1,p(1,1)=.3 Calculate the conditional probability mass function of X given that Y = 1 Ans. 2/5,3/5 Example Suppose that 15 percent of the families in a certain community have no children, 20% have 1, 35% have 2, & 30% have 3 children; suppose further that each child is equally likely (and independently) to be a boy or a girl. If a family is chosen at random from this community, then B, the number of boys, and G , the number of girls, in this family will have the joint probability mass function . j0 1 2 3 i 0 .15 .10 .0875 .0375 P(B = 1, G = 1) = P(BGorGB) 1 .10 .175 .1125 0 2 .0875 .1125 0 0 3 .0375 0 0 0 = P(BG ) + P(GB) 11 = .35 × × × 2 = .175 22 P(B = 2, G = 1) = P(BBGorBGBorGBB) 1 1 1 = P(BBG) + P(BGB) + P(GBB) = .30 × × × × 3 2 2 2 .9 = = .1125 8 If the family chosen has one girl, compute the conditional probability mass function of the number of boys in the family Ans.8/31,14/31,9/31,0 Example The joint density of X and Y is given by 12 x ( 2 − x − y ) ,0 < x < 1,0 < y < 1 f(x, y) = 5 0otherwise compute the conditional density of X , given that Y = y , where 0 < y < 1 f ( x , y) 6x ( 2 − x − y ) f ( x | y) = ans = 4 − 3y f Y ( y) Example The joint pdf of a two - dimensional RV is given by x2 f(x, y) = xy 2 + ,0 ≤ x ≤ 2,0 ≤ y ≤ 1. 8 1 1 1 Compute P(X > 1), P(Y < ), P(X > 1/Y < ), P(Y < / X > 1) 2 2 2 P(X < Y) & P(X + Y ≤ 1) (ans. 19/24,1/4,5/6,5/19,53/480,13/480) Variance of a Sum of random variables If X and Y are random variables, then the variance of their sum is Var(X + Y) = E({(X+Y) – (µ X + µ Y)}2) 2 Var(X + Y) = E({(X+Y) – (µ x+ µ y)} ) ( )( ) = E ( X − µ X ) + E ( Y − µ Y ) − 2E ( ( X − µ X ) ( Y − µ Y ) ) 2 2 = Var ( X ) + Var (Y ) − 2 E (( X − µ X )( Y − µY ) ) = Var ( X ) + Var ( Y ) − 2( E ( XY ) − µ X µY ) The covariance of X and Y is defined by Cov ( X , Y ) = E ( ( X − µ X ) ( Y − µY ) ) = E ( XY ) − µ X µY • If X and Y are mutually independent, then Cov(X,Y) = 0. Q: Is the reverse of the above true? • If X and Y are mutually independent, then Var(X + Y) = Var(X) + Var(Y) • If X1, …, Xn are mutually independent, and Sn = X1 + …+ Xn, then Var(Sn) = Var(X1) + … + Var(Xn) Q: Let Sn be a binomial random variable with parameter n and p. Show that Var(Sn) = np(1-p) Example Compute Var(X) when X represents the outcome when we roll a fair die. Solution Since P(X=i)=1/6, i = 1,2,3,4,5,6, we obtain 6 E (X ) = ∑ i P[X = i] 2 2 i =1 1 21 21 21 21 21 =1 ( ) + 2 ( ) +3 ( ) + 4 ( ) +5 ( ) + 6 ( ) 6 6 6 6 6 6 2 =91/6 Var (X) = E (X ) − E (X) 2 2 2 91 7 = − = 35 / 12 6 2 Compute the variance of the sum obtained when 10 independent rolls of a fair die are made. Ans 175/6 Compute the variance of the number of heads resulting from 10 independent tosses of a fair coin. The coefficient of correlation between X and Y denoted by ρ xy , is defined as ρ xy = C xy σxσy The correlation co-efficient is a measure of dependence between RV’s X and Y. If ρ xy = 0 , we say that X and Y are uncorrelated If E(XY) = 0 , X and Y are said to be orthogonal RV’s. ρ xy ≤ 1 or C xy ≤ σ x σ y Example Calculate the correlation coefficient for the following heights (in inches) of fathers (X) & their sons (Y): X 65 66 67 67 68 69 70 72 Y 67 68 65 68 72 72 69 71 (Ans .603) ρ xy = C xy σxσy ∑ xy = n 2 − ∑x − n n ∑x 2 ∑x ∑y n n 2 ∑y − n n ∑y 2 Example Let the random variables X and Y have the x + y 0 < x < 1,0 < y < 1 joint pdf f(x, y) = 0 elsewhere Find the correlation coefficient between X and Y. (ans. - 1/11) Example If X,Y and Z are uncorrelated RVs with zero means and standard deviations 5, 12 and 9 respectively and if U=X+Y and V=Y+Z find the correlation coefficient between U and V. (ans 48/65) Conditional Expected Values (X, Y) p ij g ( X, Y ) E{g (X, Y) / Y = y j} = ∑ g ( x i , y j )P(X = x i / Y = y j ) i = ∑ g(x i , y j ) i P{X = x i Y = y j} P{Y = y j} = ∑ g(x i , y j ) i ∞ E{g (X, Y) / Y} = ∫ g ( x , y)f ( x / y)dx −∞ ∞ E{g (X, Y) / X} = ∫ g ( x , y)f ( y / x )dy −∞ p ij p* j Conditional means ∞ µ y/x = E (Y / X) = ∫ yf ( y / x )dy −∞ Conditional variance are σ 2 y/x ∞ = E (Y − µ y / x ) = ∫ ( y − µ y / x ) f ( y / x )dy 2 −∞ 2 Example The joint pdf of (X,Y) is given by f(x,y)=24xy, x>0,y>0, x+y<=1,and 0, elsewhere, find the conditional mean and variance of Y, given X. f ( y / x ) = 2 y /(1 − x ) , E (Y / X) = 2 / 3(1 − x ), 2 var = 1 / 18(1 − x ) 2 If (X, Y) is uniformly distributed over the semicircle bounded by y = 1 - x & y = 0 , find E(X/Y) and E(Y/X) 2 Also verify that E{E(X/Y)} = E(X) and E{E(Y/X)} = E(Y) Properties (1)If X and Y are independent RV’s, then E(Y/X)=E(Y) and E(X/Y)=E(X). (2)E[E{g(X,Y)/X)=E{g(X,Y)} in particular E{E(X/Y)}=E(X) (3)E(XY)=E[X.E(Y/X)] E (X 2 Y 2 ) = E (X 2 E (Y 2 / X)] Example Three coins are tossed. Let X denote the number of heads on the first two coins,Y denote the no of tails on the last two, and z denote the number of heads on the last two. Find (a)The joint distribution of (i) X and Y (ii) X and Z (b) Conditional distribution of Y given X = 1 (c) Find covariance of x,y and x,z (d) Find E(Z/X=1) (e)Give a joint distribution , that is not the joint distribution of X and Z in (a), but has the same marginals as of (b) RVs (X1 , X 2 ,...X n ) − independent f ( x1 , x 2 ,..., x n ) = f ( x1 ) × f ( x 2 ) × ... × f ( x n ) conditional density f(x1 , x 2 , x 3 ) f(x1 , x 2 / x 3 ) = f( x 3 ) f(x1 , x 2 , x 3 ) f(x1 / x 2 , x 3 ) = f(x 2 , x 3 ) Definition Let X denote a random variable with probability density function f(x) if continuous (probability mass function p(x) if discrete) Then M(t) = the moment generating function of X () = E etX ∞ tx ∫ e f ( x ) dx if X is continuous = −∞ ∑ etx p ( x ) if X is discrete x Example xe x > 0, y > 0 If f ( x , y ) = 0otherwise Find the moment - generating function of XY. − x ( y +1) ∞∞ Solution : M XY = ∫ ∫ e xe xyt − x ( y +1) 00 ∞ −x ∞ xyt − x ( y ) = ∫ xe { ∫ e e 0 ∞ 0 −x ∞ = ∫ xe { ∫ e 0 dy}dx 0 − xy (1− t ) dy}dx =1/1-t dydx M X (t) = ∑ e f (x) tx Properties 2 ( tx ) + .....)f ( x ) = ∑ (1 + tx + 2! MX(0) = 1 2 2 tx M 'X ( t ) = ∑ ( x + + ...)f ( x ) 2! M 'X ( t ) t =0 = ∑ xf ( x ) = E(X) = µ1 3 6 tx M X ( t ) = ∑ (x + ...)f ( x ) 3! 2 2 M X ( t ) t =0 = ∑ x f ( x ) = µ 2 2 2 ( k) mX ( 0) = k th derivative of mX ( t ) at t = 0. µk k µ 2 2 µ3 3 mX ( t ) = 1 + µ1t + t + t + L + t + L. 2! 