Problem source

The random variable $U$ has a Poisson distribution with parameter $\lambda$. The random variables $X$ and $Y$ are defined as follows.
\begin{align*}
X&=
 \begin{cases} 
U & \text{ if $U$ is 1, 3, 5, 7, $\ldots\,$} \\
0 & \text{ otherwise}
\end{cases}
\\
Y&=
 \begin{cases} 
U & \text{ if $U$ is 2, 4, 6, 8, $\ldots\,$ } \\
0 & \text{ otherwise} 
\end{cases}
\end{align*}
\begin{questionparts}
\item Find $\E(X)$ and $\E(Y)$ in terms of $\lambda$, $\alpha$ and $\beta$, where
\[
\alpha = 1+\frac{\lambda^2}{2!}+\frac{\lambda^4}{4!} +\cdots\,
\text{ \ \ and \ \ }
\beta = \frac{\lambda}{1!} + \frac{\lambda^3}{3!} + \frac{\lambda^5}{5!}
+\cdots\,.
\]
\item
Show that 
\[
\var(X) = \frac{\lambda\alpha+\lambda^2\beta}{\alpha+\beta}
 - \frac{\lambda^2\alpha^2}{(\alpha+\beta)^2}
\]
and obtain the corresponding expression for $\var(Y)$. Are there any  non-zero values of $\lambda$ for which $ \var(X) + \var(Y) = \var(X+Y)\,$? 
\end{questionparts}

Solution source

\begin{questionparts}
\item \begin{align*}
\mathbb{E}(X) &= \sum_{r=1}^\infty r \mathbb{P}(X = r) \\
&= \sum_{j=1}^{\infty} (2j-1)\mathbb{P}(U=2j-1) \\
&= \sum_{j=1}^{\infty}(2j-1) \frac{e^{-\lambda} \lambda^{2j-1}}{(2j-1)!} \\
&= \sum_{j=1}^{\infty} e^{-\lambda} \frac{\lambda^{2j-1}}{(2j-2)!} \\
&= \lambda e^{-\lambda} \sum_{j=1}^{\infty} \frac{\lambda^{2j-2}}{(2j-2)!} \\
&= \lambda e^{-\lambda} \alpha
\end{align*}

Since $\mathbb{E}(X+Y) = \lambda, \mathbb{E}(Y) = \lambda(1-e^{-\lambda}\alpha) = \lambda(e^{-\lambda}(\alpha+\beta) - e^{-\lambda}\alpha) = \lambda e^{-\lambda} \beta$. Alternatively, as $\beta + \alpha = e^{\lambda}$, $\mathbb{E}(X) = \frac{\lambda \alpha}{\alpha+\beta}, \mathbb{E}(Y) = \frac{\lambda \beta}{\alpha+\beta}$

\item \begin{align*}
\textrm{Var}(X) &= \mathbb{E}(X^2) - [\mathbb{E}(X)
]^2 \\
&= \sum_{odd} r^2 \mathbb{P}(U = r) - \left [ \mathbb{E}(X)
\right]^2 \\
&= \sum_{odd} (r(r-1)+r)\frac{e^{-\lambda}\lambda^r}{r!} - \frac{\lambda^2 \alpha^2}{(\alpha+\beta)^2} \\
&= \sum_{odd} \frac{e^{-\lambda}\lambda^r}{(r-2)!}+\sum_{odd} \frac{e^{-\lambda}\lambda^r}{(r-1)!} - \frac{\lambda^2 \alpha^2}{(\alpha+\beta)^2} \\
&= e^{-\lambda}\lambda^2 \beta + e^{-\lambda}\lambda \alpha - \frac{\lambda^2 \alpha^2}{(\alpha+\beta)^2} \\
&= \frac{\lambda \alpha + \lambda^2 \beta}{\alpha+\beta}- \frac{\lambda^2 \alpha^2}{(\alpha+\beta)^2} 
\end{align*}

Similarly, \begin{align*}
\textrm{Var}(Y) &= \mathbb{E}(Y^2) - [\mathbb{E}(Y)
]^2 \\
&= \sum_{even} r^2 \mathbb{P}(U = r) - \left [ \mathbb{E}(Y)
\right]^2 \\
&= \sum_{even} (r(r-1)+r)\frac{e^{-\lambda}\lambda^r}{r!} - \frac{\lambda^2 \beta^2}{(\alpha+\beta)^2} \\
&= e^{-\lambda}\lambda^2\alpha + e^{-\lambda}\lambda \beta - \frac{\lambda^2 \beta^2}{(\alpha+\beta)^2} \\
&= \frac{\lambda \beta + \lambda^2 \alpha}{\alpha+\beta}- \frac{\lambda^2 \beta^2}{(\alpha+\beta)^2}
\end{align*}

Since $\textrm{Var}(X+Y) = \textrm{Var}(U) = \lambda$, we are interested in solving:

\begin{align*}
\lambda &=  \frac{\lambda \alpha + \lambda^2 \beta}{\alpha+\beta}- \frac{\lambda^2 \alpha^2}{(\alpha+\beta)^2}  +  \frac{\lambda \beta + \lambda^2 \alpha}{\alpha+\beta}- \frac{\lambda^2 \beta^2}{(\alpha+\beta)^2} \\
&= \frac{\lambda(\alpha+\beta) + \lambda^2(\alpha+\beta)}{\alpha+\beta} - \frac{\lambda^2(\alpha^2+\beta^2)}{(\alpha+\beta)^2} \\
&= \lambda + \lambda^2 \frac{(\alpha+\beta)^2 - (\alpha^2+\beta^2)}{(\alpha+\beta)^2} \\
&= \lambda + \lambda^2 \frac{2\alpha\beta}{(\alpha+\beta)^2}
\end{align*}

which is clearly not possible if $\lambda \neq 0$
\end{questionparts}