Let $M_{X}(t)$ be an mgf of $X$. Show that the first derivative of $\ln M_{X}(t)$ at $t=0$ is $\mathbb{E}[X]$ and the second derivative of $\ln M_{X}(t)$ at $t=0$ is $\text{Var}[X]$
I'm not entirely sure this statement is correct because what I learned in class was that if we took the $n$th derivative of $M_x(t)$ at $t=0$, then we get the expectation of $X$. But once you put in something the $\log$ function it seems like this statement doesn't really work anymore. Additionally I got something like the integrand over the integral, which doesn't seem to make any sense.
How can this be proved?
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$\begingroup$$\newcommand{\deriv}[2]{\dfrac{\text{d}}{\text{d}#1}\left[#2\right]}$
1.$$\frac{d}{dt}{\ln M_{X}(t)}|_{t=0} = \frac{M^{'}_{X}(t)}{M_{X}(t)}|_{t=0} = \frac{M^{'}_{X}(0)}{M_{X}(0)} = \frac{E[X]}{1}$$
2. $$\begin{align}\frac{d^2}{dt^2}{\ln M_{X}(t)}|_{t=0} &= \deriv{t}{\frac{M^{'}_{X}(t)}{M_{X}(t)}}|_{t=0}\\ &= \frac{M_{X}(t)M^{''}_{X}(t) - (M^{'}_{X}(t))^2}{(M_{X}(t))^2}|_{t=0}\\ &= \frac{M_{X}(0)M^{''}_{X}(0) - (M^{'}_{X}(0))^2}{(M_{X}(0))^2} \\ &= \frac{(1)E[X^2] - (E[X])^2}{(1)} \end{align}$$
$\endgroup$ $\begingroup$$\newcommand{\deriv}[2]{\dfrac{\text{d}}{\text{d}#1}\left[#2\right]}$Hint (not a complete solution): $$\deriv{t}{\ln M_{X}(t)} = \dfrac{M^{\prime}_{X}(t)}{M_{X}(t)}$$ and use the quotient rule to take the second derivative of $\ln M_{X}(t)$.
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