(adsbygoogle = window.adsbygoogle || []).push({}); 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].

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# Moment Generating Function (proof of definition)

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