Poisson random variables and Binomial Theorem I'm working on a problem from Casella and Berger's Statistical Inference. X is distributed as Poisson$(\theta)$ and Y is distributed as Poisson$(\lambda)$, with X and Y being independent. We let U = X + Y and V = Y, and the conditional pdf of V|U is: 
$\hspace{15mm}\large f(v|u) = \large \frac{f(u,v)}{f(u)} = \huge \frac{\frac{\theta^{u-v}}{(u-v!)}*\frac{\lambda^ve^{-\lambda}}{v!}}{\frac{(\theta+\lambda)^ue^{-(\theta+\lambda)}}{u!}}$
Apparently this simplifies to 
$\hspace{15mm}\Large {u \choose v} (\frac{\lambda}{\theta+\lambda})^v(\frac{\theta}{\theta+\lambda})^{u-v}$
I don't see how the $\Large(\frac{\lambda}{\theta+\lambda})^v$ is possible. I keep ending up with $\Large(\frac{\lambda}{\theta})^v$ when I simplify the problem. 
 A: It is quite likely that you were told that $X$ and $Y$ are independent random variables, but neglected to pass on this information to us.
Assuming that $X$ and $Y$ are independent Poisson random variables, $U = X+Y$ is
also a Poisson random variable with parameter $\theta+\lambda$. Thus, the
probability that $U = m$ ($m$ is a nonnegative integer here) is
$$p_{X+Y}(m) = \frac{e^{-(\theta+\lambda)}(\theta+\lambda)^m}{m!}.$$
Conditioned on $X+Y = m$, the conditional probability that $Y = n$ is $0$ if
$n >m$, while for $0 \leq n \leq m$,
$$\begin{align}p_{Y\mid X+Y=m}(n\mid m) 
&= \frac{P\left(\{Y=n\} \cap \{X+Y=m\}\right)}{P\{X+Y=m\}}
&{\scriptstyle{\text{definition of comditional probability}}}\\
&= \frac{P\{X=m-n,Y=n\}}{P\{X+Y=m\}}
&{\scriptstyle{\text{a re-write}}}\\
&= \frac{P\{X=m-n\}P\{Y=n\}}{\frac{e^{-(\theta+\lambda)}(\theta+\lambda)^m}{m!}}
&{\scriptstyle{\text{independence of}~X~\text{and}~Y}}\\
&= \frac{\frac{e^{-(\theta)}(\theta)^{m-n}}{(m-n)!}
\frac{e^{-(\lambda)}(\lambda)^{n}}{(n!}}{\frac{e^{-(\theta+\lambda)}(\theta+\lambda)^m}{m!}}\\
&= \binom{m}{n}\left(\frac{\theta}{\theta+\lambda}\right)^{m-n}
\left(\frac{\lambda}{\theta+\lambda}\right)^{n}
\end{align}$$
which shows that, conditioned on the value of $X+Y$, $Y$ is a binomial
random variable.
A: Forgetting some irrelevant factorial and exponential terms, you seem to be trying to factor the ratio $$R=\frac{\theta^{u-v}\lambda^v}{(\theta+\lambda)^u}.$$
You are not very specific about the complete expression you get for $R$, but indeed, $$R=\left(\frac{\lambda}{\theta}\right)^v\left(\frac{\theta}{\theta+\lambda}\right)^u.$$ And $R$ is also, as Casella and Berger say, $$R=\left(\frac{\lambda}{\theta+\lambda}\right)^v\left(\frac{\theta}{\theta+\lambda}\right)^{u-v}.$$ So... what is the problem, really?
