Expected Integrals with a Random Upper Limit via Tail Probabilities

Summary

The exchange derives an alternative expression for the expectation of an integral whose upper limit is a positive random variable. Starting from the indicator representation of the event that the random limit exceeds a point, it rewrites the random-bound integral as an integral over the positive half-line. Taking expectations then weights the deterministic integrand by the probability that the random bound reaches each point. For a variable with density, that tail probability can itself be written as the integral of the density above the point.

This tail-probability formulation complements direct averaging over the density of the random upper limit. It relies on being able to interchange expectation and integration, which requires suitable integrability conditions (or nonnegativity for the standard Tonelli argument); the answer does not spell these conditions out. The question’s time subscript on expectation is also not developed. The result is a probability and integration identity useful in quantitative modeling, rather than a trading strategy or empirical result.

Key ideas

• Represent an integral up to a positive random bound using an indicator over the positive half-line.
• The expected integral weights each point by the probability that the random bound exceeds it.
• With a density, the exceedance probability is the density’s upper-tail integral.
• Interchanging expectation and integration requires appropriate conditions, which the response leaves implicit.

Tags

# Expectation of integral where one of limits of integration is a random variable


# Expectation of integral where one of limits of integration is a random variable












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?

## Answer by ir7 (score 3)

https://quant.stackexchange.com/a/63478

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

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

## Answer by Brickcity (score 0)

https://quant.stackexchange.com/a/63304

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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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.