VIEWS: 20 PAGES: 13 POSTED ON: 4/8/2011
Chapter 6, Section 5 Transformations of Variables Method of Moment-Generating Functions John J Currano, 04/16/2010 1 Method of Moment-Generating Functions The crucial theorem is: Theorem 6.1 (p. 318): If X and Y are random variables which both have moment-generating functions, and if mX(t) = mY(t) for all t in some interval around t = 0, then X and Y have the same probability distribution. 2 Method of Moment-Generating Functions Some other useful facts: 1. If U = aY + b, then m U (t ) E (e tU ) E e aYt bt e bt E eY (at ) e bt m Y (at ). 2. If Y1, Y2, … , Yn are independent and U = Y1 + Y2 + + Yn, then ( m U (t ) E (e tU ) E e tY1 ) E e tY2 L tYn ( tY1 tY2 e Le ) tYn ( )( ) ( ) E etY1 E etY2 L E e tYn by independence mY (t ) mY (t ) L mY (t ) 1 2 n 3 Method of Moment-Generating Functions Some other useful facts: n 3. If Y1, Y2, … , Yn are independent and U aiYi , i 1 n then m U (t ) m Y (ai t ). i i 1 4. If Y1, Y2, … , Yn are independent and identically distributed (iid) with common distribution Y, i.e., a Random Sample from Y, and [ ]. U = Y1 + Y2 + + Yn, then m U (t ) m Y (t ) n 4 Method of Moment-Generating Functions Example 1. Suppose that Y1, Y2, … , Ym are independent binomial RVs with Yi ~ bin(ni, p) [same p]. Then for i = 1, 2, … , m, m Yi (t ) q pet ni m Let Y Yi . Then by Property 2 on slide 3, i 1 ( ) n1 n2 L nm m mY (t ) m Yi (t ) q pe t , i 1 m so Y has a binomial distribution with n ni trials and probability of i 1 success, p. Thus, the sum of independent binomials with the same probability of success, p, is also binomial. 5 Example 2. Let Y ~ NegBin(r, p), X0 = 0, and Xi = # of trial on which i th success occurs. Then Yi = Xi Xi 1 ~ Geom(p) for i = 1, 2, … , r, and Y1, Y2, … , Yr are independent. Also, r r Yi X i X i 1 X r X 0 X r 0 Y . i 1 i 1 telescoping sum r r pe t Thus, m Y (t ) m Y (t ) by Property 2 on slide 3. t 1 qe i i 1 We have found the moment-generating functions of the negative binomial distributions using those of the geometric distributions. 6 r pe t Example 2. Y ~ NegBin(r, p), m Y (t ) 1 qe t . We can now use the mgf of Y to find its mean and variance. For example, pe 0 E (Y ) mY (0) r r 1 pe 1 qe pe qe 0 0 0 0 1 qe 0 1 qe 0 2 r 1 ( ) ) p p ( q pq 1 r 1 q (1 q)2 r 1 p r ( p pq ) p p2 r . p 7 Example 3. Let Y ~ N( , 2) and Z Y 1 Y . Then m Y (t ) exp t 1 2 t 2 , 2 1 so m Z (t ) exp t m Y t by Property 1 on slide 3 1 1 2 exp t exp t 2 2 t 1 t 2 t 2 exp t 1 2 2 exp ( ) 1 2 t2 So Z ~ N(0, 1 ) : Z has a standard normal distribution. We proved this fact in Chapter 4 using the Distribution Function method – this is simpler. Transforming a random variable Y by subtracting its mean and dividing by its standard deviation is called standardizing Y. 8 Example 4. Let Z ~ N(0,1) and Y = + Z . Then m Z (t ) exp 1 2 t2 , so m Y (t ) exp t m Z t by Property 1 on slide 3 exp t exp ( 1 2 ) t 2 ( exp t 1 2 2 t) 2 So Y ~ N( , 2) : Y has a normal distribution with mean and variance 2. We also proved this in Chapter 4 using the Distribution Function method – again this method is simpler. Transforming a standard normal random variable Z in this fashion is a way of simulating other normal random variables. 9 Example 5. If Yi ~ N( i , i 2) are independent for i = 1, 2, . . . , n, i then m Y (t ) exp i t 1 2 i2 t 2 for 1 i n. Let Y ai Yi . m i 1 Then by the Property 3 on slide 4, m i 1 m i 1 ( m Y (t ) m Yi ( ai t ) exp i a i t 1 2 (a t) . i2 i) 2 m m exp t a i i 1 t 2 ai2 i2 , i 1 2 i 1 m m so Y ~ N a i i , ai2 i2 . Thus, a linear combination of i 1 i 1 independent normal random variables is also normal. 10 Special Case. Suppose Y1, Y2, . . . , Yn are a random sample from n a N( , 2) - distribution, and Y 1 Yi , the sample mean. n i 1 Then by Example 5, Y ~ N , 2 n . Thus the sample mean of a random sample of size n from a normal distribution also has a normal distribution. Its mean is the same as that of the original distribution and its variance is smaller 2/n instead of 2. This will be used often next semester. 11 Example 6. (p. 319, Example 6.11) 1 z2 2 Let Z ~ N(0,1), so f ( z ) e , and let Y Z 2 . Then 2 m Y (t ) E e tZ2 etz 2 1 2 e z 2 2 dz 1 (1 2t ) z 2 e 2 dz 2 1 e )dz z 2 ( (1 2t ) 1 2 2 12 Example 6. (p. 319, Example 6.11) Let Z ~ N(0,1) and Y Z 2 . Then m Y (t ) 1 e z2 2(12t )1 dz . 2 The integrand is proportional to the density function of the normal distribution with = 0 and 2 = (1 – 2t)1. Thus, if we multiply by 1 1 2t inside the integral and by 1 2t 1 2 outside, 12 we obtain m Y (t ) 1 2t 1 1 2t 1 2 1 2 . Therefore, Y Z 2 has a c 2 (1) - distribution. We proved this in Chapter 4 using the Distribution Function method. Once again, this method is simpler. 13