Show that
$$\frac{\sum_{i=1}^n(Y_i-\lambda_i)}{\sqrt{\sum_{i=1}^n\lambda_i}}\overset{d}{\rightarrow}\mathcal
N(0,1)$$where $\{Y_i\}\sim\mathsf{Pois}(\lambda_i)$ are independent and
$\sum_{i=1}^n\lambda_i\rightarrow\infty$
My Try:
I tried using Lyapunov CLT but could not get a sharp enough bound. I next tried moment generating functions where we'd need $M_{\frac{\sum(Y_i-\lambda_i)}{\sqrt{\sum\lambda_i}}}(t)\rightarrow e^{\frac{t^2}{2}}$ but we have
$$\begin{align*}
M_{\frac{\sum(Y_i-\lambda_i)}{\sqrt{\sum\lambda_i}}}(t)
&=M_{\frac{Y_1}{\sum\lambda_i}}(t)M_{\frac{-\lambda_1}{\sum\lambda_i}}(t)\cdots M_{\frac{Y_n}{\sum\lambda_i}}(t)M_{\frac{-\lambda_n}{\sum\lambda_i}}(t)\\\\
&=M_{Y_1}\left(\frac{t}{\sum\lambda_i}\right)\exp\left(\frac{-\lambda_1}{\sum\lambda_i}t\right)\cdots M_{Y_n}\left(\frac{t}{\sum\lambda_i}\right)\exp\left(\frac{-\lambda_n}{\sum\lambda_i}t\right)\\\\
&=\exp\left(\lambda_1\left(e^{\frac{t}{\sum\lambda_i}}-1\right)\right)\cdots\exp\left(\lambda_n\left(e^{\frac{t}{\sum\lambda_i}}-1\right)\right)e^{-t}\\\\
&=\underbrace{\exp\left(\sum\lambda_i\left(e^{\frac{t}{\sum\lambda_i}}-1\right)\right)}_{\rightarrow e^t}e^{-t}\\\\
&\rightarrow 1
\end{align*}$$
so either I made a mistake or this approach doesn't work. I think the property that $\sum_{i=1}^n Y_i\sim\mathsf{Pois}\left(\sum_{i=1}^n \lambda_i\right)$ is of importance but I'm not sure how I'd use that here. Perhaps the Lindeberg condition would be the way to go. It's used in a similar problem here for the case of independent, but non-identically distributed exponential random variables. Any help would be appreciated!
Best Answer
For simplicity, let $Y=\sum Y_i$ and $\lambda=\sum\lambda_i$. Then the MGF of $(Y-\lambda)/\sqrt{\lambda}$ is $$ \exp(\lambda(e^{t/\sqrt{\lambda}}-1))\cdot e^{-t\sqrt{\lambda}} $$ You can prove this approaches $e^{t^2/2}$ using the MacLaurin series of $e^{t/\sqrt{\lambda}}$.