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Suppose I have some random variable $W$ along with its expectation $\mathbb{E}[W]$. My goal it to compute the integral

\begin{equation} \mathbb{E}\left[\int_{0}^{W}f(t)dt\right] = \int_{0}^{\mathbb{E}[W]}f(t)dt \end{equation}

Is the equality correct? If not, then what would be the correct result? Or is there any known theorem for this? Appreciate if anyone can provide such references.

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  • $\begingroup$ Did you try with any function $f$? You are basically asking when $\mathbb{E}[F(W)] = F(\mathbb{E[W]})$ for a differentiable function $F$. $\endgroup$
    – Sasho Nikolov
    Commented Feb 6, 2017 at 6:35
  • $\begingroup$ Yes, that's what I mean basically, but I, in fact, did not come up with your representation. I have not tried any function yet, but my case will be continuous and differentiable functions with $\inf\{s > 0 : f(s) > 0\}$. $\endgroup$
    – Liäm
    Commented Feb 6, 2017 at 10:39
  • $\begingroup$ Let me be more explicit: your equality will not hold in general, and examples are abundant. For example take $f(t) = 2t$. In that case you are asking if $\mathbb{E}[W^2] = \mathbb{E}[W]^2$, which holds if and only if $W$ has variance 0. In general, I expect that the only continuous $f$ for which your equality holds are constant on the range of $W$. $\endgroup$
    – Sasho Nikolov
    Commented Feb 6, 2017 at 13:51

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\begin{align}\mathsf{E}\biggl[\int_0^{W}f(t)\mathrm{d}t\biggr] &=\mathsf{E}\biggl[\int_{-\infty}^{+\infty}f(t)(1_{\{0<t\le W\}}-1_{\{W\le t\le 0\}})\mathrm{d}t\biggr]\\ &= \int_0^{+\infty}f(t)\mathsf{P}(W\ge t)\mathrm{d}t-\int_{-\infty}^0f(t)\mathsf{P}(W\le t)\mathrm{d}t. \end{align}

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  • $\begingroup$ Thanks! But I am not sure about the range of integration and the second indicator in your first equality. $W$ is always non-negative (maybe it was my fault not indicating this fact?). Why do we need to integrate over the entire entended $\mathbb{R}$? $\endgroup$
    – Liäm
    Commented Feb 6, 2017 at 11:12
  • $\begingroup$ What is your concern? If $W$ is always non-negative, then the second term is 0 so you can ignore it. In the general case, the sign is negative since by convention $\int_{0}^w f(t) dt = -\int_{w}^0 f(t) dt$. $\endgroup$
    – James Martin
    Commented Feb 6, 2017 at 14:34
  • $\begingroup$ @JamesMartin Thanks for your comment. $\endgroup$
    – JGWang
    Commented Feb 7, 2017 at 1:59

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