# moment generating function of the sum of independent random variables

Let $X_{i}$ be independent random variables for $i=1,\ldots,n$, 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,

 $\displaystyle M_{X}(t)$ $\displaystyle=E\left(\mathrm{e}^{tX}\right)$ $\displaystyle=E\left(\mathrm{e}^{t\left(X_{1}+\cdots+X_{n}\right)}\right)$ $\displaystyle=E\left(\mathrm{e}^{tX_{1}}\cdots\mathrm{e}^{tX_{n}}\right).$

Now, since each $X_{i}$ is independent of the others, this becomes

 $\displaystyle M_{X}(t)$ $\displaystyle=E\left(\mathrm{e}^{tX_{1}}\right)\cdots E\left(\mathrm{e}^{tX_{n% }}\right)$ $\displaystyle=M_{X_{1}}(t)\cdots M_{X_{n}}(t)$ $\displaystyle=\prod_{i=1}^{n}M_{X_{i}}(t)$

as required. ∎

Title moment generating function of the sum of independent random variables MomentGeneratingFunctionOfTheSumOfIndependentRandomVariables 2013-03-22 17:17:11 2013-03-22 17:17:11 me_and (17092) me_and (17092) 5 me_and (17092) Corollary msc 60E05