Show that the first derivative at t=0 of $ln(M_{x}(t))$ is the expected value of X and its second derivative at t=0 is the Variance of X

Question

Let $M_{X}(t)$ be a moment generating function of X. So far, I know that the first derivative of $\ln M_{x}(t)$ would be $M'_{X}(t)/M_{X}(t)$ and the second derivative would be $M_{X}(t) M_{X}''(t)-M_{X}'(t)M_{X}'(t)/M_{X}'(t)^2$. I think my issue is that if you plug t=0 into these equations, I'm not sure how else to keep evaluating it. I could really use some help.

Answer 1

Your second derivative should be $$ \frac{M_{X}(t) M_{X}^{\prime \prime}(t)-M_{X}^{\prime}(t) M_{X}^{\prime}(t)}{[M_{X}(t)]^{2}}. $$ (i.e. There is no derivative in the denominator and also be careful and use brackets to enclose the entire numerator if you're going to write it on one line like that.)

Note that $M_{X}(t) = E[e^{tX}]$ $\Rightarrow$ $M_{X}^{\prime}(t) = E[X e^{tX}]$ and $M_{X}^{\prime \prime}(t) = E[X^{2} e^{tX}]$.

So, $M_{X}^{\prime}(0) = E[X]$ and $M_{X}^{\prime \prime}(0) = E[X^{2}]$.

(Also, $M_{X}(0) = E[e^{0\cdot X}] = E[1]=1$.)