If $X$ and $Y$ are two independent Poisson random variables with same parameter $\lambda$, $Z=X-Y$ is the Skellam distribution with parameters $\lambda$ and $\lambda$. This is distributed around zero with a width which scales with the value of $\lambda$.
I found numerically that the distribution of $\frac{X-Y}{\sqrt{𝑋+𝑌}}$ is independent of $\lambda$. Does someone know if this PDF has a known analytical form, or how I could try to derive it? Also interested in the normal approximation if its easier.
Best Answer
Your assertion that the distribution of $(X-Y)/\sqrt{X+Y}$ is independent of $\lambda$ is incorrect. To address the issue with the indeterminate form $0/0$ with positive probability, it is worth noting that the most reasonable choice for the function $$W(X,Y) = \frac{X - Y}{\sqrt{X + Y}}$$ at $X = Y = 0$ is $W = 0$, since the limiting behavior as $(X,Y) \to (0,0)$ in the first quadrant is $W \to 0$. Thus we will define $$W(X,Y) = \begin{cases}\frac{X - Y}{\sqrt{X + Y}}, & X \in \mathbb Z^+ \cup Y \in \mathbb Z^+ \\ 0, & \text{otherwise} . \end{cases}$$
Then $$\begin{align} \Pr[W = 1] &= \sum_{c=0}^\infty \Pr[X = (c+1)(c+2)/2] \Pr[Y = c(c+1)/2] \\ &= e^{-2\lambda} \sum_{c=0}^\infty \frac{\lambda^{(c+1)^2}}{(\frac{c(c+1)}{2})!(\frac{(c+1)(c+2)}{2})!} , \end{align}$$ since the solution to $(X-Y)^2 = X+Y$ in nonnegative integers leads to $(X,Y)$ being consecutive triangular numbers. For $\lambda = 1$, this gives approximately $$\Pr[W = 1 \mid \lambda = 1] \approx 0.157922,$$ whereas $$\Pr[W = 1 \mid \lambda = 10] \approx 0.0127439.$$ Very few terms are needed to achieve this precision since the series converges quite rapidly.