3! k! µk = E ( X k ) x k f ( x ) dx ∫ = k ∑ x p ( x) X continuous X discrete Let X be a random variable with moment generating function MX(t). Let Y = bX + a Then MY(t) = MbX + a(t) = E(e [bX + a]t) = eatMX (bt) Let X andY be two independent random variables with moment generating functionMX(t)and MY(t) . Then MX+Y(t) = MX (t) MY (t) 6. Let X and Y be two random variables with moment generating function MX(t) and MY(t) and two distribution functions FX(x) and FY(y) respectively. Let MX (t) = MY (t) then FX(x) = FY(x). This ensures that the distribution of a random variable can be identified by its moment generating function Example If X represents the outcome, when a fair die is tossed, find the MGF of X and hence find E(X) and Var(X). (Ans. 7/2,35/12) If a RV X has the MGF M(t) = 3/(3-t), obtain the standard deviation of X. (ans, 1/3) 1 t 2 2 t 3 3t 4 4 t M(t ) = e + e + e + e 10 10 10 10 Find p.d.f. 1 t 2 2 t 3 3t 4 4 t M(t ) = e + e + e + e 10 10 10 10 M(t ) = ∑ e f ( x ) tx x 1 t 2 2 t 3 3t 4 4 t at bt e + e + e + e = f (a )e + f (b)e + ... 10 10 10 10 x , x = 1,2,3,4 f ( x ) = 10 0, otherwise 1 2 ,1 < x < ∞ If X has the p.d.f f(x) = x 0, otherwise find the mean of X. The characteristic function of a random variable X is defined by φ X ( w ) = E(e iwx ) ∑ e iwx f ( x ),if X is discrete x =∞ iwx ∫ e f ( x )dx , if X is continuous − ∞ e iwx = cos wx + i sin wx = 1 =1 1/ 2 ( = cos wx + sin wx 2 2 ) 1/ 2 φ x ( w ) = E (e ∞ ≤∫e −∞ iwx iwx ∞ ) = ∫ e f ( x )dx iwx −∞ ∞ f ( x )dx = ∫ f ( x )dx = 1 −∞ Hence the characteristic function always exist even when moment-generating function may not exist. Properties of Characteristic Function iω 1.µ = E (X ) = the co - efficient of in the n! expansion of φ(ω) in series of asending powers of iω. ' n n n φ X ( w ) = E (e iwx n ) = ∑ e f (x) iwx x 2 3 ( iwx ) + ( iwx ) + ...f ( x ) = ∑ 1 + iwx + x 2! 3! ( iw ) = ∑ f ( x ) + iw ∑ xf ( x ) + x x 2! 2 ∑x x 2 f ( x ) + ..... 1 2.µ'n = n i d n φ(ω) dω ω= 0 n 3.If the characteristic function of a RV X is φ x (ω) and if Y = aX + b, then φ y (ω) = e ibωφ x (aω) 4. If X and Y are independent RVs, then φ x + y (ω) = φ x (ω) × φ y (ω). 5. If the characteristic function of a continuous RV X with 1∞ −ixω density function f(x) is φ(ω), then f(x) = dω. ∫ φ(ω)e 2π − ∞ 6.If the density function of x is known , the density function of Y = g(X) can be found from the CF of Y, provided Y = g(X) is one - one. The characteristic function of a random variable X is given by 1 − w , w ≤ 1 φx (w ) = 0, w > 1 Find the pdf of X. The pdf of X is 1∞ −iwx f(x) = dw ∫ φ x ( w )e 2π − ∞ 1 1 −iwx = dw ∫ (1 − w )e 2π −1 1 1 0 = (1 + w )e −iwx dw + ∫ (1 − w )e −iwx dw ∫ 0 2π −1 1 1 ix −ix = (2 − e − e ) = 2 (1 − cos x ) 2 2πx πx 1 sin ( x / 2 ) = ,−∞ < x < ∞ x/2 2π 2 Show that the distribution for which the characteristic function is e -ω has the density 1 1 function f(x) = × ,−∞ < x < ∞ 2 π 1+ x 1∞ − i ωx f (x) = dω ∫ φ(ω)e 2π − ∞ If X1 & X 2 are two independent RVs that follow Poisson distribution with parameters λ1 & λ 2 , prove that (X1 + X 2 ) also follows a Poisson distribution with parameter (λ1 + λ 2 ). Reproductive property of Poisson distribution φ x1 ( t ) = e λ1 ( e iω −1) φx 2 ( t ) = e λ 2 ( e iω −1) since X1 & X 2 are independent RVs, φ x1 + x 2 ( t ) = e ( λ1 + λ 2 ) ( e iω −1) Joint Characteristic Function If (X, Y) is a two - dimensional RV, then iω1X + iω2 Y E(e ) is called the joint characteristic function of (X, Y) and denoted by φ xy (ω1 , ω2 ). ∞∞ φ xy (ω1 , ω2 ) = ∫ ∫ e iω1x + iω2 y −∞ −∞ = ∑∑e i iω1x + iω2 y j f ( x , y)dxdy p( x i , y j ) (i)φ xy (0,0) = 1 ∂ m+n 1 mn (ii)E{X Y } = m + n φ xy (ω1 , ω2 ) m n i ∂ω1 ∂ω2 ω1 =0,ω2 =0 (iii)φ x (ω) = φ xy (ω ,0) & φ y (ω) = φ xy (0, ω ) (iv)If X and Y are independent φ xy (ω1 , ω2 ) = φ x (ω1 ) × φ y (ω2 ) and conversely. Compute the characteristic function of the discrete r.v.' s X and Y if the joint PMF is PXY 1 / 3, x = y = 0 1 / 6, x = ±1, y = 0 = 1 / 6, x = y = ±1 0, else. 1 1 φ XY ( w 1 , w 2 ) = ∑ ∑ e k = −1 l = −1 i ( w 1k + w 2 l ) PXY 1 iw1 0+iw 2 0 1 iw1 1 iw1 +iw 2 1 −iw1 1 −( iw1 +iw 2 ) =e +e+e +e +e 3 6 6 6 6 11 1 = + ( cos w1 + i sin w 1 ) + ( cos( w 1 + w 2 ) + i sin ( w 1 + w 2 ) ) + 36 6 1 1 ( cos w1 − i sin w1 ) + ( cos( w1 + w 2 ) − i sin ( w1 + w 2 ) ) 6 6 11 1 = + ( cos w 1 ) + ( cos( w 1 + w 2 ) ) 33 3 Example Two RVs X and Y have the joint characteristic function φ xy (ω1 , ω2 ) = e ( − 2 ω 1 2 −8 ω 2 2 ) . Show that X and Y are both zero mean RVs and also that they are uncorrelated 1 ∂ m+n mn (ii)E{X Y } = m + n φ xy (ω1 , ω2 ) m n i ∂ω1 ∂ω2 ω1 =0,ω2 =0 By the property of joint CF 1 ∂ ( − 2 ω12 −8ω2 2 ) E(X) = e i ∂ω1 ω = 0 ,ω [ ( −2ω =e 1 2 −8 ω 2 2 ) 4iω 1 1 ω1 = 0 , ω2 = 0 2 =0 =0 [ ( −2ω E(Y) = e 1 E(XY) = 2 i 1 2 −8 ω 2 2 )16iω 2 ω1 = 0 ,ω2 = 0 ∂ ( − 2ω12 −8ω2 2 ) e ∂ω1∂ω2 ω1 =0,ω2 =0 2 ∂ ( − 2 ω12 −8ω2 2 ) = e 16ω2 ∂ω1 ω1 =0,ω2 =0 ( −2ω = {−64ω1ω2 e 1 =0 2 −8 ω 2 2 )} ω1 = 0 , ω2 = 0 =0 C xy = E(XY) − E (X) × E (Y) = 0 Compute the joint characteristic function of X and Y if 1 1 2 2 f xy = exp − ( x + y ) 2π 2 Ans. 1 ∞ ∞ − ( x 2 + y2 ) iω1x + iω2 y 2 1 φ xy ( w 1 , w 2 ) = ∫ ∫e 2π − ∞ − ∞ e dxdy Random Variable Binomial Distribution The Bernoulli probability mass function is the density function of a discrete variable X having 0 and 1 as the only possible values The pmf of X is given by P(0) = P{X = 0} = 1-p P(1) = P{P = 1} = p where p, 0<=p<=1 is the probability that the trial is a success. random variable X is said to be Bernoulli random variable f pmf satisfies above equation for some p ∈ (0,1). An experiment consists of performing a sequence of subexperiments. If each subexperiment is identical, then the subexperiments are called trials. Bernoulli Trials • Each trial of an experiment that has only two possible outcomes (success or failure) is called a “Bernoulli trial.” • If p is the probability of success, then (1-p) is the probability of failure. • The probability of exactly k successes in n independent Bernoulli trials, with probability of success p and probability of failure q = 1-p, is given by a formula called the Binomial Distribution: C(n, k) pk q(n-k) Example of Bernoulli Trials • Suppose I roll a die and I consider a 3 to be success and any other number to be a failure. • What is the probability of getting exactly 5 successes if I roll the die 20 times? • Solution: C(20, 5) (1/6)5 (5/6)15 • What is the probability of getting 5 or more successes? Theorem If the probability