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Unformatted text preview: Example class 5 STAT1301 Probability and Statistics I Chan Chi Ho, Chan Tsz Hin & Shi Yun October 12, 2011 Review: Some common discrete distributions Table of Important Discrete Distributions Name ( for which ( ) ) these for () () Bernoulli One trial; 1 if success, 0 if failure Number of success in trials () Binomial Bernoulli Number of Bernoulli trials needed to get one success Geometric Pascal (Negative Binomial) ( Number of Bernoulli trials needed to get success ) Approximates binomial for large, but not large Poisson ( ) ( )( Hypergeometric ( Uniform Description ) () ) ( ) Number of red balls chosen when balls are chosen from balls, of which are red Choose one of the numbers Review: Probability Density Function (PDF) Definition of PDF dFX (x ) P (X ≤x + dx ) − P (X ≤x ) = lim dx dx (dx →0) P (X < x + dx ) − P (X ≤x ) = lim dx (dx →0) P (X ≤x + dx ) − P (X < x ) = lim dx (dx →0) P (X < x + dx ) − P (X < x ) = lim dx (dx →0) fX (x ) = Review: Probability Density Function (PDF) Properties of PDF 1 2 3 4 fX (x )≥0 +∞ −∞ fX (x )dx = 1 x FX (x ) = −∞ fX (t )dt P (a < X ≤b ) = P (a≤X ≤b ) = P (a≤X < b ) = P (a < X < b ) = b a f (x )dx = FX (b ) − FX (a) Review: Markov’s inequality Markov’s inequality If X is a nonnegative random variable with finite mean E (X ), then for any c > 0, E (X ) . P (X ≥ c ) ≤ c Proof. Note that since X ≥ 0, cI {X ≥ c } ≤ X The inequality follows by taking expectation on both sides. Review: Chebyshev’s inequality Chebyshev’s inequality If the random variable X has finite mean µ and finite variance σ 2 , then for any real number k > 0, P (|X − µ | ≥ k σ ) ≤ 1 . k2 Proof. By Markov’s inequality, P ((X − µ )2 ≥ k 2 σ 2 ) ≤ 1 E ((X − µ )2 ) =2 2σ 2 k k Problem 1 Two balls are chosen randomly from an urn containing 7 white, 4 black, and 1 orange balls. Suppose that we win $1 for each white ball drawn and we lose $1 for each orange ball drawn. Denote $X as the amount that we can win. 1 What are the possible values of X ? 2 Determine the probability mass function and cumulative distribution function of X . Solution to Problem 1 I Sample space: Ω = {WW , WB , WO , BB , BO } Note that the outcomes are not equally likely with associated probabilities: (7) 7 × × 7 2 P ({WW }) = 12 = 22 , P ({WB }) = 7664 = 14 , P ({WO }) = 7661 = 66 , 33 (2) (4) 1 × 2 2 P ({BB }) = 66 = 11 , P ({BO }) = 4661 = 33 . Solution to Problem 1 The random variable X is the following function: The the possible values of X are –1, 0, 1, 2. The probability mass function is given by 2 P (X = −1) = P ({BO }) = 33 7 1 P (X = 0) = P ({WO , BB }) = 66 + 11 = 13 66 14 P (X = 1) = P ({WB }) = 33 7 P (X = 2) = P ({WW }) = 22 The cumulative distribution function is 0 P (X = −1) = 2/33 F (x ) = P (X ≤ x ) = P (X = −1) + P (X = 0) = 17/66 P (X = −1) + P (X = 0) + P (X = 1) = 15/22 1 if x < −1 if − 1 ≤ x < 0 if 0 ≤ x < 1 . if 1 ≤ x < 2 if x ≥ 2 Problem 2 In each box of a given product there is a coupon with a number ranging from 1 to 6. If a housewife succeeds in collecting all the coupons of number 1 to 6 then she can win a prize. Assume that each number is equally likely to appear. What is the expected number of boxes she should buy in order to win the prize? Solution to Problem 2 I