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Although this question touched on MGF exists at neighborhood of 0, I still don't understand why the definition of MGF says $M_X{(t)}$ = E $e^{tX}$ for $t$ in neighborhood of 0. Why must it be 0 but not in neighborhood of an arbitrary number, say 1 for example?

Nemo
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  • Let's say I specify an MGF. How would you figure out what $\mu_2$ was? – Glen_b Jan 13 '20 at 02:45
  • Is it $\mu_2$ = $E(X^2)$?, or you meant we would take second derivative of MGF? – Nemo Jan 13 '20 at 02:58
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    Yes; sorry to be unclear, I meant ($E(X^2)$ (I left the dash off, it should have been $\mu_2^\prime$). Yes you take the second derivative, but where do you evaluate the second derivative? – Glen_b Jan 13 '20 at 07:13
  • Thanks for your confirmation, @Glen_b-ReinstateMonica. Second derivative is still a function of $x$ so it can be evaluated at any point within the domain, say $x$ = 1, for example? – Nemo Jan 13 '20 at 07:57
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    When finding a moment like $E(X^2)$ where do you evaluate the derivative? – Glen_b Jan 13 '20 at 12:12
  • @Glen_b-ReinstateMonica, I can evaluate the derivative at any value of $x$. However, I just came across this information which might explain why $M_X(t)$ needs to be evaluated at 0. Apparently, $M_X(t)$ = $E[e^{tX}]$ = $\sum_{k=0}^{\infty}E[X^k]\frac{t^k}{k!}$. Thus, the $k$th moment of $X$ is the coefficient of $\frac{t^k}{k!}$ in the Taylor series of $M_X(t)$; and that coefficient is obtained by taking the $k$th derivative of $M_X(t)$ and evaluating at $t$ = 0. Do you think that's the case? – Nemo Jan 14 '20 at 01:39
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    It sounds like you hadn't read much basic information on the MGF; sorry, but I wouldn't have guessed that from the phrasing of the question. That you evaluate the derivatives of the MGF at 0 (and multiply by a factorial) to extract the moments is central to my point - since otherwise, how is the moment generating function going to generate the moments? – Glen_b Jan 14 '20 at 01:55

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