of occurrance of an event (probability of success) in a single trial of a Bernoulli' s experiment is p, then the probability that the event occurs exactly r times out of n independent trials is equal to nC r q n − r p r , where q = 1 - p, the probability of failure of the event. Pr oof Getting exactly r successes means getting r successes and (n - r) failures simultaneously. P(getting r successes and n - r failures) =p q r n −r The trials, from which the successes are obtained are not specified. There are nC r ways of choosing r trials for successes. Once the r trials are chosen for successes, the remaining (n - r) trials should result in failures. These nC r ways are mutually exclusive. In each of these nC r ways, P(getting exactly r successes ) = p r q n − r . ∴, by the addition theorem, the required probability = nC r q n −r r p. Example If war breaks out on the average once in 25 years, find the probability that in 50 years at a strech, there will be no war. 1 24 p = , q = , n = 50; 25 25 0 50 1 24 24 P = C(50,0) = 25 25 25 50 Example Each of two persons A and B tosses 3 fair coins. What is the probability that they obtain the same number of heads? P(A&B get the same no. of heads) =P(they get no head each or 1 head each or 2 heads each or 3 heads each) = P(A gets 0 head) P(B gets o head)+-----32 32 1 1 = C(3,0) + C(3,1) 2 2 =5\16 32 32 1 1 + C(3,2) + C(3,3) 2 2 Example. A game consists of 5 matches and two players, A and B. Any player who firstly wins 3 matches will be the winner of the game. If player A wins a match with probability 2/3. Suppose matches are independent. What will be the probability for player A to win the game? Solution: Player A wins 3 matches or more out of the 5 matches. This event has probability equal to: 3 2 4 1 5 5 2 1 5 2 1 5 2 = + + 3 3 4 3 5 3 3 3 64 = 81 Example. A sequence of independent trials is to be performed. Each trial results in a success with probability p and a failure with probability 1–p. What is the probability that (a) at least 1 success occurs in the first n trials; (b) exactly k success occurs in the first n trials; Solution: (a)The probability of no success in the first n trials is (1-p)n. Thus, the answer is 1–(1–p)n. (b) n k n−k p (1 − p ) k Assuming that p remains the same for all repetitions, if we consider n independent repetitions (or trials) of E and if the random variable X denotes the number of times the event A has occurred, then X is called a binomial random variable with parameters n and p The pmf of a binomial random variable having parameters (n,p) is given by P(i) = nCi p i q n −i , i = 0,1,......, n , where p + q = 1 Example It is known that car produced by an automobile company will be defective with probability 0.01 independently of each other. The company sells the cars in packages of 10 and offers a money-back guarantee that atmost 1 of the10 cars is defective. What proportion of packages sold must the company replace? P(X > 1) = 1 − {P(X ≤ 1)} = 1 − {P(X = 0) + P(X = 1) = 1 − {10C 0 (.01) (.99) + (10C1 (.01)(.99) } = .004 0 10 9 n mean = E (X) = ∑ xnC x p (1 − p) x n = ∑ xnC x p (1 − p) n −x x =0 x n −x x =1 n (n − 1)! x n −x = ∑x p (1 − p) x!(n − x )! x =1 n n (n − 1)! x −1 n−x = ∑x pp (1 − p) x ( x − 1)![ (n − 1) − ( x − 1)]! x =1 n (n − 1)! x −1 n −x = np∑ p (1 − p) x =1 ( x − 1)![ ( n − 1) − ( x − 1) ]! n n = np∑ n − 1C x −1p x −1 (1 − p) ( n −1) −( x −1) x =1 = np(p + (1 − p)) n −1 , sin ce[ p + (1 − p)] = ∑ n C x p x (1 − p) n − x n n x =0 = np n Second moment µ1 = E(X 2 ) = ∑ x 2 nC x p x (1 − p) n − x 2 x =0 Show that variance is np(1-p) Example For a binomial distribution with mean 6, standard deviation 2 , find the first two terms of the distribution. 9 ((1 / 3) ,2 / 2187) The mean and variance of binomial distribution are 4 and 3 respectively. Find P(X ≥ 1). (1 − (3 / 4) ) = .9899 16 Poisson Distribution The probability density function of the poisson variate can be obtained as a limiting case of the Binomial probability density function under the following assumption. (i) The number of trials is increased indefinitely (n → ∞) (ii) The probability of success in a single trial is very small(p → 0) (iii) np is a finite constant say np = λ. Cosider the probability density function of a Binomial random variable X as P(X = x) = nC x p (1 − p) x n −x x n! = p x (1 − p)n − x x!(n − x )! n (n − 1)...(n − x + 1) λ λ = 1 − x! n n n −x n λ 1 − n (n − 1)....( n − x + 1) x n = λ x x x!n λ 1 − n n (n − 1) (n − x + 1) .... x nn n = λ x! n λ 1 − n x λ 1 − n For given x, as n → ∞, the terms 1 - 1/n,1 - 2/n,...1 - ( x - 1) /n, (1 - λ/n) x tends to 1. −λ Also, Lt n →∞ (1 − λ / n ) = e . n λx −λ Hence, Lt n →∞ P(X = x ) = e . x! Poisson random variable A random variable X taking on one of the values 0,1,2,... is said to be a Poisson random variable with parameter λ e -λ λx if for some λ > 0, P(x) = P(X = x) = x = 0,1,2,.... x! ∞ e − λ λx λx P( x ) =∑ = e −λ ∑ = e −λ e λ = 1 ∑ x! x =0 x =0 x = 0 x! ∞ ∞ e − λ λx ∞ e − λ λx Mean = E(X) = ∑ xP(X = x ) = ∑ x =∑ x =0 x =0 x =1 ( x − 1)! x! ∞ x −1 ∞ λ λλ −λ = λe ∑ = λe 1 + + + ... 1! 2! x =1 ( x − 1)! −λ ∞ −λ λ = λe e = λ Var (X) = λ Example The average number of radioactive particles passing through a counter during 1 millisecond is in a laboratory experiment is 4. What is the probability that 6 particles enter the counter in a given millisecond? If the probability of a defective fuse from a manufacturing unit is 2%, in a box of 200 fuses, find the probability that exactly 4 fuses are defective. −4 6 4 −4 e 4 4e , 6! 