Let X1 be the number of boxes need to be bought to get a coupon. X2 be the number of boxes need to be bought to get a different coupon from the first. X3 be the number of boxes need to be bought to get a different coupon from the first and second. Similarly for X4 , X5 , X6 . Let X be the total number of boxes need to be bought in order to collect all 6 different coupons. Then X = X1 + X2 + · · · + X6 . We need to find E (X ). Solution to Problem 2 But we know E (X ) = E (X 1 ) + E (X 2 ) + · · · + E (X6 ). Note that X1 = 1, and so E (X 1 ) = 1. For X2 , since we already obtained a coupon, the probability that we obtain a new coupon in the next trial will be 5 . 6 5 Hence, P (X2 = x ) = (1 − 5 )x −1 5 . i.e. X2 is a geometric r.v. with p = 6 . 6 6 6 Hence E (X2 ) = 5 . Similarly, X3 is a geometric r.v. with p = 4 . X4 , X5 , X6 are geometric 6 32 random variables with p = 6 , 6 and 1 respectively. 6 Finally, we can compute E (X ). 6 E (X ) = E (X 1 ) + E (X 2 ) + · · · + E (X6 ) = 1 + 5 + 6 + · · · + 6 = 14.7. 4 1 Problem 3 Suppose that airplane engines will fail, when in flight, with probability 1 − p independently from engine to engine. If an airplane needs a majority of its engines operative to make a successful flight, for what values of p is a 5-engine plane preferable to a 3-engine one? Solution to problem 3 I The number of operative engines in a 5-engine plane follows binomial distribution b (5, p ). P (5 − engine plane fail) = P (at least 3 engines fail) = P (at most 2 engines operative) = (1 − p )5 + 5(1 − p )4 p + 10(1 − p )3 p 2 = (1 − p )3 (1 + 3p + 6p 2 ) The number of operative engines in a 3-engine plane follows binomial distribution b (3, p ). P (3 − engine plane fail) = P (at least 2 engines fail) = P (at most 1 engines operative) = (1 − p )3 + 3(1 − p )2 p = (1 − p )2 (1 + 2p ) Solution to problem 3 Hence P (5 − engine plane fail) − P (3 − engine plane fail) = (1 − p )2 [(1 − p )(1 + 3p + 6p 2 ) − (1 + 2p )] = 3p 2 (1 − p )2 (1 − 2p ) This difference is negative whenever p is greater than 0.5. Hence a 5-engine plane is preferable to a 3-engine one if p is greater than 0.5, i.e. if each engine has more than half chance to be successfully functioning. Problem 4 1 Verify that Poisson distribution can approximate Binomial distribution when the number of Bernoulli trials is very large and Pr(succuss ) very small, while the mean remain finite. To be precise, suppose X has a binomial distribution with parameters n and p . If p → 0 and np → l as n → +∞ then λk . k! n = 100, p = 0.05, λ = 5, X ∼ B (n, p ) and Y ∼ Pois (λ ). Compare the values ofPX (X ≤ 5) and PY (Y ≤ 5). P (X = k ) → e −λ 2 Solution to Problem 4 I 1 The following manipulation is purely technical: as n → ∞, let λ = np , n! p k (1 − p )n−k k !(n − k )! = n! k !(n − k )! p 1−p = n! k !(n − k )! λ n−λ k 1− λ n n k e −λ = e −λ → n! k !(n − k )! p 1−p k e −λ n! λk k ! (n − k )!(n − λ )k Since as n → ∞, n! (n − k )!(n − λ )k = = n n−1 n−k +1 ··· n−λ n−λ n−λ λ λ −1 λ −k +1 1+ 1+ ··· 1+ n−λ n−λ n−λ we have n! λk p k (1 − p )n−k −→ e −λ . k !