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21-325 (Fall 2008): Homework 9 (ONE side) Due by December 3, in class Show FULL JUSTIFICATION for all your answers. 1. Let X1 , X2 , · · · be a sequence of independent and identically distributed random variables with mean µ and variance σ 2 . Let X n = 1 n Pn i=1 Xi . Show that E[(X n − µ)2 ] converges to 0 as n goes to ∞. Solution. E[(X n − µ)2 ] = E[( n1 Pn independent, E[( i=1 (Xi Pn i=1 (Xi − µ))2 ] = Pn − µ))2 ] = i=1 E[(Xi P 1 E[( ni=1 (Xi n2 − µ))2 ]. Since {Xi − µ}ni=1 are − µ)2 ] = nσ 2 . Then E[(X n − µ)2 ] = σ2 n → 0 as n → ∞. 2. Let X1 , X2 , · · · be a sequence of independent and identically distributed random variables with uniform distribution on [0, 1]. Let Yn = min{X1 , X2 , · · · , Xn } and Zn = max{X1 , X2 , · · · , Xn }. Show that Yn converges to 0 in probability and Zn converges to 1 in probability. Solution. Let ∈ (0, 1). P (Yn ≥ ) = P (Xi ≥ , i = 1, ..., n) = P (X1 ≥ )n = (1 − )n → 0 as n → ∞. So Yn converges in probability to 0. On the other hand, P (|Zn − 1| ≥ ) = P (Zn ≤ 1 − ) = P (∩ni=1 (Xi ≤ 1 − )) = P ((X1 ≤ 1 − ))n = (1 − )n → 0 as n → ∞. So Zn converges to 1 in probability. 3. Suppose that the number of units produced daily at factory A is a random variable with mean 20 and variance 9. and the number produced at factory B is a random variable with mean 18 and variance 36. Assuming independence, derive a (non-trivial) upper bound for the probability that more units are produced today at factory B than at factory A. Solution. Let X be the number of units produced daily at factory A and Y be the number of units produced daily at factory B. X and Y are independent. We want to estimate P (Y > X). We first try to use Markov inequality directly. P (Y > X) = P (Y −X > 0) = P (Y −X ≥ 1) = P (Y −X −(−2) ≥ 1−(−2)) ≤ P ((Y −X +2)2 ≥ 9), where −2 is the mean of Y − X. Then we have by the Markov inequality, P (Y > X) ≤ 91 V ar(Y − X) = V ar(Y )+V ar(X) 9 = 45 9 = 5. But 5 is not the non-trivial upper bound. We need to modify the argument. Let a > 0. P (Y > X) = P (Y −X −(−2) ≥ 1−(−2)) ≤ P ((Y −X +2+a)2 ≥ (3+a)2 ) ≤ 1 E[(Y (3+a)2 − X + 2 + a)2 ] = 1 (V (3+a)2 ar(Y − X) + a2 ) = 1 45+a2 . (3+a)2 Since P (Y > X) ≤ 45+a2 , (3+a)2 we then try to minimize 45+a2 . (3+a)2 Simple Calculus yields that 45+a2 (3+a)2 is minimized at a = 15. Plug it in to get P (Y > X) ≤ 56 , which is non-trivial. 4. The servicing of a machine requires two steps, with the time needed for the first step being an exponential random variable with mean 0.2 hour and the time for the second step being an independent exponential random variable with mean 0.3 hour. If a repair person has 20 machines to service, approximate the probability that all the work can be completed in 8 hours. (You may use the table for the standard normal distribution on the web at http://www.sjsu.edu/faculty/gerstman/EpiInfo/ztable.htm) Solution. Let Xi and Yi be the time needed for the first step and the second step for the ith machine. Then Xi is an exponential random variable with mean 0.2 and Yi is an exponential random variable with mean 0.3. Then P ( 0.26. So P ( P20 i=1 (Xi + Yi ) ≤ 8). Xi + Yi has mean 0.5 and variance P20 P20 + Yi ) ≤ 8) can be approximated by P ( i=1 (Xi P20 P20 correction. Then P ( i=1 (Xi i=1 (Xi (Xi +Yi )−10 i=1 √ 20∗0.26 + Yi ) < 8.5) = P ≤ + Yi ) < 8.5) by the continuity √8.5−10 20∗0.26 −1.5 = Φ( √ ). 5.2 5. Suppose that a fair die is rolled 100 times. Let Xi be the value obtained on the ith roll. Compute an approximation for P 100 Y ! 100 Xi ≤ a , 1 < a < 6. i=1 Solution. P Q 100 i=1 Xi ≤ a100 = P P 100 i=1 log(Xi ) ≤ 100 log(a) . Let E[log(X1 )] = 1 6 (log(1) + log(2) + log(3) + log(4) + log(5) + log(6)). So P 100 X ! log(Xi ) ≤ 100 log(a) P100 ! − 100E(log(X1 )) 100 ∗ log(a) − 100 ∗ E(log(X1 )) p p ≤ , 10 V ar(log(X1 )) 10 V ar(log(X1 )) i=1 log(Xi ) =P i=1 1 )) √ which can be estimated by Φ( 100∗log(a)−100∗E(log(X ). 10 V ar(log(X1 )) 6. Let f (x) be a continuous function defined for 0 ≤ x ≤ 1. Consider the functions n X ! n k Bn (x) = f (k/n) x (1 − x)n−k k k=0 and prove that limn→∞ Bn (x) = f (x). Solution. Let Yn be an Binomial random variable with parameters n and x. Then Bn (x) = E[f (Yn /n)]. Since Yn has the same distribution as X1 + X2 + ... + Xn , where Xi is a Bernoulli random variable with parameter x, i = 1, 2, ..., n. Then by the strong law of large numbers, we 2 have (X1 + X2 + ... + Xn )/n → x as n → ∞. Then Bn (x) = E[f ((X1 + X2 + ... + Xn )/n)] → f (x) as n → ∞. 3