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Let $X_n$ be distributed as a poisson random variable with parameter $n$. Then which of the following are true ?

1.$\underset{n\rightarrow \infty}{\lim} \mathbb{P} (X_n > n + \sqrt n)=0 $

2.$\underset{n\rightarrow \infty}{\lim} \mathbb{P} (X_n \le n)=1 $

3.$\underset{n\rightarrow \infty}{\lim}\mathbb{P} (X_n \le n)=\frac {1}{2} $

The answer is 3. I am not able to think how to start on this. Please any suggestions.

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3 Answers 3

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  • Let $\lambda$ be the lamda for your sum of independent Poisson random variables ($X_n$)
  • Since $\lambda$ approaches to infinity, we should be able to approximate it with $N(\lambda,\lambda)$.
  • The normal approximation centres on $\lambda$ and is symmetric. You get your answer.
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First of all, by elimination the first two are solvable (you can look at the variance and mean of the distribution).

But I strongly agree 3 is tricky, but I came up with an explanation: Poisson distribution has a nice property: $x_i - \operatorname{Poisson}(\lambda_i) \Rightarrow \sum_ix_i-\operatorname{Poisson}(\sum_i\lambda_i) $

Therefore:

$X_n = \sum_{i=1}^{n}X_i, X_i-\operatorname{Poisson}(1)\rightarrow_{n\rightarrow\infty}N\left(\mathbb{E}\left[\sum_iX_i\right],\sum_i\lambda_i\right)=N(n,n) \\\Rightarrow \mathbb{P}(X_n\leq n)=0.5$

The last part is true since Gaussian distribution is symmetric.

BTW - I encourage you to use the same way the calculate the actual probability in 2 and 3

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The limiting distribution of $X_{n}$ as $n\rightarrow\infty$ is normal. We can show this using the moment generating function.

Consider a standardized Poisson random variable:

$$\frac{X_{n}-n}{\sqrt{n}}$$

With moment generating function:

$$M_{X_{n}}(t)=E\big[e^{tX_{n}}\big]=e^{n(e^{t}-1)}$$

Taking the limit:

$$\begin{align} \lim_{n\rightarrow\infty}M_{(X_{n}-n)/\sqrt{n}}(t)&=\lim_{n\rightarrow\infty}E\bigg[\text{exp}\bigg(t\cdot\frac{X_{n}-n}{\sqrt{n}}\bigg)\bigg]\\ &=\lim_{n\rightarrow\infty}\text{exp}\big(-t\sqrt{n}\big)E\bigg[\text{exp}\bigg(\frac{tX_{n}}{\sqrt{n}}\bigg)\bigg]\\ &=\lim_{n\rightarrow\infty}\text{exp}\big(-t\sqrt{n}\big)\text{exp}\big(n(e^{t/\sqrt{n}}-1)\big)\\ &=\lim_{n\rightarrow\infty}\text{exp}\big(-t\sqrt{n}+n(tn^{-1/2}+t^{2}n^{-1}/2+t^{3}n^{-3/2}/6+\ldots)\big)\\ &=\lim_{n\rightarrow\infty}\text{exp}\big(t^{2}/2+t^{3}n^{-1/2}/6+\ldots\big)\\ &=\text{exp}\big(t^{2}/2\big)\\ &=M_{Z}(t) \end{align}$$

Which is the moment generating function of the standard normal random variable, $Z\sim N(0,1)$. Thus, the limiting distribution of our Poisson random variable is simply:

$$\sqrt{n}Z+n\sim N(n,n)$$

So for your options above:

  1. $$\begin{align}\mathbb{P}(X_{n}>n+\sqrt{n})&=\mathbb{P}(N(n,n)>n+\sqrt{n})\\ &=\mathbb{P}(N(0,1)>1)\\ &=1-\Phi(1)\\ &\ne 0 \end{align}$$

  2. $$\begin{align}\mathbb{P}(X_{n}<n)&=\mathbb{P}(N(n,n)<n)\\ &=\mathbb{P}(N(0,1)<0)\\ &=\Phi(0)\\ &\ne 1 \end{align}$$

  3. $$\begin{align}\mathbb{P}(X_{n}<n)&=\mathbb{P}(N(n,n)<n)\\ &=\mathbb{P}(N(0,1)<0)\\ &=\Phi(0)\\ &=0.5 \end{align}$$

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