Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The goal of this problem is to prove that $P(S_n = k)\sqrt{2\pi n} \rightarrow e^{-\frac{x^2}{2}}$ by using Stirling's formula.

Here is what is given:

1) $S_n = \sum_1^n X_i$, where $\{X_i\}$ are independent, with Poisson distribution, mean 1.

2) Stirling's formula is that $\dfrac{n!}{n^n e^{-n} \sqrt{2 \pi n}} \rightarrow 1$.

3) $\frac{k-n}{\sqrt n} \rightarrow x$

Here is what I don't understand:

1) What does it mean for $\frac{k-n}{\sqrt n} \rightarrow x$? As what approaches $\infty$? I thought that $k$ and $n$ were integers? How are they related? Are they really integer-valued functions? Or is $k$ a function of $n$, somehow? None of this is clear from the context.

Here is what I have, though I'm assuming all limits are in terms of $n$:

1) $-\frac{(n-k)^2}{2n} \rightarrow \frac{-x^2}{2}$.

2) Since $\sqrt{n} \rightarrow \infty$, we know that $(n-k)^2 \rightarrow \infty$.

3) There is a lemma which I can use to show that $(1 + (\frac{-1}{2n}))^{(n-k)^2} \rightarrow e^\frac{-x^2}{2}$. (The lemma states that if $c_ja_j \rightarrow \lambda$ where $c_j \rightarrow \infty$ and $a_j \rightarrow 0$ then $(1+a_j)^{c_j} \rightarrow e^\lambda$).

4) Putting all this together yields that $$ P(S_n = k)\sqrt{2\pi n} = \frac{e^{-n}n^k \sqrt{2\pi n} }{k!} = \frac{e^{-n} n^n \sqrt{2 \pi n}}{n!} \cdot \frac{n! n^k}{k! n^n}$$.

5) So what I would love to do is show that $\frac{n! n^k}{k! n^n} \rightarrow e^\frac{-x^2}{2}$, or equivalently, by using 3), that $|\frac{n! n^k}{k! n^n} - (\frac{2n-1}{2n})^{(n-k)^2}| \rightarrow 0$.

Any tips or insights or clarifications will be greatly appreciated.


share|improve this question
Treat $k$ as fixed. $n$=1,2,3,4,... is a sequence of integers approaching infinity. $k$ does not depend on $n$, but $P(S_n=k)$ clearly does depend on $n$. –  Silverfish Nov 6 '13 at 2:24
If what you say is true, then why does $(k-n)/\sqrt n$ not approach $-\infty$ –  user28877 Nov 6 '13 at 3:02
Bingo, that's exactly what it does! Think of it like the z score for your Sn. Let's take k as 10. When n is small, eg 3, it's unlikely Sn will be so high (add three Poisons with mean 1 and see). K is well above the expected value of $S_3$ and so has a positive z score. But when n is 50 it's very unlikely $S_{50}$ will be anywhere near as low as 10. K is now well below the mean and has a very negative z score. As n gets bigger the z score will get ever more negative.... –  Silverfish Nov 6 '13 at 3:32
Btw you can see it is like a z score when you realise $S_n$ is Poisson with parameter n so the mean and variance are both n and the standard deviation is the square root of that. –  Silverfish Nov 6 '13 at 3:46
Yes I see what you mean, though I still think that $(k-n)/\sqrt n $ cannot converge to $x \in \mathbb{R}$ unless $k$ is a function of $n$. Is this some kind of shorthand? –  user28877 Nov 6 '13 at 4:31
show 2 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.