hw3_sol

Stats 120B Homework 3: Solutions 1. Let X 1 , . . . , X n be a random sample from Geometric distribution with pmf P ( X = x ) = (1 - p ) x p. The mean of this distribution is (1 - p ) /p . (a) Find the estimator for p using method of moments. Match (1 - p ) /p with ¯ X n , we get ˆ p = 1 ¯ X n + 1 . (b) Write down the likelihood function of X 1 , . . . , X n . Find the MLE of p . The likelihood function is L ( p ) = n Y i =1 (1 - p ) x i p = (1 - p ) ∑ i x i p n . Taking log and the derivative, we get log L ( p ) = log(1 - p ) X i x i + n log p. ⇒ d dp log L ( p ) = ∑ i x i p - 1 + n p = 0 . ⇒ ˆ p = 1 ¯ X n + 1 . (c) Consider a Beta prior on p , i.e., p ∼ Beta( α, β ). Find the posterior distribu- tion of p . Take the product of prior and the likelihood, we get ξ ( β | x ) ∝ p α - 1 (1 - p ) β - 1 × (1 - p ) ∑ i x i p n = p n + α - 1 (1 - p ) ∑ i x i + β - 1 . So the posterior distribution is Beta( n + α, ∑ i X i + β ). (d) What is the Bayes estimator of p under squared error loss? Denote it by ˆ p B . The Bayes estimator under squared error loss is the posterior mean of Beta( n + α, ∑ i X i + β ), which is ˆ p B = n + α n + α + ∑ i X i + β 1

(e) What happens to ˆ p B if both α and β goes to 0? Let α, β → 0, then ˆ p B becomes MLE/MOM. This result can be viewed as prior contributing no weight, which makes the data likelihood solely deter- mines the posterior distribution. 2. The mgf of χ 2 ( m ) is (1 - 2 t ) - m/ 2 for any t < 1 / 2. (a) In the textbook, Theorem 8.2.2 states that if the random variables X 1 , . . . , X k are independent with X i ∼ χ 2 ( m i ) for i = 1 , ..., k , then their sum ∑ k i =1 X i ∼ χ 2 ( ∑ k i =1 m i ). Prove this theorem using mgf. Since independence, the mgf of ∑ X i is k Y i =1 M X i ( t ) = k Y i =1 (1 - 2 t ) - m i / 2 = (1 - 2 t ) - ∑ k i =1 m i / 2 , which is the mgf of χ 2 ( ∑ k i =1 ( m i ). (b) Theorem 8.2.3 states that if Z ∼ N (0 , 1), then Y = Z 2 ∼ χ 2 (1). Prove this theorem using mgf. (Hint: Make the integral look like a known pdf.) M Y ( t ) = E ( e Y t ) = E ( e Z 2 t ) = Z ∞ -∞ e z 2 t 1 √ 2 π e - z 2 / 2 dz = Z ∞ -∞ 1 √ 2 π e - ( √ 1 - 2 tz ) 2 / 2 dz let y = √ 1 - 2 tz = Z ∞ -∞ 1 √ 2 π e - y 2 / 2 1 √ 1 - 2 t dy pdf of N(0,1) for y = (1 - 2 t ) - 1 / 2 (c) If Y ∼ χ 2 (1), does X = Y 1 / 2 follow the normal distribution? Explain. No, the support of X is [0 , ∞ ). (d) Prove the following result using CLT: Let X ∼ χ 2 ( n ), then X - n √ 2 n d → N (0 , 1) as n → ∞ . We can write X = Y 1 + · · · + Y n with Y 1 , . . . , Y n i.i.d ∼ χ 2 (1). By CLT: ¯ Y n - 1 p 2 /n d → N (0 , 1) . 2

