Question

6. Suppose random variables Z, are exponentially distributed: ZiExp(2) for i 1,2,..., n. Assume that the random variables Z, are independent. For each of the following functions of the Zi, find the expectation E] and variance Var[ ]. (a) 3 Z1- (b) 1.5Z1 +222-3 (c) i 32 (simplify, but final answer is an expression)

Answer 1

$If~X\sim Exp(\theta)~then\\ E(X^r)=\int_0^\infty x^r\theta e^{-\theta x}dx=\int_0^\infty(y/\theta)^re^{-y} dy~~take,~\theta x=y~or,~\theta dx=dy\\ =\frac{\Gamma(r+1)}{\theta^r}=\frac{r!}{\theta^r}\\ E(X)=\frac{1}{\theta},~E(X^2)=\frac{2}{\theta^2},~Var(X)=E(X^2)-E^2(X)=\frac{1}{\theta^2}\\ Now,~Z_i\sim Exp(2^i)~then~\theta=2^i\\ E(Z_i)=\frac{1}{2^i},~Var(Z_i)=\frac{1}{2^{2i}},~i=1,2,...,n.\\ (a)~E(3Z_1-5)=3E(Z_1)-5=\frac{3}{2}-5=-\frac{7}{2}\\ Var(3Z_1-5)=9Var(Z_1)=\frac{9}{4}\\ (b)~E(1.5Z_1+2Z_2-3)=1.5E(Z_1)+2E(Z_2)-3=1.5*(1/2)+2*(1/2^2)-3\\ =\frac{1.5+1-3}{2}=-\frac{15}{20}=-\frac{3}{4}\\ Var(1.5Z_1+2Z_2-3)=1.5^2Var(Z_1)+2^2Var(Z_2)~since~Z_i's~are~independent\\ =1.5^2*(1/2^2)+2^2*(1/2^4)=(9/4+1)(1/4)=\frac{13}{16}\\ (c)~E(\sum_{i=1}^n3Z_i)=\sum_{i=1}^n3E(Z_i)=3\sum_{i=1}^n\frac{1}{2^i}=\frac{3}{2}\times \frac{1-\frac{1}{2^{n}}}{1-\frac{1}{2}}=3\left(1-\frac{1}{2^n} \right )\\$

$Var(\sum_{i=1}^n3Z_i)=\sum_{i=1}^n3^2Var(Z_i)~~since~Z_i's~are~independent\\ =\frac{9}{4}\sum_{i=1}^n\frac{1}{4^{i-1}}=\frac{9}{4}\times \frac{1-\frac{1}{4^n}}{1-\frac{1}{4}}=3\left(1-\frac{1}{4^n} \right )$