sum and difference between two independent Poisson random variables [closed]
The first problem can be tackled heads on: $$ \begin{eqnarray} \Pr(X+Y=k) &=& \sum_{p=0}^\infty \sum_{q=0}^\infty \mathrm{e}^{-a-b} \frac{a^p}{p!} \frac{b^q}{q!} \mathbf{1}_{p+q=k} = \mathrm{e}^{-a-b}\sum_{p=0}^{k} \frac{a^p}{p!} \frac{b^{k-p}}{(k-p)!} \\ &=& \mathrm{e}^{-a-b} \frac{1}{k!} \cdot \sum_{p=0}^{k} \binom{k}{p} a^p b^{k-p} = \mathrm{e}^{-a-b} \frac{1}{k!} \left(a+b\right)^k \end{eqnarray} $$ The last equality follows by the binomial theorem. The expression is the p.m.f. of a Poisson distribution, hence $X+Y \sim \operatorname{Poi}\left(a+ b\right)$.
The second follows from the first problem. Assuming $0 \leqslant k \leqslant n$: $$\begin{eqnarray} \Pr\left(X=k \mid X+Y=n\right) &=& \frac{\Pr\left(X=k, Y=n-k\right)}{\Pr(X+Y=n)} \\ &=& \frac{\exp(-a) a^k/k! \cdot \exp(-b) b^{n-k}/(n-k)!}{\exp(-a-b) \cdot (a+b)^n/n!} \\ &=& \binom{n}{k} \left(\frac{a}{a+b}\right)^k \left(1-\frac{a}{a+b}\right)^{n-k} \end{eqnarray} $$ The latter is the pmf of a binomial distribution, i.e. $X \mid X+Y=n \sim \operatorname{Bin}\left(n, \frac{a}{a+b} \right)$.
One has to evaluate $\Pr(X-Y=n)$. The difference of $X-Y$ is known to follow Skellam distribution: $$ \begin{eqnarray} \Pr(X-Y=n) &=& \sum_{p=0}^\infty \sum_{q=0}^\infty \mathrm{e}^{-a-b} \frac{a^p}{p!} \frac{b^q}{q!} \mathbf{1}_{p-q=n} = \mathrm{e}^{-a-b}\sum_{q=\max(0,-n)}^{\infty} \frac{a^{n+q}}{(n+q)!} \frac{b^{q}}{q!} \\ &=& \begin{cases} a^n \left(a b\right)^{-n/2} I_n\left(2 \sqrt{a b}\right) & n \geqslant 0 \cr b^{-n} \left(a b\right)^{n/2} I_{-n}\left(2 \sqrt{a b}\right) & n < 0\end{cases} = \left(\frac{a}{b}\right)^{n/2} I_{\vert n\vert} \left(2 \sqrt{a b}\right) \end{eqnarray} $$ where $I_n(x)$ is the modified Bessel function of the first kind.
Characteristic functions were mentioned, but maybe more insight comes from other points of view. \begin{align} & \Pr(X+Y=w) \\[8pt] = {} & \Pr([X=0\ \&\ Y=w]\text{ or }[X=1\ \&\ Y=w-1]\text{ or }[X=2\ \&\ Y=w-2\cdots{}\cdots \\[8pt] & {}\qquad\qquad\qquad\cdots{}\cdots\text{ or }X=w\ \&\ Y=0]) \\[8pt] = {} & \sum_{k=0}^w \Pr(X=k\ \&\ Y=w-k) = \sum_{k=0}^w \Pr(X=k)\Pr(Y=w-k) \\[8pt] = {} & \sum_{k=0}^w \frac{a^ke^{-a}}{k!}\cdot\frac{b^{w-k}e^{-b}}{(w-k)!} = e^{-(a+b)} \sum_{k=0}^w \frac{a^k}{k!}\cdot\frac{b^{w-k}}{(w-k)!} \\[8pt] = {} & \underbrace{\frac{e^{-(a+b)}}{w!} \sum_{k=0}^w \frac{w!}{k!(w-k)!} a^k b^{w-k} = \frac{e^{-(a+b)}}{w!} (a+b)^w}_{\text{binomial theorem}}, \end{align} So the sum has a Poisson distribution with expected value $a+b$.
\begin{align} \Pr(X=x\mid X+Y=n) & = \frac{\Pr(X=x\ \&\ X+Y=n)}{\Pr(X+Y=n)} \\[8pt] & = \frac{\Pr(X=x\ \&\ Y=n-x)}{\Pr(X+Y=n)} \\[8pt] & = \frac{\Pr(X=x)\cdot\Pr(Y=n-x)}{\Pr(X+Y=n)}. \end{align} Then use the formula for the distribution of $X+Y$ found above, and the formulas for the distributions of $X$ and $Y$ separately. You should get a binomial distribution.
Hint: for (a), use the characteristic functions (and the fact that if $X$ and $Y$ are independent, $\mathbb{E}[e^{itX}e^{itY}]=\mathbb{E}[e^{itX}]\mathbb{E}[e^{itY}]$).