# CLT for independent, but non-identically distributed Poisson random variables

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!

• I believe you dropped the square root in the denominator in your first equality. Using $\sum Y_i\sim \text{Pois}(\sum \lambda_i)$ will simplify the computation a bit. Commented Nov 1, 2020 at 18:44
• Whoops, silly mistake. Thank you!
– Remy
Commented Nov 1, 2020 at 19:41

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 $$e^{t/\sqrt{\lambda}}$$.