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 McGrawHill.
2. J. I. Aunon & V. Chandrasekhar. Introduction to Probability
and Random Processes. McGrawHill International Ed.
3. A. Papoulis & S. U. Pillai. Probability, Random Variables
and Stochastic Processes. Tata WcGrawHill.
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 averagecase
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 6faced cubic die, the no. of telephone
calls received in a board in a 5min. 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(EF).
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(E3E1E2)p(E4E1E2E3)
p(E1) = C(4,1)C(48,12)/C(52,13)
p(E2E1) = C(3,1)C(36,12)/C(39,13)
p(E3E1E2) = C(2,1)C(24,12)/C(26,13)
p(E4E1E2 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(EF)p(F) +p(EFc)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(EF) = 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 multiplechoice
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 multiplechoice
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 lefthanded. If 20
percent of the population is lefthanded,and it turns out that the
suspect is also lefthanded, 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 lefthanded. If 20
percent of the population is lefthanded,and it turns out that
the suspect is also lefthanded, then does this change the
guilty probability of the suspect? By how much? Solution:
G = the suspect is guilty
LH = the suspect is lefthanded 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 TwoDimensional 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. Twodimensional continuous RV.
Twodimensional 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 twodimensional 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(aX)}. 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 (revisit)
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(1p) 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 coefficient 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 =1/1t 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/(3t), 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 momentgenerating 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} = 1p
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 (1p) 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 = 1p, is given by a formula called
the Binomial Distribution: C(n, k) pk q(nk) 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 (1p)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 moneyback 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(1p) 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)
negativebinomially 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 negativebinomially 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 nonsixes 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
β1 − 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
6065
6570
7075
7580
8085
8590
9095
95100 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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 Spring '11
 KUMAR
 Probability, Probability theory, Random Processes

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