moment generating function of the sum of independent random variables

Let $X_i$ be independent random variables for $i=1,\mathrm{\dots },n$, let each $X_i$ have moment generating function $M_{X_i}(t)$, and let $X={\sum }_{i=1}^n{X}_i$. Then the moment generating function of $X$ is

$$M_X(t)=\prod _{i=1}^n{M}_{X_i}(t).$$

Proof.

By definition,

$M_X(t)$$=E\left({\mathrm{e}}^{tX}\right)$

$=E\left({\mathrm{e}}^{t\left(X_1+\mathrm{\cdots }+X_n\right)}\right)$

$=E\left({\mathrm{e}}^{t{X}_1}\mathrm{\cdots }{\mathrm{e}}^{t{X}_n}\right).$

Now, since each $X_i$ is independent of the others, this becomes

$M_X(t)$$=E\left({\mathrm{e}}^{t{X}_1}\right)\mathrm{\cdots }E\left({\mathrm{e}}^{t{X}_n}\right)$

$=M_{X_1}(t)\mathrm{\cdots }{M}_{X_n}(t)$

$={\displaystyle \prod _{i=1}^n}{M}_{X_i}(t)$

as required.∎