Could you please help me to prove the following equation:
$$E(x^{-1})=\int_{0}^{\infty}M_{x}(-t)dt$$
Where $M_{x}(-t)$ is the moment-generating function. I think the following equation will be useful: $$x^{-1}=\int_{-\infty}^{0}e^{ux}du$$ for $x>0$
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$\begingroup$ If this question is for homework, please note that. If it is for self-study, please tag it as self-study and let us know where you got the question from. $\endgroup$– R CarnellCommented Mar 7, 2023 at 20:43
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$\begingroup$ I think applying a definition of the mgf would be the most useful way to start. But aren't you making some strong assumptions about the distribution of $x$? For instance, suppose $x$ is a constant random variable almost surely equal to $-1,$ so that $M_x(t)=E[e^{tx}] = e^{-t}.$ The integral of $M_x(-t)=e^t$ diverges, but $E[x^{-1}]=1.$ $\endgroup$– whuber ♦Commented Mar 7, 2023 at 21:11
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2$\begingroup$ "You think" doesn't mean what you listed is correct. Please consider other's comments and suggestions seriously before making the claim, or even accepted a wrong answer! $\endgroup$– ZhanxiongCommented Mar 11, 2023 at 3:48
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1 Answer
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$$E\left(\frac{1}{x}\right)=\int_{-\infty}^{0}{e^{ux}du}=\int_{0}^{\infty}{e^{-ux}du}$$
$$E\left(\frac{1}{x}\right)=\int_{0}^{\infty}{x^{-1}f\left(x\right)dx=\int_{0}^{\infty}{\left(\int_{0}^{\infty}{e^{-ux}du}\right)f\left(x\right)dx=}}$$
$$\int_{0}^{\infty}{\left(\int_{0}^{\infty}{e^{-ux}f(x)dx}\right)du\ =\int_{0}^{\infty}{M_X(-u)du\ =\ \int_{0}^{\infty}{M_X(-t)dt}}\ }$$
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1$\begingroup$ The first line should not begin with "E()." The second line is valid only when $x$ has non-negative support, which is not a condition included in your question. $\endgroup$– whuber ♦Commented Mar 10, 2023 at 22:22