Finding Standard Deviation from MGF

Question

I a working on a problem and am a little bit confused at how to approach solving it.

The problem: Given the MGF F(t) = $1\over(1-2500t)^4$

Calculate the SD.

Do we need to do some substitution with like $\mathscr N$ ~ (0,1) with sigma and mu? Or find the Var and square root it?

More broadly, how do we find the Standard Deviation of a Moment Generating Function?

Answer 1

You don't find SD of a MGF ... you find SD of a random variable, cmon.

Let $X$ be your random variable, then MGF of $X$ is basically $M_X(t) = E[e^{Xt}]$

If you check your textbook, you will find a nice property for the MGF, namely the n-th derivative of M evaluated at zero is equal to $E[X^n]$

Now, SD = $\sqrt{Var(X)} = \sqrt{E[X^2] - E[X]^2}$.

Can you take it from there?

Answer 2

Another way is to recognize that the MGF belongs to the Gamma distribution with shape parameter $\alpha=4$ and scale parameter $\beta=2500$. The variance follows directly from the variance formula $\alpha\beta^2$ of the Gamma distribution. A more direct approach is to consider the cumulant generating function $\kappa(t)=\log F(t)=-4\log(1-2500t)$ and realize that the variance is the 2nd derivative of $\kappa(t)$ evaluated at zero.