For example, the proof that the sum of two Poisson Random Variables is also a Poisson Random Variable, is much easier if we can invoke this result.

**Sum of Two Poisson Distributions:**

We can now use the result that:

Which is the $MGF$ of a Poisson Distribution with distribution $\lambda_1 + \lambda_2$ proving the result.

**Do we need a proof?**

Call me pedantic, but I never liked the fact that this uniqueness result is normally stated without proof. For me, there was always an elegance to a textbook that began with definitions and axioms and proved everything that it required on the way.

There also seems to be something disingenuous to me in using a very powerful result to prove a fairly trivial special case without really understanding the result you are using. You never really understand why anything is true, just that it is true.

Not only is this proof not often given, I had to look pretty hard to find it anywhere..

The general proof for all random variables requires measure theory and is quite complex, therefore I thought I'd just write up the result for discrete random variables.

So here is a proof of the uniqueness of $MGFs$ for discrete random variables.

**Uniqueness of MGFs**

Suppose $X$ and $Y$ are discrete random variables over $\{ x_0 , x_1, x_2, ... \} $.

Further suppose that:

$$MGF_X = MGF_Y$$

That is:

$$\forall t \in \mathbb{R} , \sum_{i=0}^{\infty} e^{tx_i} f_X (x_i) = \sum_{j=0}^{ \infty } e^{tx_j} f_Y (x_j)$$

Then,

$$\sum_{i=0}^{\infty} e^{tx_i} f_X (x_i) - \sum_{j=0}^{ \infty } e^{tx_j} f_Y (x_j) = 0$$

Changing the range for the second sum: $$\sum_{i=0}^{\infty} e^{tx_i} f_X (x_i) - \sum_{i=0}^{ \infty } e^{tx_i} f_Y (x_i) = 0 $$

Bringing the two sums together: $$\sum_{i=0}^{\infty} ( e^{tx_i} f_X (x_i) - e^{ty_i} f_Y (x_i) ) = 0 $$

Rearranging:

$$\sum_{i=0}^{\infty} e^{tx_i}(f_X (x_i) - f_Y (x_i) ) = 0 $$

Since each $e^tx$ is strictly greater than 0, and the sum is equal to 0, this implies that:

Which proves the result.