Ratio Distribution: Poisson Random Variables Suppose two Poisson processes. For example, during the time interval, $\Delta t_{1} = t_{1} - t_{o} = 50\mu s$ , $x$ photons are incident on a detector with rate $\lambda_{1} = 10$x$10^4 s^{-1}$. At time point, $t_{1}$, a second process begins in which, during the time interval, $\Delta t_{2} = t_{2} - t_{1} = 50\mu s$ , $y$ photons are incident on the same detector with rate $\lambda_{2} = 6$x$10^4 s^{-1}$. 
Let $X$ and $Y$ be two independent Poisson random variables described by $X$ ~ Pois($\lambda_{1}\Delta t_1$) and $Y$ ~ Pois($\lambda_{2}\Delta t_2$). And let $Z$ be a ratio distribution defined as $Z = X/(X+Y)$.
[1] What is the general distribution of $Z$ for $X+Y>0$? its standard deviation? and how are both derived?
Next, suppose we know the total number of photons, $n=x+y$ , over the time interval $\Delta t = \Delta t_{1} + \Delta t_{2} = 100\mu s$ ; e.g., $n=10$. 
We would like to predict the probability distribution for observing an $(x,y)$ pair given $n$ and the knowledge that both $x$ and $y$ were drawn from Poisson distributions with rates $\lambda_1 \Delta t_1$ and $\lambda_2 \Delta t_2$, respectively. 
[2] What is the new distribution for $Z|n$? its standard deviation? and how are both derived?
 A: $Z$ is not well-defined; consider $X=Y=0$ which occurs with probability $> 0$, when $Z = 0/0$.  Perhaps you should consider the distribution only for $n>0$?  In that case, there is no standard distribution with a name, you just have to work through the math yourself.  
Note that since X and Y are integers, Z is, except for the $0/0$ case, defined on the set of nonnegative rationals.  If $\lambda_1$ and $\lambda_2$ are large, you could use a Beta distribution as a continuous approximation, based on the following:
1) The Poisson distribution approaches the Gamma distribution (with scale parameter $= 1$) as $\lambda \to \infty$, with the shape parameter $r$ of the Gamma equal to $\lambda$ (you can get to this by a) moment matching of the Poisson and Gamma, and b) observing that both distributions approach the same Gaussian as $r= \lambda \to \infty$, therefore they approach each other as well.)
2) If $x_1, x_2$ are distributed $\Gamma$ with identical scale parameters and shape parameters equal to $r_1, r_2$ respectively, then $x_1/(x_1+x_2)$ is distributed $\beta(r_1,r_2)$.  
Edit (edited again) in response to followup question by OP:
The Beta approximation seems good except perhaps in the tails at $\lambda=20$, and quite good at $\lambda=50$.  Quantile-quantile plots of a sample of size 10,000 from the Poisson ratio vs. the approximating Beta distribution for $\lambda=20,50$ are below.  The actual distribution appears to have a slightly fatter lower tail than the Beta approximation at $\lambda=20$, and seems to be a very good fit at $\lambda=50$ (your definitions may vary.)  Depending on the application, I'd say somewhere in the 20 - 50 range the Beta approximation would start to work quite well.   
N <- 10000
Lambda <- 50

x1 <- rpois(N, Lambda)
x2 <- rpois(N, Lambda)
y <- sort(x1/(x1+x2))

z <- ((1:N)-0.5)/N
qqplot(qbeta(z,Lambda,Lambda), y,
       xlab = paste("Beta(",Lambda,",",Lambda,")",sep=""),
       ylab = paste("X/(X+Y) (N = ",N,")"))
abline(c(0,1),lwd=2,col=2)



A: Let $X$ and $Y$ be independent Poisson random variables with respective rates $a$ and $b$. Then $X+Y$ is Poisson with rate $a+b$. For every $0\leqslant k\leqslant n$,
$$
\mathrm P(X=k\mid X+Y=n)=\frac{\mathrm P(X=k,X+Y=n)}{\mathrm P(X+Y=n)}=\frac{\mathrm P(X=k)\mathrm P(Y=n-k)}{\mathrm P(X+Y=n)},
$$
that is,
$$
\mathrm P(X=k\mid X+Y=n)=\frac{p_a(k)p_b(n-k)}{p_{a+b}(n)},
\quad \text{where}\quad 
p_c(i)=\mathrm e^{-c}\frac{c^i}{i!}.
$$
After some easy simplifications, one gets
$$
\mathrm P(X=k\mid X+Y=n)={n\choose k}p^k(1-p)^{n-k},
\quad \text{where}\quad
p=\frac{a}{a+b}.
$$
This means that, conditionally on $X+Y=n$, the distribution of $X$ is binomial $(n,p)$. In particular, the conditional expectation of $X$ is $np$ and its conditional variance is $np(1-p)$.
A: Mathematica claims that there is an analytical solution for this: 
$$
\sum_{x=0}^\infty \sum_{y=0}^\infty P(\lambda_x,x) P(\lambda_y,y)\frac{x}{x+y}
=\frac{\lambda_x+e^{\lambda_x+\lambda_y}\lambda_y}{\lambda_x+\lambda_y}
$$
although I have not yet managed to prove it. 
