Moment Generating Function (proof of definition)by Oxymoron Tags: definition, function, generating, moment, proof 

#1
Mar2411, 06:45 AM

P: 867

1. The problem statement, all variables and given/known data
Prove that for a random variable [tex]X[/tex] with continuous probability distribution function [tex]f_X(x)[/tex] that the Moment Generating Function, defined as [tex] M_X(t) := E[e^{tX}] [/tex] is [tex] M_X(t) = \int_x^{\infty}e^{tx}f_X(x)dx [/tex] 2. Relevant equations Above and [tex] E[X] = \int_{\infty}^{\infty}xf_X(x)dx [/tex] 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: [tex] M_X(t) = E[e^{tX}] [/tex] [tex] = E[1+tX+\frac{t^2}{2!}X^2+\frac{t^3}{3!}X^3+...] [/tex] Since [tex]E[1] = 1[/tex] and the [tex]E[t^n/n!]=t^n/n![/tex] because they are constant and the expectation of a constant is itself you get: [tex] = 1+tE[X]+\frac{t^2}{2!}E[X^2]+\frac{t^3}{3!}E[X^3]+... [/tex] ...also using the linearity of E. Now, writing the series as a sum: [tex] =\sum_{t=0}^{\infty}\frac{t^n}{n!}E[X^n] [/tex] And extracting the exponential: [tex] =e^t\sum_{n=0}^{\infty}E[X^n] [/tex] Now I am stuck! I know that I am meant to use [tex] E[X] = \int_{\infty}^{\infty}xf_X(x)dx [/tex] but I have [tex]E[X^n][/tex] and I also have [tex]e^t[/tex] and not [tex]e^{tx}[/tex]. 



#2
Mar2411, 07:02 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,881

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
[tex]E(u(x))= \int_{\infty}^\infty u(x)f(x)dx[/itex] Replace [itex]u(x)[/itex] with [itex]e^{tx}[/itex] 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
Mar2411, 07:17 AM

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 [/tex] and then substitute [tex]u(x)[/tex] with [tex]e^{tx}[/tex] as you said and I'm done. But once again there is nothing to prove because it is a definition. 


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