(e) Y = 1 2 ( X 1 + X 2 ) 2 + 1 2 ( X 3 + X 4 ) 2 By Theorem 5.6.7, X 1 + X 2 ∼ N (0 , 2) and X 3 + X 4 ∼ N (0 , 2). Thus, ( X 1 + X 2 - 0) / √ 2 ∼ N (0 , 1) and ( X 3 + X 4 - 0) / √ 2 ∼ N (0 , 1). By Corollary 8.2.1, Y = X 1 + X 2 √ 2 2 + X 3 + X 4 √ 2 2 ∼ χ 2 (2) . (f) Y = 2 X 1 √ V 1 By Definition 8.4.1, and the fact that X 1 independent with V 1 , Y = X 1 p V 1 / 4 ∼ t (4) . (g) Y = ¯ X √ 32 √ V 1 + V 2 As shown in (d), 2 ¯ X ∼ N (0 , 1). Also, by Theorem 8.2.2, V 1 + V 2 ∼ χ 2 (8). Also, ¯ X is independent with V 1 + V 2 . Thus, by Definition 8.4.1, Y = 2 ¯ X p ( V 1 + V 2 ) / 8 ∼ t (8) . (h) Y = V 1 V 2 Note that V 1 and V 2 are independent, and Y = V 1 / 4 V 2 / 4 ∼ F (4 , 4). (i) Y = X 2 1 X 2 2 Since X 2 1 ∼ χ 2 (1) and X 2 2 ∼ χ 2 (1), since X 2 1 is independent with X 2 2 , so Y = X 2 1 / 1 X 2 2 / 1 ∼ F (1 , 1). (j) Y = W - 5 3 X 1 Since ( W - 5) / 3 ∼ N (0 , 1) and X 1 ∼ N (0 , 1), and they’re independent, so Y follows the standard Cauchy distribution, which is also the t (1) distribu- tion. (k) Y = 32 ¯ X 2 V 1 + V 2 We know T = ¯ X √ 32 √ V 1 + V 2 ∼ t (8) . 4

Since Y = T 2 , Y ∼ F (1 , 8). 4. Suppose that a random sample of eight observations is taken from the normal dis- tribution with unknown mean μ and unknown variance σ 2 , and that the observed values are 3.1, 3.5, 2.6, 3.4, 3.8, 3.0, 2.9, and 2.2. Find the shortest confidence interval for μ with each of the following three confidence coefficients: (a) 0.90, (b) 0.95, and (c) 0.99. R code and output: > dat = c(3.1,3.5,2.6,3.4,3.8,3.0,2.9,2.2) > m = mean(dat) > m [1] 3.0625 > s = sd(dat) > s [1] 0.5125218 > n = 8 > > # (a) 90% CI for mu: > q = qt(.95,7) > q [1] 1.894579 > m + c(-1,1)*q*s/sqrt(n) [1] 2.719195 3.405805 > > # (b) 95% CI for mu: > q = qt(.975,7) > q [1] 2.364624 > m + c(-1,1)*q*s/sqrt(n) [1] 2.634021 3.490979 > > # (c) 99% CI for mu: > q = qt(.995,7) > q [1] 3.499483 > m + c(-1,1)*q*s/sqrt(n) [1] 2.42838 3.69662 (a) 3 . 0625 ± (1 . 8946) 0 . 5125 √ 8 = (2 . 719 , 3 . 406) (b) 3 . 0625 ± (2 . 3646) 0 . 5125 √ 8 = (2 . 634 , 3 . 491) 5

(d) A Jan. 31, 2014, Gallup article reported that “Russians See Gold in Sochi Olympic Games Yet Many Concerned About Corruption” (http://www.gallup.com/poll/167138/russians-gold-sochi-olympic-games.aspx). The article reports on results of a poll of a random sample of 2,000 adults in Russia. In the sample, 66% responded that they thought the olympic games would increase corruption. i. Using your formula from part (c), construct an approximate 95% confi- dence interval for the true proportion of all adult Russians that believe the olympic games will increase corruption. R code and output: > .66+c(-1,1)*qnorm(.975)*sqrt(.66*(1-.66)/2000) [1] 0.6392392 0.6807608 An approximate 95% confidence interval for p is . 66 ± 1 . 96 r . 66(1 - . 66) 2000 = . 66 ± . 01059 = ( . 6392 , . 6808) . ii. Give an interpretation of your interval in context, i.e., “We are 95% con- fident that .” We are 95% confident that the true proportion of adults in Russia that believe the olympic games will increase corruption (at the time of the sur- vey) is between .6392 and .6808. iii. Explain what the phrase “95% confident” means in your interpretation. By “95% confident”, we mean that if we were to repeat this study many times, in 95% of random samples of 2,000 adults in Russia, the confidence interval produced by the sample would capture the true proportion of adult Russians that believe the olympic games will increase corruption. 7