Section 5.4 is n  a i X i i = 1 n  M i (a i t). i = 1 M 1 (a 1 t) M 2 (a 2 t) … M n (a n t) = Y = a 1 X 1 + a 2 X 2 + … + a n X n = If X 1, X 2, …, X n comprise a random sample from a distribution with moment generating function M(t), then the moment generating function Corollary is n  X i i = 1 Y = X 1 + X 2 + … + X n = ofM Y (t) = Important Theorems in the Text: If X 1, X 2, …, X n are independent random variables with respective moment generating functions M 1 (t), M 2 (t), …, M n (t), and a 1, a 2, …, a n are constants, then the moment generating function of M Y (t) = Theorem n  M(t) = [M(t)] n, i = 1 and the moment generating function ofthe sample mean X = is n  X i i = 1 n M X (t) = n  M i (t/n) = [M(t/n)] n. i = 1

If X 1, X 2, …, X n are independent chi-square random variables with r 1, r 2, …, r n degrees of freedom respectively, then the random variable Y = has a distribution. Theorem n  X i i = 1  2 (r 1 + r 2 + … + r n ) If Z 1, Z 2, …, Z n are independent random variables each with a standard normal (N(0, 1)) distribution, then the random variable W = has a distribution. Corollary n  Z i 2 i = 1 2(n)2(n) If X 1. X 2. …, X n are independent random variables with respective normal distributions N(  i,  i 2 ) for i = 1, 2, …, n, then the random variable W = has a distribution. Corollary n  i = 1 2(n)2(n) X i –  i ———  i 2

1. (a) Let X 1, X 2, …, X n be a random sample of size n from an exponential distribution with mean . From Corollary 5.4-1, we have that M Y (t) = 1 ——— = 1 –  t n 1 ——— (1 –  t) n From this m.g.f., we recognize that Y must have a distribution. gamma(n,  ) [M(t)] n = Find the m.g.f. of random variable Y = X 1 + X 2 + … + X n = Is it possible to tell from the m.g.f. what the distribution of Y is? (Note: This is essentially Text Exercise ) n  X i. i = 1

(b) Find the m.g.f. of random variable X = (the sample mean). Is it possible to tell from the m.g.f. what the distribution of X is? n  X k k = 1 n From Corollary 5.4-1(b), we have that M X (t) = 1 ———– = 1 –  (t/n) n 1 ————— [1 – (  /n)t] n From this m.g.f., we recognize that X must have a distribution. gamma(n,  /n) n  M i (t/n) = i = 1

2.Suppose the joint p.m.f. of random variables X and Y could be graphically displayed as follows: 2 y 1 12 x 1/41/3 1/61/4 Can X and Y be treated as a random sample of size n = 2? Why or why not? The random variables X and Y cannot be treated as a random sample, because they are not independent. f 1 (x) = if x = 1 if x = 2 5/12 7/12 f 2 (y) = if y = 1 if y = 2 5/12 7/12 f 1 (1) = f 2 (1) =f(1,1) =5/12 1/6

Suppose W, X, and Y are mutually independent random variables each having exponential distribution where E(W) = 1, E(X) = 5, and E(Y) = (a) (b) Can W, X, and Y be treated as a random sample of size n = 3? Why or why not? The random variables W, X, and Y cannot be treated as a random sample, because they do not have identical distributions. Find the m.g.f. of random variable U = 10W + 2X + Y. Is it possible to tell from the m.g.f. what the distribution of U is? From Theorem 5.4-1, we have that M U (t) = 3  M i (a i t) = i = 1 1 ——— 1 – 10t 3 M W ( )M X ( )M Y ( ) =10t2t2tt 1 ——— 1 – ( ) 1 ———— 1 – 5( ) 1 ——— = 1 – 10t 10t2t2t From this m.g.f., we recognize that U must have a distribution. gamma(3, 10)

(c)Find the m.g.f. of random variable V = 8W + 5X – 3Y. Is it possible to tell from the m.g.f. what the distribution of V is? From Theorem 5.4-1, we have that M V (t) = 3  M i (a i t) = i = 1 1 ——— 1 – ( ) 1 ———— 1 – 5( ) 1 ————— = 1 – 10( ) Since we do not recognize this m.g.f., we cannot tell what type of distribution V has. 1 —————————— (1 – 8t)(1 – 25t)(1 + 30t) M W ( )M X ( )M Y ( ) =8t8t5t5t–3t 8t8t5t5t