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The MGF, $M_x(t)$ is a function of $t$. It has the property that $\lim_{t\to 0} M_x(t)=1$. It can be shown that:

$\lim_{t\to 0}\frac{d}{dt} \log[M_x(t)]=E[X^1]=E[X]$

Find an expression for

$\lim_{t\to 0}\frac{d^2}{dt^2} \log[M_x(t)]$

That is the problem that is presented. I think the best way to show this is by, $E[X^r]= \lim_{t\to 0}\frac{d^r}{dt^r} [M_x(t)] = \lim_{t\to 0}\frac{d^2}{dt^2} log[M_x(t)]$

but from there I'm not sure where the next steps would be. We are just trying to find an expression that will help us solve for $\frac {d^2}{dt^2}$

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  • $\begingroup$ @ThomasAndrew and ideas of how to attack this beast of a problem? $\endgroup$ Commented Dec 15, 2021 at 6:02

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$$\frac{d}{dt}\log[M_X(t)]=\frac{M_X'(t)}{M_X(t)}$$

$$\frac{d^2}{dt^2}\log[M_X(t)]=\frac{d}{dt}\frac{M_X'(t)}{M_X(t)}=\frac{M_X(t)M_X''(t)-M_X'(t)^2}{M_X(t)^2}$$

Now, try to take the limit.

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