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Given that $E[g(X)] = \int_{-\infty}^\infty g(x)f(x) dx$, how to prove $E[g(X)] = \int_{0}^\infty g'(x)S(x)dx$, where $S(x) = 1-F(x)$?

By integration by parts, I can get the following: \begin{align} E[g(X)] &= \int_{-\infty}^\infty g(x)f(x) dx \\ &= \int_{-\infty}^\infty g(x)F'(x) dx \\ &= g(x)F(x)|_{-\infty}^\infty - \int_{-\infty}^\infty g'(x)F(x) dx \\ &= g(x)F(x)|_{-\infty}^\infty + \int_{-\infty}^\infty g'(x)[1-S(x)] dx \\ &= g(x)F(x)|_{-\infty}^\infty - \int_{-\infty}^\infty g'(x)dx +\int_{-\infty}^\infty g'(x)S(x) dx\\ \end{align}

Then I am stuck.

Jango
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2 Answers2

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I think what you are trying to prove is not exactly correct. I will work through the problem and show the correct answer at the end.

You need to split the integral into two, one for the region $x\ge 0$ and one for $x\le 0$.

In the positive region, instead of writing $f(x)=F'(x)$ and integrating by parts, you need to use $f=(F(x)-1)'$. This is the only choice which ensures that the resulting integral actually exists; note that $F(x)-1$ goes to zero as $x\to+\infty$, but $F(x)$ does not.

\begin{align} \int_0^\infty g(x)f(x)\,dx &=\int_0^\infty g(x)[F(x)-1]'\,dx \\&=g(x)[F(x)-1]\Big|^\infty_0-\int_0^\infty g'(s)[F(x)-1]\,dx \\&=\underbrace{\Big(\lim_{x\to\infty}g(x)[F(x)-1]\Big)}_{=0}-g(0)[F(0)-1]+\int_0^\infty g'(x)S(x)\,dx \end{align} It takes some doing, but you can show that that limit actually is zero, as long as $E[g(X)]$ is finite. See Is it true that $\lim\limits_{x\to\infty}{x·P[X>x]}=0$? for some inspiration. Edit: Actually, I am not quite sure this is true without some additional assumptions on $g.$ It is true as long as $g$ is either bounded or monotonic.

For the negative region, you do want to use $F(x)$ as the antiderivate of $f(x)$, because $\lim_{x\to-\infty}F(x)=0$. \begin{align} \int_{-\infty}^0 g(x)f(x)\,dx &=\int_{-\infty}^0 g(x)F(x)'\,dx \\&=g(x)F(x)\Big|^0_{-\infty}-\int_{-\infty}^0 g'(s)F(x)\,dx \\&=g(0)F(0)-\underbrace{\Big(\lim_{x\to-\infty}g(x)F(x)\Big)}_{=0}-\int_{-\infty}^0 g'(x)F(x)\,dx \end{align} Putting this all together, we get $$ \bbox[7px,border:2px solid red]{\int_{-\infty}^\infty g(x)f(x)\,dx=g(0)+\int_0^\infty g'(x)S(x)\,dx-\int_{-\infty}^0g'(x)F(x)\,dx} $$ This is correct for any function $g$ such that $E[g(X)]$ is finite. Edit: Well, as long as $g$ is either bounded or monotonic.

If we make further assumptions about $g$, we can write this more nicely. Assuming $g(-\infty):=\lim_{x\to-\infty}g(x)$ exists, you would have $$ g(0)=g(-\infty)+\int_{-\infty}^0 g'(x)\,dx $$ so $$ \bbox[7px,border:2px solid green]{\int_{-\infty}^\infty g(x)f(x)\,dx=g(-\infty)+\int_{-\infty}^\infty g'(x)S(x)\,dx} $$ You cannot in general avoid the presence of some constant $g(a)$. This is because the $\int g'(x)S(x)\,dx$ part of the formula does not change when $g$ is shifted by a constant (this does not affect $g'$), but $E[g(x)]$ does change when $g$ is shifted by a constant.

Mike Earnest
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  • Thank you very much. I did not think it is so difficult since I read this from "Loss Model" other than analysis. Anyway, your work is clear and easy to understand. Thanks. – Jango Apr 02 '19 at 14:34
  • See this question for confirmation that the formula works when $g$ is bounded or monotonic, but that it does not hold for $g$ in general. https://math.stackexchange.com/questions/4419069/the-conditions-for-lim-x-to-inftygx-barfx-0/ – Mike Earnest Nov 04 '24 at 23:03
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Write $1-F(x)=\int_{(x,\infty)} d\mu(y)$ where $\mu$ is the measure corresponding to $F$. Now apply Fubini/Tonelli Theorem to the right hand side.

Kavi Rama Murthy
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