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Second Moment of a Brownian-Motion Time Integral

Article Quant Q&A · Author: Bogaso
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Summary

The document derives the second moment of a time integral formed by multiplying a deterministic function by Brownian motion. Expanding the square produces a double integral involving the product of Brownian values at two times. Moving expectation inside the integral leaves the Brownian cross-moment, which is the smaller of the two time points, yielding the stated double-integral expression.

The question also asks why the integral is normally distributed, but the included answer does not address that part. The derivation relies on exchanging expectation and integration, which requires suitable integrability conditions on the deterministic function and process. It provides a calculation rather than an application to trading, and does not give further assumptions or examples.

Key ideas

  • Squaring the time integral produces a double integral over pairs of time points.
  • The Brownian cross-moment at two times equals the earlier time.
  • Under suitable integrability conditions, expectation can be moved inside the double integral.
  • The answer derives the second moment but does not explain the normality claim.

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# Calculate Variance of a function of Stochastic Process


# Calculate Variance of a function of Stochastic Process












I was studying an example given in Variance of a time integral with respect to a Brownian Motion function

There we need to calculate the Variance of $I_t = \int\limits_{0}^{t} f \left(s \right) W_s ds$.

Then, it said that $E \left[ I_t^2 \right] = \int\limits_{0}^{t}\int\limits_{0}^{t} f \left(s \right) f \left(u \right) \min \left(s,u\right) dsdu$

Can someone please help to understand the detailed calculation on how it can be arrived?

It is also stated that $I_t$ follows a Normal distribution. What is the reason for that?

## Answer by StackG (score 0, accepted)

https://quant.stackexchange.com/a/57112

This uses the autocorrelation of the Weiner process (proved in this post), ${\mathbb E}[W_s W_u] = \min(s,u)$

From your expression, \begin{align} {\mathbb E}[I^2_t] &= {\mathbb E}[\int_0^t f(s)W_sds \int_0^t f(u)W_u du]\\ &= {\mathbb E}[\int_0^t \int_0^t f(s)W_s f(u)W_u ds du]\\ &= \int_0^t \int_0^t f(s) f(u) {\mathbb E}[W_s W_u] ds du \end{align} where we moved the expectation into the integral as the Weiner terms are the only non-deterministic things

Then we substitute in the autocorrelation, and bingo! \begin{align} {\mathbb E}[I^2_t] &= \int_0^t \int_0^t f(s) f(u) {\mathbb E}[W_s W_u] ds du\\ &= \int_0^t \int_0^t f(s) f(u) \min(s,u) ds du \end{align}

as required

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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