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Is it correct to write

\begin{equation} E_t \int_0^{X_T} f(z) dz = \int_0^\infty \left(\int_0^x f(z) dz \right) p(x)dx \,\,? \end{equation}

Here $X_T$ is a positive random variable with density $p(x)$, and $f(z)$ is a deterministic function. Are there any other ways to calculate the integral?

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  • $\begingroup$ You can try using the law of iterated expectations, with the inner expectation being conditional on $X_T$. $\endgroup$
    – Daneel Olivaw
    Apr 9, 2021 at 13:40
  • $\begingroup$ Arguably my previous comment does not add much, this technique is already embedded in your expression. Not sure I understand your question then, when you write "other ways". What did you had in mind? $\endgroup$
    – Daneel Olivaw
    Apr 9, 2021 at 15:03
  • $\begingroup$ Yes, so I was thinking of writing $E_t \int_0^{X_T} f(z) dz = E_t \left[ E_T \int_0^{X_T} f(z) dz | \mathcal{F}_T \right]$ which I think leads to the right hand side of the equality above since $\int_0^{X_T} f(z) dz = F(X_T) - F(0) $ $\endgroup$
    – user34971
    Apr 9, 2021 at 15:03
  • $\begingroup$ Yes indeed, I added a comment, my initial one was not very useful. $\endgroup$
    – Daneel Olivaw
    Apr 9, 2021 at 15:04
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    $\begingroup$ Our posts crossed. I can only think of applying Fubini, $\int_0^\infty \left(\int_0^x f(z) dz \right) p(x)dx = \int_0^\infty \left( \int_z^\infty p(x)dx \right) f(z) dz$. Other than this I cannot see another way to write the expectation. $\endgroup$
    – user34971
    Apr 9, 2021 at 15:09

2 Answers 2

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This is related to the integrated tail probability expectation formula:

$$ X= \int_0^X dx = \int_0^\infty 1_{X>x} dx,$$

followed by

$$ E[X] = E \left[ \int_0^\infty 1_{X>x} dx \right] = \int_0^\infty E[1_{X>x}] dx $$ $$=\int_0^\infty P(X>x) dx = \int_0^\infty \left( \int_x^\infty p(z) dz\right) dx$$

Similarly, for deterministic $f$, we have:

$$ \int_0^X f(x) dx = \int_0^\infty 1_{X>x} f(x) dx,$$

followed by

$$ E \left[ \int_0^X f(x) dx \right] = E \left[ \int_0^\infty 1_{X > x} f(x) dx \right] = \int_0^\infty E[1_{X> x} f(x)] dx $$ $$ =\int_0^\infty P(X> x) f(x) dx = \int_0^\infty \left( \int_x^\infty p(z) dz\right) f(x) dx $$

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I think you can't definitely solve the integral, but there should be some way to solve for a distribution on the integral. Provided the integral is smooth enough for all values the random variable could take on. Good luck!

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