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I am having trouble understanding the rationale of the following related to the MGF:

Function Mx(t)=E[exp(tX)], the expectation exists for all t in a neighbourhood of zero, and X has mean mx, show that （1）logMx(t)/t>=mx for t>0 (2) the above inequality becomes equality as t tends to 0.

Here is what I have so far: For (1), I applied Jensen's inequality, so that LogE[(exp(tX)] >= E[log(exp(tX)) = E[tX] = tE[X] = tmx. So the first problem was solved.

I think I stuck here, so I have two questions: (1) was I on the right path to solve the first problem? (2) how should I proceed to solve the second problem?

Thank you very much.

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    $\begingroup$ Please tag as self-study and read its wiki! Then, you can use $\LaTeX$ on this site, please do so, your question will be easier to read! $\endgroup$
    – kjetil b halvorsen ♦
    Commented Oct 5, 2020 at 21:30

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You'll take the limit of the expression and apply L'hopital's rule:

$$\lim_{t\rightarrow0}\frac{\log M_X(t)}{t}=\lim_{t\rightarrow0} \frac{M_X'(t)}{M_X(t)}=\frac{M'_X(0)}{M_X(0)}=\frac{\mathbb E[X]}{1}=mx$$

P.S. Please use $\LaTeX$ as suggested in the comments.

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  • $\begingroup$ Thank you so much for the help @gunes! Could you please elaborate on the part after your first equal sign? I am still confused about how it was obtained. $\endgroup$
    – QIHUN CHEN
    Commented Oct 6, 2020 at 5:18
  • $\begingroup$ just differentiated both numerator and denominator separately as the l’hopital rule requires $\endgroup$
    – gunes
    Commented Oct 6, 2020 at 6:31

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