4! Example At a busy traffic intersection the probability p of an Individual car having an accident is very small say p=0.0001. However, during a certain peak hours of the day say between 4 p.m. and 6 p.m., a large number of cars (say 1000) pass through the intersection. Under these conditions what is the probability of two or more accidents occurring during that period. In a component manufacturing industry, there is a small probability of 1/500 for any component to be defective. The components are supplied in packets of 10. Use Poisson distribution to calculate the app no of packets containing (i) no defective (ii)one defective components in a consignment of 10,000 packets..9802 ×10000,.0196 ×10000 Binomial Distribution P(X = r )= C r p q n r n −r ; r = 0,1,2,..., n If we assume that n trials constitute a set and if we consider N sets, the frequency function of the binomial distribution is given by f(r)=N p(r) = N Cr p q n r n −r Example Fit a binomial distribution for the following data and hence find the theoretical frequencies: x:0 1 2 3 4 f: 5 29 36 25 5 The following data are the number of seeds germinating out of 10 on damp filter paper for 80 set of seeds. Fit a binomial distribution to these data: x 0 1 2 3 4 5 6 7 8 9 10 y 6 20 28 12 8 6 0 0 0 0 Ans. 7,26,37,34,6 6.89,19.14,23.94,17.74,8.63,2.88,0.67,0.1,0.01,0,0 0 Fit a Poisson distribution for the following distribution: x: 0 12345 f: 142 156 69 27 5 1 Fit a Poisson distribution to the following data which gives the number of yeast cells per square for 400 squares No. of cells per square (x) 0 1 No. of squares (f) 103 143 Ans. 147 147 74 25 6 1 107,141,93,41,4,0,0,0,0,0 2 3 45 6 7 8 9 10 98 42 8 4 2 0 0 00 GEOMETRIC DISTRIBUTION Suppose that independent trials, each having a probability p of success, are performed until a success occurs. If we let X = n, then P(X = n) = (1 - p) n -1 p, n = 1,2,... Since in order that X = n, it is necessary and sufficient that the first (n - 1) trials are failures and the nth trial is a success. ∞ ∞ n =1 n =1 ∑ P( X = n ) = ∑ (1 − p) n −1 Geometric Random Variable p p= = 1. 1 − (1 − p) In a chemical engineering process industry it is known that , on the average, 1 in every 100 items is defective. What is the probability that the fifth item inspected is the first defective item found. ans(.01)(.99) 4 = .0096 At busy time, a telephone exchange will be working busy with full capacity . So people cannot get a line to use immediately. It may be of interest to know the number of attempts necessary in order to get a connection. Suppose that p = 0.05, then find the probability that 5 attempts are necessary for a successful call connection. Ans .041 Mean ∞ E (X) = ∑ n (1 − p) n −1 n =1 p = p ∑ nt n =1 = p[1 + 2 t + 3t + 4 t + ....] 2 ∞ 3 n −1 , t = 1− p (sin ce t < 1) p1 = p[1 − t ] = p[1 − (1 − p)] = 2 = p p −2 −2 ∞ E(X ) = ∑ n (1 − p) 2 2 n −1 n =1 ∞ = ∑ [n (n − 1) + n ]t = ∑ [n (n − 1)]t n =1 ∞ n −1 p = ∑n t n −1 p ∞ p + ∑ nt = pt ∑ [n (n − 1)]t n =2 2 n −1 n =1 n =1 ∞ ∞ n =1 n −2 n −1 p +1/ p p = pt[2 + 3.2 t + 4.3t + 5.4 t + ....] + 1 / p 2 3 −3 = 2pt[1 + 3t + 6 t + ...] + 1 / p = 2pt[1 − t ] + 1 / p 2 2p(1 − p) 2 − p = =2 3 p p 1− p Var (X) = 2 p Negative Binomial Distribution Trials repeated until a fixed number of success occur. Instead of finding the probability of r success in n trials when n is fixed Probability that rth success occurs on the xth trial. Negative binomial experiment The number of x trials to produce r success in a negative binomial experiment is called a negative binomial random variable and its probability distribution is called the negative binomial distribution The negative binomial distribution is used when the number of successes is fixed and we're interested in the number of failures before reaching the fixed number of successes. An experiment which follows a negative binomial distribution will satisfy the following requirements 1.The experiment consists of a sequence of independent trials. 2.Each trial has two possible outcomes, S or F. 3.The probability of success,is constant from one trial to another. 4.The experiment continues until a total of r successes are observed, where r is fixed in advance Suppose we repeatedly throw a die, and consider a "1" to be a "success". The probability of success on each trial is 1/6. The number of trials needed to get three successes belongs to the infinite set { 3, 4, 5, 6, ... }. That number of trials is a (displaced) negative-binomially distributed random variable. The number of failures before the third success belongs to the infinite set { 0, 1, 2, 3, ... }. That number of failures is also a negative-binomially distributed random variable. A Bernoulli process is a discrete time process, and so the number of trials, failures, and successes are integers. For the special case where r is an integer, the negative binomial distribution is known as the Pascal distribution. A further specialization occurs when r = 1: in this case we get the probability distribution of failures before the first success (i.e. the probability of success on the (k+1)th trial), which is a geometric distribution. Let the random variable y denotes the no of failures before the occurrence of the rth success. Then y+r denotes the number of trials necessary to produce exactly r success and y failures with the rth success occurring at the (y+r)th trial. pmf of y = g(y) = = = [ ( y +r −1) C ( r −1) p ( y + r −1) r −1 [ ( n −1) C ( r −1) p r −1 q p, q = 1 − p y r C ( r −1) p q y y qp Example Pat is required to sell candy bars to raise money for the 6th grade field trip. There are thirty houses in the neighborhood, and Pat is not supposed to return home until five candy bars have been sold. So the child goes door to door, selling candy bars. At each house, there is a 0.4 probability of selling one candy bar and a 0.6 probability of selling nothing. What’s the probability that Pat finishes on the tenth house? A fair die is cast on successive independent trials until the second six is observed. The probability of observing exactly ten non-sixes before the second six is cast is …….. Find the probability that a person tossing three coins will get either all heads or all tails for the second time on the fifth toss. 2 11 10 2 1 5 4 1 3 C1 , C1 6 6 4 4 3 Relationship between the binomial and negative binomial Let X have a binomial distribution with parameters n,p. Let Y have a negative binomial distribution with parameters r and p. (That is Y = no of trials required to obtain r successes with probability of success p). Then (i)P(Y ≤ n ) = P(X ≥ r ) (ii)P(Y > n ) = P(X < r ) The probability that an experiment will succeed is 0.8. If the experiment is repeated until four successful outcomes have occurred, what is the expected number of repetitions required? Ans. 1 ∞ ∞ y =0 y =0 ∑ g(y) = ∑ [ ( y + r −1) r C ( r −1) p q y ( y + r − 1)! r y =∑ pq y =0 ( r − 1)! y! ∞ ∞ =∑ y =0 [ ( y + r −1) r∞ Cyp q = p ∑ r y y =0 ( y + r − 1)! y =p ∑ q y =0 ( r − 1)! y! r∞ [ ( y + r −1) C yq y ( r − 1)! ( r + 1)! q 2 + ... r! =p + q+ ( r − 1)!2! ( r − 1)! ( r − 1)! r r ( r + 1) 2 = p 1 + rq + q + ... 