4.2.6 problem 30

(a)

$$\begin{array}{cccc}\text{E}\left(\mathrm{Xg}\right(X\left)\right)& ={\sum }_{x=0}^{\infty }\mathrm{xg}\left(x\right)\frac{{\text{e}}^{-\lambda }{\left(\lambda \right)}^x}{x!}& & \\ & ={\sum }_{x=1}^{\infty }\mathrm{xg}\left(x\right)\frac{{\text{e}}^{-\lambda }{\left(\lambda \right)}^x}{x!}& & \\ & =\lambda {\sum }_{x=1}^{\infty }g\left(x\right)\frac{{\text{e}}^{-\lambda }{\left(\lambda \right)}^{x-1}}{(x-1)!}& & \\ & =\lambda {\sum }_{x=0}^{\infty }g(x+1)\frac{{\text{e}}^{-\lambda }{\left(\lambda \right)}^x}{\left(x\right)!}=\lambda \text{E}\left(g\right(X+1\left)\right)& & \end{array}$$

(b)

$$\begin{array}{cccc}\text{E}\left(X^3\right)& =\text{E}\left(X{X}^2\right)& & \\ & =\lambda \text{E}\left({(X+1)}^2\right)& & \\ & =\lambda \left(\text{E}\right(X^2)+\text{E}(2X)+1)& & \\ & =\lambda \left(\lambda \text{E}\right(X+1)+2\lambda +1)=\lambda \left(\lambda \right(\lambda +1)+2\lambda +1)& & \\ & =\lambda ({\lambda }^2+3\lambda +1)& & \end{array}$$