If I have the mean, variance, PDF and CDF of a random variable $X$ how to easily construct the MGF.
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3You only need the CDF; the MGF is $\int_{-\infty}^\infty e^{tx}dF(x)$. If the PDF exists, you can get it from that instead. – J.G. Dec 17 '23 at 13:28
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1Thanks for the comment, you mean like this: $M_X(t) = \mathbb{E}[e^{tX}] = \int^{\infty}_{0} e^{tx}f(x)dx$, where $X$ is positive random variable – Math Explorer Dec 17 '23 at 13:48
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1That's it exactly. By contrast, even infinitely many moments might not fix the distribution, so don't determine the MGF. (Knowing $M_X$ on $\Bbb Q$ is enough, provided $M_X$ is everywhere finite, but knowing it on $\Bbb N$ isn't.) – J.G. Dec 17 '23 at 16:33