2! r = p [1 − q ] = p p = 1. r −r r −r rq rq Mean = E(X) = Variance = 2 p p PROBABILITY DISTRIBUTIONS Discrete Distributions Binomial Distribution P(X = r )= C r p q n r n −r ; r = 0,1,2,..., n If we assume that n trials constitute a set and if we consider N sets, the frequency function of the binomial distribution is given by f(r)=N p(r) = N Cr p q n r n −r Example Fit a binomial distribution for the following data and hence find the theoretical frequencies: x:0 1 2 3 4 f: 5 29 36 25 5 The following data are the number of seeds germinating out of 10 on damp filter paper for 80 set of seeds. Fit a binomial distribution to these data: x 0 1 2 3 4 5 6 7 8 9 10 y 6 20 28 12 8 6 0 0 0 0 Ans. 7,26,37,34,6 6.89,19.14,23.94,17.74,8.63,2.88,0.67,0.1,0.01,0,0 0 Fit a Poisson distribution for the following distribution: x: 0 12345 f: 142 156 69 27 5 1 Fit a Poisson distribution to the following data which gives the number of yeast cells per square for 400 squares No. of cells per square (x) 0 1 No. of squares (f) 103 143 Ans. 147 147 74 25 6 1 107,141,93,41,4,0,0,0,0,0 2 3 45 6 7 8 9 10 98 42 8 4 2 0 0 00 Probbility Density Function 0.0016 0.0012 f X (x ) 0.0008 0.0004 0.0000 -100 0 100 200 300 400 500 600 700 800 0 if x < 0 1 f X ( x) = if 0 ≤ x ≤ u u 0 if u < x x F X (x ) Cumulative Distribution Function 1.0 0.8 0.6 0.4 0.2 0.0 -100 0 100 200 300 400 500 600 700 800 0 if x < 0 x x FX ( x) = if 0 ≤ x ≤ u u 1 if u < x If X is uniformly distributed over the interval [0,10] compute the probability (a) 2<X<9(b) 1<X<4 (c)X<5 (d) X>6 Ans. 7/10,3/10,5/10,4/10 1 mean = E (X) = ∫ xf ( x )dx = ∫ x dx a a b−a b 1 mean = (b + a ) 2 b If X has uniform distribution in (-3,3), find P( X - 2 < 2) Ans. 1/2 If X has uniform distribution in (-a, a), a > 0, find a such that P( X < 1) = P( X > 1) a=2 Buses arrive at a specific stops at 15min. Intervals starting at 7 A.M., that is , they arrive at 7,7:15,7:30 and so on. If a passenger arrives at the stop at a random time that is uniformly distributed between 7 and 7:30A.M., find the probability that he waits (a) less than 5 min (b) at least 12 min. for a bus. Ans. 1/3,1/5 Example: The total time it takes First Bank to process a loan application is uniformly distributed between 3 and 7 days. What is the probability that the application will be processed in less than 4 days? Total Area 1 b− a = 25 . (b − )( a 0 5 3 a 1 ) b− a 7 4 6.5 P (3 ≤ ≤ ) = 4 − )( x 4 ( 3 b 1 ) =25 . 7− 3 What is the probability that it will take more than 6.5 days? P (6.5 ≤ ≤ ) = 7 − .5)( x 7 ( 6 1 ) =125 . 7− 3 var iance = E(X 2 ) − E (X) 2 a+b E(X) = 2 1 3 E (X ) = ∫ x f ( x )dx = x a 3(b − a ) b 2 (b = 2 − a ) b + ab + a = 3(b − a ) 3 3 3 2 b a 2 a + ab + b a + 2ab + b var = − 3 4 2 2 2 a + b − 2ab ( b − a ) = = 12 12 2 2 2 2 n +1 n +1 b −a moments = E(X ) = ( n + 1) (b − a ) n central moments µ r = E{(X - E(X)) } r r 1 b a + b = ∫ x − dx b−a a 2 a +b x − 1 2 = b−a r +1 r +1 b a r +1 b−a a −b − 2 2 = ( b − a ) ( r + 1) r +1 0 if r is odd r = 1 b − a if r is even r +1 2 2n µ 2 n −1 = 0, µ 2 n 1 b−a = for n = 1,2,3,.. 2n + 1 2 mean deviation = E( X - E(X) a+b 1 = ∫ x− * dx a 2 b−a b 1 = (b − a) 4 The gamma function, denoted by Γ is defined as ∞ Γ(x) = ∫ e t - t x −1 dt , x > 0 0 Pr operties [ + ∫ ( x − 1)t − t x −1 ∞ 0 1.Γ( x ) = − e t ∞ x −2 − t 0 ∞ = ( x − 1) ∫ e − t t ( x −1) −1dt = ( x − 1)Γ( x − 1) 0 ∞ 2.Γ(1) = ∫ e dt = 1 0 −t e dt 3.put x = n Γ(n) = ( n - 1) Γ( n - 1) = ( n − 1) ( n − 2 ) Γ(n − 2) = ...... = (n − 1)(n − 2)....2.1 Γ(1) = 0!= 1 = (n − 1)! Exponential Distribution If the occurances of events over nonoverlapping intervals are independent, such as arrival times of telephone calls or bus arrival times at a bus stop, then the waiting time distribution of these events can be shown to be exponential • Time between arrivals to a queue (e.g. time between people arriving at a line to check out in a department store. (People, machines, or telephone calls may wait in a queue) • Lifetime of components in a machine Exponential distribution λe -λx x ≥ 0 f(x) = otherwise 0 parameter λ > 0 Exponential distribution λe -λx x ≥ 0 f(x) = otherwise 0 parameter λ > 0 ∞ moments = µ'r = E (X ) = ∫ x λe r r − λx 0 1 ∞ r −y Γ(r + 1) r! = r ∫ y e dy = =r r λ0 λ λ dx Example Let X have an exponential distribution with mean of 100 . Find the probability that X<90 Ans. 0.593 Customers arrive in a certain shop according to an approximate Poisson process at a mean rate of 20 per hour. What is the probability that the shopkeeper will have to wait for more than 5 minutes for his first customer to arrive? ans. e -15 Memoryless Property of the Exponential Distribution If X is exponentially distributed, then P(X > s + t/X > s) = P(X > t), for any s, t > 0 ∞ P ( X > k ) = ∫ λe − λx ( dx = − e ) − λx ∞ k =e − λk P{ X > s + t & X > s} Now P(X > s + t/X > s) = P{ X > s} −λ (s+ t ) − λt P{ X > s + t} e = e = P(X > t ). = = − λs P{ X > s} e k If x represents the lifetime of an equipment then above property states that if the equipment has been working for time s, then the probability that it will survive an additional time t depends only on t (not on s) and is identical to the probability of survival for time t of a new piece of equipment. Equipment does not remember that it has been in use for time s. A crew of workers has 3 interchangeable machines, of which 2 must be working for the crew to its job. When in use, each machine will function for an exponentially distributed time having parameters λ before breaking down. The workers decide to initially use machines A and B and keep machine C in reserve to replace whichever of A or B breaks down first. They will then be able to continue working until one of the remaining machines breaks down. When the crew is forced to stop working because only one of the machines has not yet broken down, what is the probability that the still operable machine is machine C? Suppose the life length of an appliance has an exponential distribution with λ = 10 years. A used appliance is bought by someone. What is the probability that it will not fail in the next 5 years? 