Central Limit Theorem to evaluate $\frac{1}{(n-1)!} \int_0^n e^{-x}x^{n-1}dx$

Question

Use the central limit theorem to evaluate $$ \frac{1}{(n-1)!} \int_0^n e^{-x}x^{n-1}dx$$

From what I understand, the CLT says that the sequence of partial sums of i.i.d random variables $S_n = \sum_{i=1}^n$ converges in distribution (once it's standardized by subtracting the mean and dividing by the standard deviation) to the CDF of a $N(0,1)$.

If we consider $P(S_n \le x)$ where $S_n$ is the sum of i.i.d exp(1) RVS, then by the CLT $P(\frac{S_n - n}{\sqrt{n}} \le x) \Rightarrow \int_{-\infty}^x \frac{e^{-u^2/2}}{\sqrt{2\pi}}du$. Or equivalently, $P(S_n \le x) \approx N(\frac{x-n}{\sqrt{n}})$.

However, this isn't even remotely close to what I am suppose to find. I'm confused because I'm not quite sure how to to approach this problem.

Answer 1

I thought that it might be instructive to present a simple approach that does not rely on the Central Limit Theorem and Probability theory, but rather uses real analysis only. To that end, we proceed.

We can assert that

$$\frac1{n!}\int_0^{n+1}e^{-x} x^n\,dx=\frac1{n!}\int_0^n e^{-x}x^n\,dx+\underbrace{\frac1{n!}\int_n^{n+1}e^{-x}x^n\,dx}_{=O\left(n^{-1/2}\right)}\tag1$$

Next, enforce the substitution $x\mapsto n-\sqrt n x$ to find that

$$\begin{align} \frac1{n!}\int_0^{n+1}e^{-x} x^n\,dx&=\frac{\sqrt{n}}{n!}\int_0^{\sqrt n} e^{-(n-\sqrt n x)}\left(n-\sqrt n x\right)^n\,dx+O\left(n^{-1/2}\right)\\\\ &=\frac{\sqrt{n}(n/e)^n}{n!}\int_0^{\sqrt n}e^{\sqrt n x}\left(1-\frac{x}{\sqrt n}\right)^n\,dx+O\left(n^{-1/2}\right) \end{align}$$

Next, we have the estimate

$$e^{\sqrt n x}\left(1-\frac{x}{\sqrt n}\right)^n=e^{\sqrt n x}e^{n\log\left(1-\frac{x}{\sqrt n}\right)}\le e^{-x^2/2}$$

Inasmuch as $\int_0^\infty e^{-x^2/2}\,dx<\infty$, the Dominated Convergence Theorem along with Stirling's Formula guarantee that

$$\begin{align} \lim_{n\to \infty}\frac{1}{n!}\int_0^{n+1}e^{-x}x^n\,dx&=\lim_{n\to\infty}\left(\frac{\sqrt{n}(n/e)^n}{n!}\int_0^{\sqrt n}e^{\sqrt n x}\left(1-\frac{x}{\sqrt n}\right)^n\,dx+O\left(n^{-1/2}\right)\right)\\\\ &=\lim_{n\to\infty}\left(\frac{\sqrt{n}(n/e)^n}{n!}\int_0^{\infty}\xi_{[0,\sqrt n]}(x)e^{\sqrt n x}\left(1-\frac{x}{\sqrt n}\right)^n\,dx\\+O\left(n^{-1/2}\right)\right)\\\\ &=\frac1{\sqrt{2\pi}}\int_0^\infty \lim_{n\to \infty }\left(\xi_{[0,\sqrt n]}(x)e^{\sqrt n x}\left(1-\frac{x}{\sqrt n}\right)^n\right)\,dx\\\\ &=\frac1{\sqrt{2\pi}}\int_0^\infty e^{-x^2/2}\,dx\\\\ &=\frac12 \end{align}$$

And we are done!

Answer 2

We're actually using the large-$n$ approximation $S_n\approx N(n,\,n)$ to evaluate $P(S_n\le\color{blue}{n})$, so the probability is approximated as $\tfrac12$.