Problem source

In the game of endless cricket the scores
$X$ and $Y$ of the two sides are such that 
\[
\P (X=j,\ Y=k)=\e^{-1}\frac{(j+k)\lambda^{j+k}}{j!k!},\]
for some positive constant $\lambda$, where $j,k = 0$, $1$, $2$, $\ldots$.
\begin{questionparts}
\item Find $\P(X+Y=n)$ for each $n>0$.
\item Show that $2\lambda \e^{2\lambda-1}=1$.
\item Show that $2x \e^{2x-1}$ is an increasing function
of $x$ for $x>0$ and deduce that the equation in (ii) has
at most one solution and hence determine $\lambda$.
\item Calculate the expectation $\E(2^{X+Y})$.
\end{questionparts}

Solution source

\begin{questionparts}
\item \begin{align*}
&& \mathbb{P}(X+Y = n) &= \sum_{i = 0}^n \mathbb{P}(X = i, Y = n-i) \\
&&&= \sum_{i = 0}^n e^{-1} \frac{n \lambda^n}{i! (n-i)!} \\
&&&=e^{-1} n \lambda^n \sum_{i = 0}^n\frac{1}{i! (n-i)!} \\
&&&=\frac{e^{-1} n}{n!} \lambda^n \sum_{i = 0}^n\frac{n!}{i! (n-i)!} \\
&&&= \frac{n\lambda^n}{e n!} 2^n \\
&&&= \frac{n (2 \lambda)^n}{e \cdot n!}
\end{align*}

\item \begin{align*}
&& 1 &= \sum_{n = 0}^{\infty} \mathbb{P}(X+Y =n ) \\
&&&= \sum_{n = 0}^{\infty}\frac{n (2 \lambda)^n}{e \cdot n!} \\
&&&= \sum_{n = 1}^\infty \frac{ (2 \lambda)^n}{e \cdot (n-1)!} \\
&&&= \frac{2 \lambda}{e}\sum_{n = 0}^\infty \frac{ (2 \lambda)^n}{n!} \\
&&&= \frac{2 \lambda}{e} e^{2\lambda} \\
&&&= 2 \lambda e^{2\lambda - 1}
\end{align*} \\
\item Consider $f(x) = 2xe^{2x-1}$, then
\begin{align*}
&& f'(x) &= 2e^{2x-1} + 2xe^{2x-1} \cdot 2 \\
&&&= e^{2x-1} (2 + 4x) > 0
\end{align*}
Therefore $f(x)$ is an increasing function of $x$, which means $f(x) = 1$ has at most one solution for $\lambda$. Therefore $\lambda = \frac12$

\item
\begin{align*}
\mathbb{E}(2^{X+Y}) &= \sum_{n = 0}^\infty \mathbb{P}(X+Y = n) 2^n \\
&= \sum_{n = 1}^\infty \frac{1}{e(n-1)!} 2^{n} \\
&= \frac{2}{e} \sum_{n=0}^\infty \frac{2^n}{n!} \\
&= \frac{2}{e} e^2 \\
&= 2e
\end{align*}

\end{questionparts}