0.368 Suppose that the amount of waiting time a customer spends at a restaurant has an exponential distribution with a mean value of 5 minutes Then find the probability that a customer will spend more than 10 minutes in the restaurant 0.1353 Example Suppose that the length of a phone call in minutes is an exponenetial random variable with parameter λ= 1 10. If A arrives immediately ahead of B at a public telephone booth, find the probability that B will have to wait (i) more than 10 minutes, and (ii) between 10 and 20 minutes. (ans. 0.368,0.233) Erlang distribution or General Gamma distribution A continuous RV X is said to follow an Erlang distribuion or General Gamma distribuion with parameter λ > 0, k > 0, if its pdf is given by λk x k −1e −λx , for x ≥ 0 f(x) = Γ(k ) 0, otherwise ∞ ∫ f ( x )dx 0 =1 λ = 1, k −1 − x xe pdf = , x ≥ 0; k > 0. Γ(k ) Gamma distribution or simple gamma distribution with parameter k. When k = 1, the Erlang distribution reduces to the exponential distribution with parameter λ > 0. Example The random variable X has the gamma distribution with density function given by 0 for x ≤ 0 f(x) = -2x 2e for x > 0 Find the probability that X is not smaller than 3. ∞ P(X ≥ 3) = 2 ∫ e 3 −2 x e −6 dx = 2 =e −2 −2 x Suppose that an average of 30 customers per hour arrive at a shop in accordance with a Poisson Process That is, if a minute is our unit , then λ =1 / 2 What . is the probability that the shopkeeper wait more than 5 minutes before both of the first two customer arrive? Solution. If X denotes the waiting time in minutes until the second customer arrives, then X has Erlang(Gamma) distribution with k = 2, λ =1 / 2 λ ( λx ) − x P(X > 5) = ∫ e dx 5 Γ(k ) ∞ =0.287 k −1 Mean and Variance of Erlang Distribution moments = µ'r = E (X ) r λ k + r −1 −λx =∫ x e dx 0 Γ(k ) k λ 1 ∞ k + r −1 − t = e dt ∫t k +r Γ(k ) λ 0 ∞ k k mean = E(X) = λ 1 Γ( k + r ) =r λ Γ(k ) k var(X) = 2 λ m.g.f . = M X ( t ) = E (e ) tx k∞ λ k −1 − λx tx λ k −1 − ( λ − t ) x =∫ x e e dx = dx ∫x e 0 Γ(k ) Γ(k ) 0 ∞ k λ 1 ∞ k −1 − y = y e dy k∫ Γ(k ) ( λ − t ) 0 k t = 1 − λ −k λ = k (λ − t) k Reproductive Property M ( X1 + X 2 +...+ X n ) ( t ) = M X1 ( t )M X 2 ( t ).....M X n ( t ) M X1 ( t )M X 2 ( t ).....M X n ( t ) k = 1 − λ − k1 k 1 − λ − ( k1+ k 2 ...+ k n ) t = 1 − λ = M ( X1 + X 2 +...+ X n ) ( t ) −k 2 k .....1 − λ −k n The sum of a finite number of Erlang variables is also an Erlang variable. if X1 , X 2 ,...., X n are independent Erlang variables with parameters (λ, k1 ), (λ, k 2 ),......(λ, k n ), then X1 + X 2 + .... + X n is also an Erlang variable with parameter (λ, k1 + k 2 + ..... + k n ) If a company employes n sales persons, its gross sales in thousands of rupees may be regarded as RV having an Erlang distribuition with λ = 1/2, and k = 80 n . If the sales cost is Rs. 8000 per person, how many salespersons should the company employ to maximise the expected profit? Let X represent the gross sales (in Rupees) by n Sales persons X will have Erlang distribution Weibull Distribution β −1 − αx β f ( x ) = αβ x e parameter α, β ,x > 0 α, β > 0 when β = 1, Weibull distribution reduces to the exponential distribution with parameter α. ∞ µ r ' = E(X ) = αβ ∫ x r r +β −1 − αx β e dx 0 y = ∫ 0 α ∞ r 1 +1− β β y e α −y 1 −1 β dy 1 − β 1 Mean = E (X) = µ'1 = α Γ + 1 β 2 2 1 −2 / β Var (X) = α Γ + 1 − Γ + 1 β β Each of the 6 tubes of a radio set has a life length(in years) which may be considered as a RV that follows a Weibull distribution with parameters α = 25 and β = 2. If these tubes function independently of one another, what is the probability that no tube will have to be replaced during the first 2 months of service? If X represents the life length of each tube, then β-1 − αx β its density function f(x) is given by f(x) = αβ x e ,x > 0 f ( x ) = 50 xe −25 x 2 ,x > 0 P( a tube is not to be replaced during the first 2 months) ∞ = p(X > 1 / 6) = ∫ 50 xe 1/ 6 =e − 25 x 2 ( dx = − e − 25 x 2 − 25 / 36 P(all the 6 tubes are not to be replaced during the first 2 months) ( =e ) - 25/36 6 = 0.0155 ) ∞ 1 6 Properties of a Normal Distribution • Continuous Random Variable • Symmetrical in shape (Bell shaped) • The probability of any given range of numbers is represented by the area under the curve for that range. • Probabilities for all normal distributions are determined using the Standard Normal Distribution. Probability for a Continuous Random Variable Probability Density Function for Normal Distribution x −µ 1 −1 ( ) 2 f (x ) = eσ σ 2π 2 − ∞ < x < ∞ ,−∞ < µ < ∞ , σ > 0 N(µ, σ) x +7 1 1 −( ) f (x) = e2 4 32π 2 − ∞ < x < ∞,−∞ < µ < ∞, σ > 0 N(−7,4) ∞ ∫ f ( x )dx −∞ =1 Standard Normal Distribution N(0,1) 1 φ(z) = e 2π z2 − 2 ,−∞ < z < ∞. µ = 0, σ = 1 & by changing x and f respectively into z and φ. X -µ If X has distribution N(µ, σ) and if Z = , σ then Z has distribution N(0,1) z values of φ(z), ∫ φ(z)dz are tabulated. 0 X − N(µ, σ) ∞ E(X) = ∫ xf ( x )dx −∞ 1 ∞ − ( x −µ ) 2 / 2 σ 2 = dx ∫ xe σ 2π − ∞ 1∞ x −µ −t 2 = t = ∫ µ + 2σt e dt π −∞ σ 2 ( ) µ ∞ −t 2 2 ∞ −t 2 = σ ∫ te dt ∫ e dt + π −∞ π −∞ µ = π = µ. π similarly, var iance = σ 2 Figure 6.3 Figure 6.5 Determining the Probability for a Standard Normal Random Variable • P(-∞≤ Z ≤ 1.62) = .5 + .4474 = .9474 • P(Z > 1.62) = 1 - P(-∞≤ Z ≤ 1.62) = 1 - .9474 = .0526 z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0190 0.0239 0.0279 0.0319 0.0359 0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753 0.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141 0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517 0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879 0.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224 0.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.2549 0.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852 0.8 0.2881 0.2910 0.2939 0.2969 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133 0.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389 1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3513 0.3554 0.3577 0.3529 0.3621 1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830 1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015 1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177 1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319 1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441 1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545 1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633 1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706 1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767 2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817 2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857 2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890 2.