6

Suppose $X_n\sim \text{Poisson}(n)$. Show that $\sqrt{X_n}-\sqrt{n}\overset{d}\to \mathcal N(0,1/4)$.

I already know that $(X_n-n)/\sqrt{n}\overset{d}\to \mathcal N(0,1)$. How to do next to go through the proof?

J.Mike
  • 671
  • 3
    For a one-line proof, note that $$\sqrt{X_n}-\sqrt{n}=\frac{X_n-n}{\sqrt{n}}\cdot\frac{1}{\sqrt{\frac{X_n}n}+1}$$ where the first term of the product on the RHS converges to $N(0,1)$ in distribution by the usual CLT and the second term converges to $\frac12$ by the usual LLN, qed. – Did Jan 02 '17 at 00:16
  • @Did The following shows "the second term converges to 1/2". Are there other faster ways? https://math.stackexchange.com/questions/1832338/find-the-almost-sure-limit-of-x-n-n-where-each-random-variable-x-n-has-a-po – Mathman Mar 04 '24 at 06:04

3 Answers3

8

Set $Y_n = \frac{X_n}{n}$, then you already know $\sqrt{n}(Y_n - 1) \to \mathcal{N}(0, 1)$.

Setting $g(x) = \sqrt{n}$, the delta method yields $$\sqrt{n}(g(Y_n) - g(1)) \to \mathcal{N}(0, g'(1)^2).$$

Now note that $g'(x) = \frac{1}{2\sqrt{x}}$, i.e. $g'(1)^2 = \frac{1}{4}$ and $$\sqrt{n}(g(Y_n) - g(1)) = \sqrt{n} \left(\sqrt{\frac{X_n}{n}} - 1\right) = \sqrt{X_n} - \sqrt{n}.$$

Dominik
  • 19,963
7

Hint: Apply the Delta method with $g(x)=\sqrt{x}$, on $$(X_n-n)/\sqrt{n}=\sqrt{n}\left(\frac{X_n}n-1\right) \overset{d}\to \mathcal N(0,1)$$

Jimmy R.
  • 35,868
2

$\DeclareMathOperator{\Pr}{\mathbb{P}}$ $\newcommand{\Fl}[1]{\lfloor #1 \rfloor}$

For any real $t$ we have $$ \begin{align} \Pr(\sqrt{X_n} - \sqrt{n} \leq t) &= \Pr(X_n \leq (t+ \sqrt{n})^2)\\ &= \Pr( \dfrac{X_n - n}{\sqrt{n}}\leq 2t + \frac{t^2}{\sqrt{n}})\\ &= \Pr( \dfrac{X_n - n}{\sqrt{n}}\leq 2t ) + \Pr\left( 2t < \dfrac{X_n - n}{\sqrt{n}}\leq 2t + \frac{t^2}{\sqrt{n}}\right) \end{align} $$

You already know $\lim_{n \to \infty} \Pr( \dfrac{X_n - n}{\sqrt{n}}\leq 2t ) = \Phi(2t)$ where $\Phi$ is the cdf of the standard normal distribution.

We will now show $\displaystyle\lim_{n\to\infty} \Pr\left( 2t < \frac{X_n - n}{\sqrt{n}}\leq 2t + \frac{t^2}{\sqrt{n}}\right) = \lim_{n\to\infty} \Pr\left(n + 2t\sqrt{n} < X_n \leq n + 2t\sqrt{n} + t^2 \right) =0.$

Let $U$ be any integer valued random variable, given real $ a \leq b$ we have $$\Pr( a < U \leq b) = \Pr( \Fl{a} < U \leq \Fl{b} ) = \sum_{k = \Fl{a} + 1}^{\Fl{b}} P(U = k) \leq p^{*} (\Fl{b} - \Fl{a}) \leq p^{*} ( b - a + 1)$$ where $p^{*} = \sup\{ \Pr(U=k) :k \in \mathbb{Z} \}.$

If $Y$ is Poisson with mean $m \in \mathbb{Z}$ with $m > 1$, it is easy to verify $\Pr(Y=0) \leq \Pr(Y=1) \leq \dots \leq \Pr(Y=m-1) = \Pr(Y=m) \leq \Pr(Y=m+1) \\ \leq \Pr(Y=m+2) \dots $

so the maximum value of $\Pr(Y=k)$ for $ k = 0,1,2,\dots$ is $\Pr(Y=m).$

This means for $n > 1$ $$\Pr\left(n + 2t\sqrt{n} < X_n \leq n + 2t\sqrt{n} + t^2 \right) \leq P(X_n = n) (t^2 + 1). $$

From Stirlings approximation we know $P(X_n = n) = \exp(-n) \dfrac{n^n}{n!} \approx \exp(-n) \dfrac{n^n}{\sqrt{2\pi} \exp(-n) n^{n+1/2} } = \text{O}(\dfrac{1}{\sqrt{n}})$ which implies $$\displaystyle\lim_{n\to\infty} \Pr\left( 2t < \frac{X_n - n}{\sqrt{n}}\leq 2t + \frac{t^2}{\sqrt{n}}\right) = \lim_{n\to\infty} \Pr\left(n + 2t\sqrt{n} < X_n \leq n + 2t\sqrt{n} + t^2 \right) =0.$$

Hence $\lim_{n\to\infty}\Pr(\sqrt{X_n} - \sqrt{n} \leq t) = \Phi(2t)$ and we are done.

Jimmy R.
  • 35,868
Arin Chaudhuri
  • 6,349