An actuary determines that the claim size for a certain class of accidents is a random variable, X, with moment generation function: $$M_X(t)=\frac{1}{(1-2500t)^4}.$$ Calculate the standard deviation of the claim size for this class of accidents.
(Note: the shortcut for this question is seeing that it's a gamma distribution function with parameters 4 and 2500. Therefore, Var(X) will be $4\cdot 2500^2 = 25,000,000$ and the standard deviation will then be $\sqrt{25,000,000} = 5,000$. However, let's assume we cannot see this.)
From I understand, the SD can be found by doing this calculation:
$$SD(X)=\sqrt{M''(0)-[M'(0)]^2}$$ The reason to that is because $SD(X)=\sqrt{Var(X)}=\sqrt{E(X^2)-[E(X)]^2}$ and $E(X^n) = M^n(0)$.
($M^n(0)$ is the derivative of $M(X)$ to the $n$-th time at $0$)
Why is $E(X^n) = M^n(0)$ true?
By definition, $M(t)=\int_\Omega e^{tX}dP$. Then, $M^{(n)}(t)=\int_\Omega X^ne^{tX}dP$ (it follows by differentiated $n$ times) Then, $M^{(n)}(0)=\int_\Omega X^ndP$, and the last is precisly $E[X^n]$.