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916 2.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936 2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952 2.6 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964 2.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.4974 2.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.4981 2.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.4986 3.0 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990 3.1 0.4990 0.4991 0.4991 0.4991 0.4992 0.4992 0.4992 0.4992 0.4993 0.4993 3.2 0.4993 0.4993 0.4994 0.4994 0.4994 0.4994 0.4994 0.4995 0.4995 0.4995 3.3 0.4995 0.4995 0.4995 0.4996 0.4996 0.4996 0.4996 0.4996 0.4996 0.4997 3.4 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4998 Determining the probability of any Normal Random Variable Interpreting Z • In figure Z = -0.8 means that the value 360 is .8 standard deviations below the mean. • A positive value of Z designates how many standard deviations (σ) X is to the right of the mean (µ). • A negative value of Z designates how may standard deviations (σ) X is to the left of the mean (µ). Example: A group of achievement scores are normally distributed with a mean of 76 and a standard deviation of 4. If one score is randomly selected what is the probability that it is at least 80. σ =4 76 80 x − µ 80 − 76 Z= = =1 σ 4 P( x ≥ 80) = P(z ≥ 1) = .5 − P(0 ≤ z ≤ 1) = .5 − .3413 = .1587 σ =4 76 80 x − u 80 − 76 Z= = =1 σ 4 P ( x ≥ 80) = P ( z ≥ 1) = .5 − P (0 ≤ z ≤ 1) = .3413 .5 − .3413 = .1587 0 .1587 1 Continuing, what is the probability that it is less than 70. 70 76 70 − 76 Z= = = −1.5 σ 4 P ( x ≤ 70) = P ( z ≤ −1.5) = .5 − P ( −1.5 ≤ z ≤ 0.0) = .5 −.4332 = .0668 .4332 x −u .0668 -1.5 0 -1.5 .4878 .4332 What proportion of the scores occur within 70 and 85. 0 2.25 x −u 70 −76 Z= = = − .5 1 σ 4 x −u 85 −76 Z= = = 2.25 σ 4 P (70 ≤ x ≤85) = P ( − .5 ≤ z ≤ 2.25) = 1 P ( − .5 ≤ z ≤ 0.0) + P (0.0 ≤ z ≤ 2.25) = 1 .4332 +.4878 =.9210 Time required to finish an exam is known to be normally distributed with a mean of 60 Min. and a Std Dev. of 12 minutes. How much time should be allowed in order for 90% of the students to finish? σ =12 .9 x −µ z= σ zσ = x − µ zσ + µ = x 1.28(12) + 60 = x x = 75.36 60 x An automated machine that files sugar sacks has an adjusting device to change the mean fill per sack. It is now being operated at a setting that results in a mean fill of 81.5 oz. If only 1% of the Sacks filled at this setting contain less than 80.0 oz, what is the value of the variance for this population of fill weights. (Assume Normality). .01 -2.33 µ = .5 81 Z= 0 x= 80 PROB = 01 . x− u σ 80.0 − .5 81 − .33 = 2 σ (80.0 − .5) 81 = 6437 . − .33 2 σ= σ =.4144 2 Moment generating function of N(0,1) ∞ M Z ( t ) = E(e tZ ) = ∫ e tz φ(z)dz −∞ 1 ∞ tz − z 2 / 2 1 ∞ −( z 2 − 2 tz ) / 2 = dz = ∫e e dz ∫e 2π − ∞ 2π − ∞ 1 ∞ −( z − t ) = e ∫ 2π − ∞ =e t2 / 2 2 −t 2 / 2 dz = e 1 ∞ − u du = e ∫e 2π − ∞ 2u t2 / 2 t2 / 2 1∞ ∫e 2π − ∞ ( z−t ) 2 − 2 1 t2 / 2 Γ(1 / 2) = e π dz The moment generating function of N(µ, σ) = M X ( t ) = M σZ+µ ( t ) = E (e t ( σz +µ) µt ) µt = e M Z (σt ) µt σ 2 t 2 / 2 =e e tσz = e E (e ) (( )) =e t µ+ σ2 t / 2 t σt t σt2 = 1 + (µ + ) + (µ + ) + ..... + ∞ 1! 2 2! 4 2 2 2 If X has the distribution N(µ, σ) then Y = aX + b has the distribution N(aµ + b, aσ) M X (t) = e ( ( ) ) t µ+ σ2 t / 2 M Y ( t ) = M aX + b ( t ) = e M X (at ) bt (( =e e )) bt t aµ + a 2 σ 2 t / 2 ( =e ( )) t ( aµ + b ) + a 2 σ 2 t / 2 MGF N(aµ + b, aσ) X -µ If X has distribution N(µ, σ), then Z = σ 1 µ1 has the distribution N( µ − , .σ) = N(0,1) σ σσ Additive property of normal distribution If X i (i = 1,2,...., n ) be n independent normal RVs with mean µ i and variance σ , then 2 i n ∑ a i Xi i =1 n is also a normal RV with mean ∑ a i µ i i =1 n 2 & var iance ∑ a i σ i . 2 i =1 Mn ∑ a iXi i=1 ( t ) = M a1X1 ( t )M a 2 X 2 ......M a n X n , (by independence) =e =e 22 a1µ1t + a1 σ1 t 2 / 2 .e a 2µ 2 t + a 2 σ 2 t 2 / 2 22 ( ∑ a iµi ) t + ∑ a i2σi2 t 2 / 2 ........ When n is very large and neither p nor q is very small X—B(n,p) s tan dard binomial variable Z is given by X - np Z= npq as X varies from 0 to n with step size 1, Z varies - np np 1 from to with step size . npq npq npq ∞ P(z) = F(z) = ∫ −∞ 1 −t 2 / 2 e dt ,−∞ < z < ∞ 2π Let X be the number of times that a fair coin, flipped 40 times, land heads. Find P(X=20). Use normal approximation and compare it to the exact solution. P(X=20)=P(19.5<X<20.5) 19.5 − 20 X − 20 20.5 − 20 = P < < 10 10 10 = φ(.16) − φ(−.16) = .1272 0123 4 5 6 7 8 9 10 11 12 01 2 3 4 5 6 7 8 9 10 11 12 If 20% of the momory chips made in a certain plant are defective, what are the probabilities that in a lot of 100 randomly chosen for inspection (a) at most 15 will be defective? (b) exactly 15 will be defective ? µ = 100(.20) = 20, σ = 4 15.5 − 20 F( ) = 0.1292 4 15.5 − 20 14.5 − 20 F − F = 0.0454 4 4 Fit a normal distribution to the following distribution and hence find the theoretical frequencies: Class 60-65 65-70 70-75 75-80 80-85 85-90 90-95 95-100 Freq 3 21 150 335 336 135 26 4 ------------1000 β −1 − αx β f ( x ) = αβ x e ,x > 0 The special case of Weibull with α = 1/σ & β = 2 is known as the Rayleigh Distribution. 2 Thus Rayleigh has linear rate . x − x 2 / 2a 2 If the pdf of a continuous RV X is f(x) = 2 e ,0 < x < ∞, σ then X follows a Rayleigh distribution with parameter σ. In communication systems, the signal amplitude values of a randomly received signal usually can be modeled as a Rayleigh distribution. Let X have a gamma distribution with λ = 1/2 and k = r/2 where r is a positive integer. x ( r/2 ) −1 −x / 2 e ,0 ≤ x ≤ ∞ r/2 The pdf of X is f(x) = Γ(r / 2)2 0 x<0 X has a chi - square distribution χ (r ) with r degrees of freedom. 2 k ( r / 2) E(X) = = =r λ (1 / 2 ) k ( r / 2) Var (X) = 2 = = 2r (1 / 4 ) λ Mean equals the number of degrees of freedom and the variance equals twice the number of degrees of freedom. M X ( t ) = (1 − 2t ) −r / 2 , t < 1/ 2 df \ p .005 .01 .025 1 . 0000 4 . 0001 6 .05 .10 .90 .95 .975 .99 .995 . . 003 00098 9 .0158 2.71 3.84 5.02 6.63 7.88 .2107 4.61 5.99 7.38 9.21 10.60 2 .0100 .0201 .0506 . 