Let $$ and $$ be two independent Poisson random variables with same lambda parameter. What is the distribution of $\frac{X-Y}{\sqrt{+}}$? 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 $\frac{X-Y}{\sqrt{+}}$ 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.
 A: 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.
A: I will here be giving a normal approximation, which will hold for large $\lambda$. Using the central limit theorem we can conclude, that
$$
\sqrt{\lambda}(\frac{1}{\lambda}\begin{pmatrix} X \\\ Y \end{pmatrix}-\begin{pmatrix} 1 \\\ 1 \end{pmatrix}) \xrightarrow{\sim}
N_2(0, \begin{pmatrix} 1 & 0 \\\ 0 & 1 \end{pmatrix}) \quad \text{as $\lambda \rightarrow \infty$}.
$$
If we let $g(x,y)=\frac{x-y}{\sqrt{x+y}}$, then the asymptotic distribution of $g(X/\lambda,Y/\lambda) = \frac{1}{\sqrt{\lambda}}\frac{X-Y}{\sqrt{X+Y}}$ can be found by the delta method. The delta method gives us that
$$\frac{X-Y}{\sqrt{X+Y}} = \sqrt{\lambda}(g(X/\lambda , Y \lambda)-g(1,1)) \xrightarrow{\sim} N(0, || \nabla g(1,1) ||^2).$$
Indeed we can see that the asymptotic distribution does not depend on $\lambda$. Let us calculate the asymptotic variance. We note that
$$\nabla g(x,y) = \begin{pmatrix} \frac{x+3y}{2(x+y)^{(3/2)}}    & \frac{-3x-y} {2(x+y)^{(3/2)}})\end{pmatrix},$$
yielding that
$$||\nabla g(1,1)||^2 = \big(\frac{1+3}{2(1+1)^{(3/2)}}\big)^2 + \big(\frac{-3-1}{2(1+1)^{(3/2)}}\big)^2 = 1,$$
We may thus for large values of lambda conclude that $\frac{X-Y}{\sqrt{X+Y}} \sim N(0,1)$ approximately.
