4.2.7 problem 31

(a)

$$P\left(X\right)=\{\begin{array}{cc}p+(1-p)\text{Poiss}(X=k)& k=0\\ (1-p)\text{Poiss}(X=k)& k>0\end{array}$$

(b)

First, notice that $(1-I)Y\in \{0,1,2,\dots \}$. $(1-I)Y=0$ if $I=1$, or $Y=0$. Thus $P\left(\right(1-I)Y=0)=p+(1-p)P(Y=0)$. For any other value $k$ of $(1-I)Y$, it is achieved if $I=0$ and $Y=k$. Thus, $P\left(\right(1-I)Y=k)=(1-p)P(Y=k)$.

(c)

$\text{E}\left(X\right)=(1-p){\sum }_{k=1}^{\infty }k\frac{e^{-\lambda }{\lambda }^k}{k!}=(1-p)\text{E}\left(\text{Poss}\right(\lambda \left)\right)=(1-p)\lambda$.

$\text{E}\left(X\right)=\text{E}\left(\right(1-I\left)Y\right)=\text{E}(1-I)\text{E}\left(Y\right)=(1-p)\lambda$.

(d)

$\text{Var}\left(X\right)=\text{E}\left(X^2\right)-{\left(\text{E}\right(X\left)\right)}^2$. $\text{E}\left(X^2\right)=(1-p)e^{-\lambda }{\sum }_{k=1}^{\infty }{k}^2\frac{{\lambda }^k}{k!}=(1-p)\lambda (1+\lambda )$. Thus, $\text{Var}\left(X\right)=(1-p)\lambda (1+\lambda )-{\left(\right(1-p\left)\lambda \right)}^2=(1-p)\lambda (1+\mathrm{p\lambda })$.