102 6 3 .0717 .115 .216 .352 .584 6.25 7.81 9.35 11.34 12.84 4 .207 .297 .484 .711 1.064 7.78 9.49 11.14 13.28 14.86 5 .412 .554 .831 1.15 1.61 9.24 11.07 12.83 15.09 16.75 .6 .676 .872 1.24 1.64 2.20 10.64 12.59 14.45 16.81 18.55 7 .989 1.24 1.69 2.17 2.83 12.02 14.07 16.01 18.48 20.28 8 1.34 1.65 2.18 2.73 3.49 13.36 15.51 17.53 20.09 21.96 9 1.73 2.09 2.70 3.33 4.17 14.68 16.92 19.02 21.67 23.59 10 2.16 2.56 3.25 3.94 4.87 15.99 18.31 20.48 23.21 25.19 11 2.60 3.05 3.82 4.57 5.58 17.28 19.68 21.92 24.73 26.76 12 3.07 3.57 4.40 5.23 6.30 18.55 21.03 23.34 26.22 28.30 df \ p .005 .01 .025 .05 14 4.07 4.66 5.63 6.57 7.79 21.06 23.68 26.12 29.14 31.32 15 4.6 5.23 6.26 7.26 8.55 22.31 25 27.49 30.58 32.80 16 5.14 5.81 6.91 7.96 9.31 23.54 26.30 28.85 32.00 34.27 18 6.26 7.01 8.23 9.39 10.86 25.99 28.87 31.53 34.81 37.16 20 7.43 8.26 9.59 10.8 12.44 5 28.41 31.41 34.17 37.57 40.00 24 9.89 10.86 12.40 13.8 15.66 5 33.20 36.42 39.36 42.98 45.56 30 13.79 14.95 16.79 18.4 20.60 9 40.26 43.77 46.98 50.89 53.67 40 20.71 22.16 24.43 26.5 29.05 1 51.81 55.76 59.34 63.69 66.77 60 35.53 37.48 40.48 43.1 46.46 9 74.40 79.08 83.30 88.38 91.95 120 83.85 86.92 91.58 95.7 100.6 0 2 140.2 146.5 3 7 152.2 158.9 1 5 163.6 4 .10 .90 .95 .975 .99 .995 Let X be χ (10). Find P(3.25 ≤ X ≤ 20.5). 2 Ans.0.95 If (1 - 2t ) , t < 1 / 2, is the m.g.f. of the −6 random variable χ, find P(X < 5.23). 0.05 If X is χ (5) , determine the constant c and d so that P(c < X < d) = 0.95 & P(X < c) = 0.025 2 0.831, 12.8 Beta Distribution The random variable x is said to have beta distribution with nonnegative parameters α & β 1 if f(x) = x α −1 (1 − x )β−1 ,0 < x < b B(α, β) 0 otherwise 1 α -1 where B(α, β) = ∫ x (1 − x ) 0 2π = 2 ∫ ( sin θ) 0 2 α −1 ( cos θ) 2β −1 β −1 dx Γ( α ) Γ( β ) dθ = Γ( α + β ) when α = β = 1, beta distribution is uniform distribution on (0,1) beta function provides greater flexibility than uniform distribution on (0,1) Depending on the values of α & β , the beta distribution takes a variety of shapes. α αβ 2 µ= ,σ = 2 α+β ( α + β) ( α + β + 1) In a certain country, the proportion of highway requiring repairs in any given year is a random variable having the beta distribution with α = 3 and β = 2. Find (a) on the average what percentage of the highway sections require repairs in any given year : (b) the probability that at most half of the highway sections will require repairs in any given year. 3 (a )µ = = 0.60, i.e.60% of the high way 3+ 2 sections require repairs in any given year. 1/2 (b)P(X ≤ 1/2) = ∫ 12x (1 − x )dx = 5 / 16 0 2 The lognormal distribution If X = log T follows a normal distribution N(µ, σ), then T follows a lognormal distribution whose pdf is given 1 1 by f(t) = exp − 2 st 2π 2s t log tm 2 , t ≥ 0 where s = σ is a shape parameter and t m , the median time(or mean time) to failure is the location parameter, given by log t m = µ. 2 s MTTF = E ( t ) = t m exp( ) 2 2 2 2 2 var(T ) = σ T = t m exp(s ) exp(s ) − 1 [ F( t ) = P(T ≤ t ) = P{ log T ≤ log t} log T − µ log t − µ = P ≤ σ σ log T − log t m 1 t = P ≤ log σ s tm 1 t = P Z ≤ log s tm We can compute R(T) and λ(t). Fatigue wearout of a component has a log normal distribution with t M = 5000 hours and s = 0.20. (a) Compute the MTTF and SD. (b) Find the reliability of the component for 3000 hours. (c) Find the design life of the component for a reliability of 0.95. 5101hours,1030hours ∞ ∞ 3000 1 3000 log( ) 0.2 5000 R (3000) = ∫ f ( t )dt = ∫ φ( z )dz ∞ = ∫ φ(z)dz = 0.9946 − 2.55 3598.2 z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.000 0.004 0.008 0.012 0.016 0.019 0.023 0.027 0.031 0.035 0 0 0 0 0 0 9 9 9 9 0.1 0.039 0.043 0.047 0.051 0.055 0.059 0.063 0.067 0.071 0.075 8 8 8 7 7 6 6 5 4 3 0.2 0.079 0.083 0.087 0.091 0.094 0.098 0.102 0.106 0.110 0.114 3 2 1 0 8 7 6 4 3 1 0.3 0.117 0.121 0.125 0.129 0.133 0.136 0.140 0.144 0.148 0.151 9 7 5 3 1 8 6 3 0 7 0.4 0.155 0.159 0.162 0.166 0.170 0.173 0.177 0.180 0.184 0.187 4 1 8 4 0 6 2 8 4 9 0.5 0.191 0.195 0.198 0.201 0.205 0.208 0.212 0.215 0.219 0.222 5 0 5 9 4 8 3 7 0 4 0.6 0.225 0.229 0.232 0.235 0.238 0.242 0.245 0.248 0.251 0.254 7 1 4 7 9 2 4 6 7 9 0.7 0.258 0.261 0.264 0.267 0.270 0.273 0.276 0.279 0.282 0.285 0 1 2 3 4 4 4 4 3 2 0.8 0.288 0.291 0.293 0.296 0.299 0.302 0.305 0.307 0.310 0.313 1 0 9 9 5 3 1 8 6 3 0.9 0.315 0.318 0.321 0.323 0.326 0.328 0.331 0.334 0.336 0.338 9 6 2 8 4 9 5 0 5 9 1.0 0.341 0.343 0.346 0.348 0.350 0.351 0.355 0.357 0.352 0.362 3 8 1 5 8 3 4 7 9 1 1.1 0.364 0.366 0.368 0.370 0.372 0.374 0.377 0.379 0.381 0.383 3 5 6 8 9 9 0 0 0 0 1.2 0.384 0.386 0.388 0.390 0.392 0.394 0.396 0.398 0.399 0.401 9 9 8 7 5 4 2 0 7 5 1.3 0.403 0.404 0.406 0.408 0.409 0.411 0.413 0.414 0.416 0.417 2 9 6 2 9 5 1 7 2 7 1.4 0.419 0.420 0.422 0.423 0.425 0.426 0.427 0.429 0.430 0.431 2 7 2 6 1 5 9 2 6 9 1.5 0.433 0.434 0.435 0.437 0.438 0.439 0.440 0.441 0.442 0.444 2 5 7 0 2 4 6 8 9 1 1.6 0.445 0.446 0.447 0.448 0.449 0.450 0.451 0.452 0.453 0.454 2 3 4 4 5 5 5 5 5 5 1.7 0.455 0.456 0.457 0.458 0.459 0.459 0.460 0.461 0.462 0.463 4 4 3 2 1 9 8 6 5 3 1.8 0.464 0.464 0.465 0.466 0.467 0.467 0.468 0.469 0.469 0.470 1 9 6 4 1 8 6 3 9 6 1.9 0.471 0.471 0.472 0.473 0.473 0.474 0.475 0.475 0.476 0.476 3 9 6 2 8 4 0 6 1 7 2.0 0.477 0.477 0.478 0.478 0.479 0.479 0.480 0.480 0.481 0.481 2 8 3 8 3 8 3 8 2 7 2.1 0.482 0.482 0.483 0.483 0.483 0.484 0.484 0.485 0.485 0.485 1 6 0 4 8 2 6 0 4 7 2.2 0.486 0.486 0.486 0.487 0.487 0.487 0.488 0.488 0.488 0.489 1 4 8 1 5 8 1 4 7 0 2.3 0.489 0.489 0.489 0.490 0.490 0.490 0.490 0.491 0.491 0.491 3 6 8 1 4 6 9 1 3 6 2.4 0.491 0.492 0.492 0.492 0.492 0.492 0.493 0.493 0.493 0.493 8 0 2 5 7 9 1 2 4 6 2.5 0.493 0.494 0.494 0.494 0.494 0.494 0.494 0.494 0.495 0.495 8 0 1 3 5 6 8 9 1 2 2.6 0.495 0.495 0.495 0.495 0.495 0.496 0.496 0.496 0.496 0.496 2.7 0.496 0.496 0.496 0.496 0.496 0.497 0.497 0.497 0.497 0.497 5 6 7 8 9 0 1 2 3 4 2.8 0.497 0.497 0.497 0.497 0.497 0.497 0.497 0.497 0.498 0.498 4 5 6 7 7 8 9 9 0 1 2.9 0.498 0.498 0.498 0.498 0.498 0.498 0.498 0.498 0.498 0.498 1 2 2 3 4 4 5 5 6 6 3.0 0.498 0.498 0.498 0.498 0.498 0.498 0.498 0.498 0.499 0.499 7 7 7 8 8 9 9 9 0 0 3.1 0.499 0.499 0.499 0.499 0.499 0.499 0.499 0.499 0.499 0.499 0 1 1 1 2 2 2 2 3 3 3.2 0.499 0.499 0.499 0.499 0.499 0.499 0.499 0.499 0.499 0.499 3 3 4 4 4 4 4 5 5 5 3.3 0.499 0.499 0.499 0.499 0.499 0.499 0.499 0.499 0.499 0.499 5 5 5 6 6 6 6 6 6 7 3.4 0.499 0.499 0.499 0.499 0.499 0.499 0.499 0.499 0.499 0.499 7 7 7 7 7 7 7 7 7 8 ...
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