Question

3. For each of the following distributions, find the expected value of X (a) f(z,8) = θ(1-9)"-, z = 1, 2, . . . . 0 (b) f(z,8) = (θ + 1)2 , z > 1,0 > 0 (e) f(z,8) = θ2ze-e", z > 0, θ > 0 9s 1 -0-2

Answer 1

a)

$E(X)=\sum_{x=0}^{\infty }x*\theta * (1-\theta )^{x-1}$

$\Rightarrow E(X)=\theta \sum_{x=0}^{\infty }x* (1-\theta )^{x-1}$

$\Rightarrow E(X)=\theta [1+2*(1-\theta )+3*(1-\theta )^2+...]$

$\Rightarrow E(X)=\theta *(1-(1-\theta ))^{-2}$

$\Rightarrow E(X)=\theta *\frac{1}{\theta ^2}=\frac{1}{\theta }$

$\Rightarrow E(X)=\frac{1}{\theta }$

b) $E(X)=\int_{1}^{\infty }x*(\theta+1) * x^{-\theta -2}dx$

$\Rightarrow E(X)=(\theta+1) \int_{1}^{\infty }x^{-\theta -1}dx$

$\Rightarrow E(X)=(\theta+1)*[\frac{x^{-\theta }}{-\theta }]_{1}^{\infty }$

$\Rightarrow E(X)=(\theta+1)*[\frac{1}{\theta }]$

$\Rightarrow E(X)=1+\frac{1}{\theta }$

c) $E(X)=\int_{0}^{\infty }x*\theta ^2*xe^{-\theta x}dx$

$\Rightarrow E(X)=\theta ^2\int_{0}^{\infty }x^2*e^{-\theta x}dx$

$\Rightarrow E(X)=\theta ^2*\frac{\Gamma (3)}{\theta^3}=\frac{2}{\theta }$

$\Rightarrow E(X)=\frac{2}{\theta }$