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Conditional Brownian Bridge Integral: A Proposed Moment Calculation

Article Quant Q&A · Author: Emmy
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Summary

The question asks for the conditional second moment of the time integral of Brownian motion given its values at the interval endpoints. The proposed route expands the square as a double integral of conditional products, then considers the conditional distribution at intermediate times. The accepted response instead identifies the conditioned process as a Brownian bridge and shifts the interval to start at zero.

It rewrites the bridge integral as a stochastic integral and uses quadratic variation to obtain a variance term depending on the interval length. However, the response's final expression omits the square of the conditional mean of the bridge integral, so it does not generally give the requested second moment when the endpoints are nonzero. The method is useful for recognizing the bridge structure, but the displayed calculation needs correction; the source provides no independent verification or discussion of that caveat.

Key ideas

  • Conditioning Brownian motion on both endpoint values gives a Brownian bridge over the interval.
  • The square of the time integral can be expressed through conditional moments of the bridge.
  • The response uses a stochastic integral representation and quadratic variation to compute a variance term.
  • The stated final result omits the squared conditional mean when endpoints are nonzero.

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# Conditional expectation of integral of brownian motion


# Conditional expectation of integral of brownian motion












I am trying to calculate $$\mathbb{E}\biggl[\biggl(\int_s^t W_u du\biggl)^2 \biggl|W_s=x, W_t=y\biggl] $$ where $W$ is a Standard Brownian Motion and $s\leq u \leq t$. Any help or tips would be greatly appreciated :)

My approach is the following \begin{align} \mathbb{E}\biggl[\biggl(\int_s^t W_u du\biggl)^2 \biggl|W_s=x, W_t=y\biggl] &=\mathbb{E}\biggl[\int_s^t \int_s^t W_v W_u du dv\; \biggl| \; W_s = x, W_t= y \biggl]\\ &=\int_s^t \int_s^t \mathbb{E}[W_v W_u | \; W_s = x, W_t= y]du dv \end{align} For $v\leq u$ I can rewrite this to \begin{align} \mathbb{E}[W_v W_u | \; W_s = x, W_t= y] &= \mathbb{E}[W_v ((W_u-W_v)+W_v) | \; W_s = x, W_t= y] \\ &=\underbrace{\mathbb{E}[W_v (W_u-W_v) | \; W_s = x, W_t= y]}_{=0}+\mathbb{E}[W_v^2 | \; W_s = x, W_t= y]\\ &=\mathbb{E}[W_v^2 | \; W_s = x, W_t= y]\\ &= \frac{(t-v)(v-s)}{t-s} - \biggl(\frac{t-v}{t-s}x+\frac{v-s}{t-s}y\biggl)^2 \end{align} Where I used in the last equation that $(W_v | \; W_s = x, W_t= y) \sim \mathcal{N}( \frac{t-v}{t-s}x+\frac{v-s}{t-s}y, \frac{(t-v)(v-s)}{t-s} )$. I end up with this aweful calculation \begin{align} &\int_s^t \int_s^t \mathbb{E}[W_v W_u | \; W_s = x, W_t= y]du dv \\ = &\int_s^t \int_s^u \mathbb{E}[W_v W_u | \; W_s = x, W_t= y]du dv + \int_s^t \int_u^t \mathbb{E}[W_v W_u | \; W_s = x, W_t= y]du dv \\ = &\int_s^t \int_s^u \frac{(t-v)(v-s)}{t-s} - \biggl(\frac{t-v}{t-s}x+\frac{v-s}{t-s}y\biggl)^2du dv + \int_s^t \int_u^t \frac{(t-u)(u-s)}{t-s} - \biggl(\frac{t-u}{t-s}x+\frac{u-s}{t-s}y\biggl)^2du dv \end{align} I am sure there must be a better solution than this endless calculation but I cannot think of one...

## Answer by R. Rayl (score 2, accepted)

https://quant.stackexchange.com/a/63966

I found this to be a very interesting question, and I took a different approach to your working. Here's my attempt:

Instead of considering the integral $\int_s^t W_u du \rvert W_s=x, W_t=y$, we can consider the integral $\int_s^tB_u du$ where $B_u$ is a Brownian bridge process with $B_s = x$, $B_t = y$.

Furthermore, we can shift the limits of the integral from $[s, t]$ to $[0, T]$ where $T := t-s$. In this case, we define $B_0 = x$, $B_T = y$. So we want to find: \begin{equation} \mathbb{E}\bigg[ \bigg(\int_0^T B_u du\bigg)^2 \bigg]. \end{equation}

We can re-write our integral as follows \begin{align} \int_0^T B_u du &= \int_0^T(T-u)dB_u. \end{align}

Then, \begin{align} \mathbb{E}\bigg[ \bigg(\int_0^T(T-u)dB_u\bigg)^2 \bigg] &= \mathbb{E}\bigg[ \int_0^T (T-u)^2 d[B]_u \bigg] \\ &= \int_0^T (T-u)^2 du \\ &= \frac{(t-s)^3}{3} \end{align}

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.