(n − k )! k! −→ 1 Solution to Problem 4 II 2 This part gives you the practical explanation of the term “Poisson approximation to Binomial”. It also shows how possibly tedious can a statistical problem be and by this way motivates you to use software packages such as the free R language (cran.r-project.org). pX (0) = (0.95)100 = 0.005920529 pX (1) = (100)(0.05)(0.95)99 = 0.031160680 00 pX (2) = 12 (0.05)2 (0.95)98 = 0.081181772 100 pX (3) = 3 (0.05)3 (0.95)97 = 0.139575678 00 pX (4) = 14 (0.05)4 (0.95)96 = 0.178142642 100 pX (5) = 5 (0.05)5 (0.95)95 = 0.1800178 PX (X ≤5) = 0.6159991 pY (0) = e −5 = 0.006737947 pY (1) = 5e −5 = 0.033689735 pY (2) = 25 e −5 = 0.084224337 2 pY (3) = 125 e −5 = 0.140373896 6 4 pY (4) = 5 ! e −5 = 0.175467370 4 5 pY (5) = 5 ! e −5 = 0.175467370 5 PY (Y ≤ 5) = 0.6159607 Problem 5 The number of times that an individual contracts a cold in a given year is a Poisson random variable with mean θ = 6. Suppose a new wonder drug (based on large quantities of vitamin C) has just been marketed that reduces the Poisson mean to θ = 4 for 60 percent of the population. For the other 40 percent of the population the drug has no appreciable effect on colds. If an individual tries the drug for a year and has 3 colds in that time, how likely is it that the drug is beneficial for him/her? (HINT: Use the Bayes’ rule) Solution to Problem 5 Let X be the no. of colds the individual contracts within a year. Then X ∼ P (4) if the drug is effective and X ∼ P (6) if the drug is not effective. Denote E as the event that the drug is effective, then e −4 43 32e −4 = , 3! 3 P (X = 3|E c ) = P (E ) = 0.6, P (X = 3|E ) = e −6 63 = 36e −6 3! P (E c ) = 0.4 Hence using law of total probability and Bayes’ theorem, P (X = 3) = 0.6 × P (E |X = 3) = 32e −4 + 0.4 × 36e −6 = 57.6e −4 + 14.4e −6 3 P (X = 3|E )P (E ) 57.6e −4 = = 0.9673 P (X = 3) 57.6e −4 + 14.4e −6 Problem 6 Let f (x ) denote the density function of the random variable X . Suppose m has a symmetric distribution about a, that is f (x + a) = f (a − x ) for every x . Show that E (X ) = a if it exists. Solution to Problem 6 I +∞ E (X ) = xf (x )dx −∞ a = +∞ xf (x )dx + −∞ 0 = xf (x )dx a 0 −∞ 0 =a −∞ 0 f (a − u )du f (u + a)du + a −∞ a =a −∞ +∞ f (x )dx + a −∞ +∞ =a f (x )dx a f (x )dx −∞ =a (a − u )f (a − u )du (u + a)f (u + a)du + Problem 7 Denote MX (t ) as the moment generating function of a random variable X . 1 Show that for any real number a, P (X ≥ a) ≤ e −ta MX (t ) for all t > 0; and P (X ≤ a) ≤ e −ta MX (t ) for all t < 0. 2 Let X be a Poisson random variable with mean θ . Show that P (X ≥ x ) ≤ e −θ ( eθ x ) x for any x > θ . Solution to Problem 7 I 1 First assume t > 0. Then I {X ≥ a} = I {e tX ≥ e ta } = I {1 ≤ e tX e tX } ≤ ta e ta e Taking expectation, E {I {X ≥ a}} ≤ e −ta E (e tX ) P (X ≥ a) ≤ e −ta MX (t ), for all t > 0 Hence we have proved the first inequality. To prove the second inequality, replace X by −X and a by −a in the first inequality, then by the first inequality, P (−X ≥ −a) ≤ e ta M−X (t ) for all t > 0 ta P (X ≤ a) ≤ e MX (−t ) for all t > 0 Hence, replacing −t by t , we have P (X ≤ a) ≤ e −ta MX (t ) for all t < 0 Solution to Problem 7 II 2 By the first inequality in part (a), P (X ≥ x ) ≤ e −tx MX (t ) = e −tx e θ (e t −1) = e −θ e θ e t −tx x Since θ e t − tx is minimized at t = ln( ) > 0 (since x > θ > 0) and θ the inequality is true for all t > 0, we have P (X ≥ x ) ≤ e −θ e x −x ln(x /θ ) = e −θ ( eθ x ) x for all x > θ . ...
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