Solutions - 5220-02

(1) State the PMFs of $X$ and $Y$:

$$\begin{array}{ccc}{f}_X(x)=\{\begin{array}{ccc}\frac{e^{-\lambda }{\lambda }^k}{k!}& & k\in {\mathbb{Z}}_{\ge 0}\\ 0& & \text{else}\end{array}& & f_Y(y)=\{\begin{array}{ccc}p& & y=1\\ 1-p& & y=0\\ 0& & \text{else}\end{array}\end{array}$$

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(2) Derive the PMF of $S=X+Y$:

$$P[S=s]=P[X+Y=s]=P[X=s-y,Y=y]=P[X=s-y]P[Y=y]$$

Since $P[Y=y]>0$ only if $y=\mathrm{0,1}$, we have $P[S=s]=P[X=s]P[Y=0]+P[X=s-1]P[Y=1]$. Thus:

$$P[S=s]=\{\begin{array}{ccc}\frac{e^{-\lambda }{\lambda }^s}{s!}(1-p)+\frac{e^{-\lambda }{\lambda }^{s-1}}{(s-1)!}p& & s>0\\ e^{-\lambda }(1-p)& & s=0\\ 0& & \text{else}\end{array}$$

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(3) Evaluate at $\lambda =2$, $p=0.3$, $s=7$:

$$P_S(7)\approx 0.006015$$