Limiting distribution of $\frac {X_n -Y_m -(n-m)}{\sqrt{X_n+Y_m}}$ where $X_n,Y_m$ are independent Poisson Let $X_n$ , $Y_m$ be independent Poisson variables with means $n$, $m$.
$$\frac {X_n -Y_m -(n-m)}{\sqrt{X_n+Y_m}}$$
Find the limiting distribution as $n,m \to \infty$
I know that $\frac{X_n-n}{\sqrt{n}} \to N(0,1)$ but don't know what the next step would be.
 A: First prove that 
$$
Z(n,m)=\frac {X_n -Y_m -(n-m)}{\sqrt{n+m}} \xrightarrow{d} \mathcal N(0,1).
$$
Use characteristic functions and Lévy's continuity theorem. Since for Poisson distribution $X\sim Poiss(\lambda)$ characteristic function is $\varphi_X(t)=e^{\lambda(e^{it}-1)}$ and $\varphi_{X-\lambda}(t)=e^{\lambda(e^{it}-1-it)}$, then
$$
\varphi_{Z(n,m)}(t) = \varphi_{X_n-n-(Y_m-m)}\left(\frac{t}{\sqrt{n+m}}\right) = \varphi_{X_n-n}\left(\frac{t}{\sqrt{n+m}}\right)\cdot \varphi_{Y_m-m}\left(-\frac{t}{\sqrt{n+m}}\right) = e^{n\left(e^{i\frac{t}{\sqrt{n+m}}}-1-i\frac{t}{\sqrt{n+m}}\right)}\cdot e^{m\left(e^{-i\frac{t}{\sqrt{n+m}}}-1+i\frac{t}{\sqrt{n+m}}\right)}
$$
Consider power of exponent: 
$$
n\left(e^{i\frac{t}{\sqrt{n+m}}}-1-i\frac{t}{\sqrt{n+m}}\right) + m\left(e^{-i\frac{t}{\sqrt{n+m}}}-1+i\frac{t}{\sqrt{n+m}}\right) = 
n\left(\color{green}{1}+ \color{red}{i\frac{t}{\sqrt{n+m}}}-\frac{t^2}{2(n+m)}+o\Bigl(\frac{t^2}{n+m}\Bigr)-\color{green}{1}-\color{red}{i\frac{t}{\sqrt{n+m}}}\right) + m\left(\color{green}{1}-\color{red}{i\frac{t}{\sqrt{n+m}}}-\frac{t^2}{2(n+m)}+o\Bigl(\frac{t^2}{n+m}\Bigr)-\color{green}{1}+\color{red}{i\frac{t}{\sqrt{n+m}}}\right) = -\frac{t^2}{2} + t^2\cdot o(1)
$$
We use Taylor series $e^x=1+x+\frac{x^2}{2}+o\bigl(\frac{x^2}{2}\bigr)$ as $x\to 0$.
So when $n+m\to\infty$
$$
\varphi_{Z(n,m)}(t) = e^{-\frac{t^2}{2} + t^2\cdot o(1)} \to e^{-\frac{t^2}{2}} = \varphi_{\mathcal N(0,1)}(t). 
$$
Next, $X_n+Y_m$ have Poisson distribution with parameter $n+m$, therefore
$$
\frac{X_n+Y_m}{n+m} \xrightarrow{p} 1.
$$
Use then continuity of square root and Slutsky's theorem to conclude that 
$$
\frac {X_n -Y_m -(n-m)}{\sqrt{X_n+Y_m}} = \frac {X_n -Y_m -(n-m)}{\sqrt{n+m}} \cdot \frac{1}{\sqrt{\frac{X_n+Y_m}{n+m} }}\xrightarrow{d} \mathcal N(0,1).
$$
