Moment Generating Function (proof of definition)

by Oxymoron
Tags: definition, function, generating, moment, proof
 P: 867 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 $$e^{tx}$$.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,565 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 $$E(u(x))= \int_{-\infty}^\infty u(x)f(x)dx[/itex] Replace $u(x)$ with $e^{tx}$ and you have it. It is true that the whole point of the "moment generating function" is that the coefficients of the powers of x in a power series expansion are the "moments" of the probability distribution, but that doesn't seem to me to be relevant to this question. I see no reason to write its Taylor series.  P: 867 Good, okay that makes sense. Then I suppose all I had to do was prove [tex] E(u(x))= \int_{-\infty}^\infty u(x)f(x)dx$$ 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.