Moment Generating Function (proof of definition)

[Oxymoron]

Homework Statement

Prove that for a random variable X with continuous probability distribution function f_X(x) that the Moment Generating Function, defined as

<br /> M_X(t) := E[e^{tX}]<br />

is

<br /> M_X(t) = \int_x^{\infty}e^{tx}f_X(x)dx <br />

Homework Equations

Above and

<br /> E[X] = \int_{-\infty}^{\infty}xf_X(x)dx<br />

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:

<br /> M_X(t) = E[e^{tX}] <br />

<br /> = E[1+tX+\frac{t^2}{2!}X^2+\frac{t^3}{3!}X^3+...]<br />

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:

<br /> = 1+tE[X]+\frac{t^2}{2!}E[X^2]+\frac{t^3}{3!}E[X^3]+...<br />...also using the linearity of E. Now, writing the series as a sum:

<br /> =\sum_{t=0}^{\infty}\frac{t^n}{n!}E[X^n]<br />

And extracting the exponential:

<br /> =e^t\sum_{n=0}^{\infty}E[X^n]<br />

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

<br /> E[X] = \int_{-\infty}^{\infty}xf_X(x)dx<br />

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

 

[HallsofIvy]

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]<br /> <br /> Replace u(x) with e^{tx} and you have it. It is true that the whole point of the &quot;moment generating function&quot; is that the coefficients of the powers of x in a power series expansion are the &quot;moments&quot; of the probability distribution, but that doesn&#039;t seem to me to be relevant to this question. I see no reason to write its Taylor series.

 

[Oxymoron]

Good, okay that makes sense.

Then I suppose all I had to do was prove

<br /> E(u(x))= \int_{-\infty}^\infty u(x)f(x)dx<br />

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.