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This is an exercise of the Central Limit Theorem:

Let $Y^{\lambda}$ be a Poisson random variable with parameter $\lambda>0$. Prove that $\frac{Y^{\lambda}-\lambda}{\sqrt{\lambda}}\to Z\sim N(0,1)$ in distribution as $\lambda\to\infty$.

I've done that $$ Z_n\to Z\sim N(0,1) $$ in distribution using the CLT, where $Z_n=(Y^n-n)/\sqrt{n}$. Some naive attempt to go is considering $$ Y^{n}\leq Y^{\lambda}\leq Y^{n+1}\tag{*} $$ where $n\leq\lambda\leq n+1$ and somehow use the squeeze theorem. But both (*) and the squeeze theorem in convergence in distribution are NOT justified. How can I go on? Or do I need an alternative direction?

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Since only a narrow special case of the CLT would be needed, I wouldn't be surprised if it can be done directly without knowing the CLT is true in general. –  Michael Hardy Nov 27 '12 at 2:12
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Your proposition labeled $(*)$ is of course not true if it's an inequality on the values of the random variables. But if $\le$ means "is stochastically smaller than" then it's true. However, if you can show $\Pr(Y^\lambda=y)$ is between the corresponding probabilities with $n$ and $n+1$ in place of $\lambda$, then maybe it will work. –  Michael Hardy Nov 27 '12 at 2:14
    
Proposition $(\ast)$ can be made fully rigorous using the natural coupling of a whole family of Poisson random variables. –  Did Nov 28 '12 at 21:08

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up vote 3 down vote accepted

The squeeze theorem in convergence in distribution can be made fully rigorous in the situation you describe--but the shortest proof here might be through characteristic functions.

Recall that if $Y^\lambda$ is Poisson with parameter $\lambda$, $\varphi_\lambda(t)=\mathbb E(\mathrm e^{\mathrm itY^\lambda})$ is simply $\varphi_\lambda(t)=\mathrm e^{-\lambda(1-\mathrm e^{\mathrm it})}$. Thus $\mathbb E(\mathrm e^{\mathrm itZ^\lambda})=\mathrm e^{-\mathrm it\sqrt{\lambda}}\varphi_\lambda(t/\sqrt{\lambda})=\mathrm e^{-g_\lambda(t)}$ with $$ g_\lambda(t)=\mathrm it\sqrt{\lambda}+\lambda-\lambda\mathrm e^{\mathrm it/\sqrt{\lambda}}. $$ Expanding the exponential up to second order yields $$ g_\lambda(t)=\mathrm it\sqrt{\lambda}+\lambda-\lambda\cdot(1+\mathrm it/\sqrt{\lambda}-t^2/2\lambda)+o(1)\to\tfrac12t^2. $$ Thus, for every $t$, $\mathbb E(\mathrm e^{\mathrm itZ^\lambda})\to\mathrm e^{-t^2/2}=\mathbb E(\mathrm e^{\mathrm itZ})$ where $Z$ is standard normal, hence $Z^\lambda\to Z$ in distribution.

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Thanks! Are you using Levy's Continuity Theorem to conclude that convergence of characteristic function here implies convergence in distribution? –  Jack Nov 28 '12 at 22:04
    
Yes, since the pointwise limit itself is a characteristic function. –  Did Nov 29 '12 at 6:02

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