**What is Stirling’s formula? **

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

where denotes that

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 be i.i.d. (independent and identically distributed) observations from the standard exponential distribution which has mean and variance both equal to Define for . It is well known that follows distribution, which has the density . We can standardize and define . Then by the standard Central Limit Theorem (CLT) we have as where Here denotes convergence in distribution.

**Connection with Stirling’s formula?**

Suprizingly enough, the expected value of relates to the Stirling’s formula. Let us do the calculation ourselves. First note that Substituting it becomes

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

Now integrating by parts, we get

and

Observe how it drastically simplifies the required integral and gives

On the other hand,

(check!).

Since it is no surprize that which gives us nothing but the Stirling’s formula.

**Are we done yet?**

We wrote above that implies which actually does **not** hold in general. However, in this case we are fortunate, because we can apply the following result [1]:

**Result:** If as and if then we can conclude that as

Note that in our case is the standardized so for every Hence applying the above result to we are through. A proof of the above result can be found here.

**References**

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

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