# A Probabilistic Proof of Stirling’s Formula

### What is Stirling’s formula?

It is familiar that the quantity $n!$ grows very quickly with $n$. Stirling’s formula gives an approximate value of $n!$ which can be computed more easily and is quite accurate even for small values of $n$. The formula states that

$\displaystyle n!\sim \sqrt{2\pi} e^{-n} n^{n+1/2},$

where $a_n \sim b_n$ denotes that $\displaystyle \lim_{n\to\infty} a_n/b_n = 1.$

Here I am going to discuss a probabilistic proof of Stirling’s formula. The proof is originally due to Rasul A. Khan [2].

### Setup of the probabilistic proof

Let $X_1, X_2, \dots$ be i.i.d. (independent and identically distributed) observations from the standard exponential distribution which has mean and variance both equal to $1.$ Define $S_n = \sum_{i=1}^n X_i,$ for $n\geq 1$. It is well known that $S_n$ follows $\text{Gamma}(n, 1)$ distribution, which has the density $\displaystyle \frac{e^{-x}x^{n-1}}{(n-1)!},\ x>0$. We can standardize $S_n$ and define $\displaystyle Z_n = \frac{S_n - n}{\sqrt{n}}$. Then by the standard Central Limit Theorem (CLT) we have $Z_n \stackrel{d}{\rightarrow} Z$ as $n\to\infty,$ where $Z\sim N(0, 1).$ Here $\stackrel{d}{\rightarrow}$ denotes convergence in distribution.

### Connection with Stirling’s formula?

Suprizingly enough, the expected value of $|Z_n|$ relates to the Stirling’s formula. Let us do the calculation ourselves. First note that $\displaystyle E|Z_n| = E\left|\frac{S_n-n}{\sqrt{n}}\right|=\int_{0}^\infty \left|\frac{x-n}{\sqrt{n}}\right|\frac{e^{-x}x^{n-1}}{(n-1)!}dx.$ Substituting $u = x/n,$ it becomes

$\displaystyle \frac{n^{n+1/2}}{(n-1)!}\int_{0}^\infty \left|u-1\right|e^{-nu}u^{n-1}du.$

Keep aside the constant factor and split the last integral in two parts:

$\displaystyle \int_{0}^1 (1-u)e^{-nu}u^{n-1}du+\int_{1}^\infty (u-1)e^{-nu}u^{n-1}du.$

Now integrating by parts, we get

$\displaystyle \int_0^1 u^{n-1}e^{-nu}du = e^{-nu}\frac{u^n}{n}\Big|_0^1 +\int_0^1 u^n e^{-nu}du$

and

$\displaystyle \int_1^\infty u^{n-1}e^{-nu}du = e^{-nu}\frac{u^n}{n}\Big|_1^\infty +\int_1^\infty u^n e^{-nu}du.$

Observe how it drastically simplifies the required integral and gives

$\displaystyle E|Z_n| = \frac{n^{n+1/2}}{(n-1)!}\left( e^{-nu}\frac{u^n}{n}\Big|_0^1 - e^{-nu}\frac{u^n}{n}\Big|_1^\infty\right) = \frac{2e^{-n}n^{n+1/2}}{n!}.$

On the other hand,

$\displaystyle E|Z| = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}|x|e^{-x^2/2}dx = \sqrt{\frac{2}{\pi}}$ (check!).

Since $Z_n \stackrel{d}{\rightarrow} Z,$ it is no surprize that $E|Z_n| \rightarrow E|Z|,$ which gives us nothing but the Stirling’s formula.

### Are we done yet?

We wrote above that $Z_n \stackrel{d}{\rightarrow} Z$ implies $E|Z_n| \rightarrow E|Z|,$ which actually does not hold in general. However, in this case we are fortunate, because we can apply the following result [1]:

Result: If $Z_n \stackrel{d}{\rightarrow} Z$ as $n\to\infty,$ and if $\sup_{n\ge 1} E(Z_n^2) <\infty,$ then we can conclude that $E(Z_n) \rightarrow E(Z)$ as $n\to\infty.$

Note that in our case $Z_n$ is the standardized $S_n,$ so $E(Z_n^2) = 1$ for every $n\ge 1.$ Hence applying the above result to $|Z_n|$ we are through. A proof of the above result can be found here.

### References

1. Billingsley, Patrick. Probability and measure. John Wiley & Sons, 2008.
2. Khan, Rasul A. “A probabilistic proof of Stirling’s formula.” The American Mathematical Monthly 81.4 (1974): 366-369.