# Moment Generating Function (proof of definition)

1. Mar 24, 2011

### Oxymoron

1. The problem statement, all variables and given/known data
Prove that for a random variable $$X$$ with continuous probability distribution function $$f_X(x)$$ that the Moment Generating Function, defined as

$$M_X(t) := E[e^{tX}]$$

is

$$M_X(t) = \int_x^{\infty}e^{tx}f_X(x)dx$$

2. Relevant equations

Above and

$$E[X] = \int_{-\infty}^{\infty}xf_X(x)dx$$

3. The attempt at a solution

This expression is given in so many textbooks and the ones that I have read all skip over this derivation. I want to be able to prove (1) to myself.

Proof:
Write the exponential function as a Maclaurin series:

$$M_X(t) = E[e^{tX}]$$

$$= E[1+tX+\frac{t^2}{2!}X^2+\frac{t^3}{3!}X^3+...]$$

Since $$E[1] = 1$$ and the $$E[t^n/n!]=t^n/n!$$ because they are constant and the expectation of a constant is itself you get:

$$= 1+tE[X]+\frac{t^2}{2!}E[X^2]+\frac{t^3}{3!}E[X^3]+...$$

...also using the linearity of E. Now, writing the series as a sum:

$$=\sum_{t=0}^{\infty}\frac{t^n}{n!}E[X^n]$$

And extracting the exponential:

$$=e^t\sum_{n=0}^{\infty}E[X^n]$$

Now I am stuck! I know that I am meant to use

$$E[X] = \int_{-\infty}^{\infty}xf_X(x)dx$$

but I have $$E[X^n]$$ and I also have $$e^t$$ and not $$e^{tx}$$.

Last edited: Mar 24, 2011
2. Mar 24, 2011

### HallsofIvy

Staff Emeritus
There isn't really much to prove. For any continuous probability distribution with density function f(x), the Expected value of any function u(x) is defined to be
$$E(u(x))= \int_{-\infty}^\infty u(x)f(x)dx[/itex] Replace $u(x)$ with $e^{tx}$ and you have it. It is true that the whole point of the "moment generating function" is that the coefficients of the powers of x in a power series expansion are the "moments" of the probability distribution, but that doesn't seem to me to be relevant to this question. I see no reason to write its Taylor series. 3. Mar 24, 2011 ### Oxymoron Good, okay that makes sense. Then I suppose all I had to do was prove [tex] E(u(x))= \int_{-\infty}^\infty u(x)f(x)dx$$

and then substitute $$u(x)$$ with $$e^{tx}$$ as you said and I'm done.

But once again there is nothing to prove because it is a definition.