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Conditional Distribution of Integrated Brownian Motion Given Its Endpoint

Article Quant Q&A · Author: Alicia
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Summary

The answer derives the conditional distribution of the time integral of standard Brownian motion given the Brownian motion’s value at the same time. It rewrites the integral as a stochastic integral, then uses joint normality of the integral and endpoint to identify the conditional mean as one half of time multiplied by the endpoint.

The residual after subtracting that linear conditional mean is independent of the endpoint, and its variance is one twelfth of time cubed. Thus the conditional variable is normal, with that mean and variance. The derivation relies on standard Brownian motion and a fixed time horizon; it is a probability result rather than a trading strategy, and the document gives no market-data application or empirical validation.

Key ideas

  • The Brownian time integral and its endpoint are jointly normally distributed.
  • The conditional mean of the integral given the endpoint is half the time horizon times that endpoint.
  • The conditional variance is the time horizon cubed divided by twelve.
  • The result assumes standard Brownian motion and does not itself specify a market application.

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# Conditional distribution of $X_t = \int_0^t W_s \mathrm{d}s$


# Conditional distribution of $X_t = \int_0^t W_s \mathrm{d}s$












What is the conditional distribution of $$X_t = \int_0^t W_s \mathrm{d}s$$with respect to $W_t = x$?

## Answer by Gordon (score 7, accepted)

https://quant.stackexchange.com/a/53511

Note that \begin{align*} X_t = tW_t -\int_0^t sdW_s = \int_0^t (t-s)dW_s, \end{align*} and \begin{align*} W_t = \int_0^t dW_s. \end{align*} Then, for any real numbers $a$ and $b$, \begin{align*} aX_t + b W_t = \int_0^t (at-as+b)dW_s, \end{align*} is normal. That is, $W_t$ and $X_t$ are jointly normal. Moreover. note that \begin{align*} E\bigg(W_t\bigg(X_t - \frac{Cov(X_t, W_t)}{Var(W_t)}W_t\bigg)\bigg) = 0. \end{align*} That is, $\frac{Cov(X_t, W_t)}{Var(W_t)}W_t= \frac{1}{2}tW_t$ and $X_t-\frac{1}{2}tW_t$ are independent. Given that \begin{align*} E(X_t^2) &= \int_0^t(t-s)^2 ds = \frac{1}{3}t^3, \end{align*} then \begin{align*} Var\big(X_t - \frac{1}{2}tW_t\big) = \frac{1}{12}t^3. \end{align*} Therefore, \begin{align*} P(X_t \le y \mid W_t) &= P\bigg(X_t - \frac{1}{2}tW_t + \frac{1}{2}tW_t \le y \mid W_t \bigg)\\ &= P\big(X_t - \frac{1}{2}tW_t \le y - \frac{1}{2}tW_t \mid W_t \big)\\ &=\int_{-\infty}^{y-\frac{1}{2}tW_t} \frac{1}{\sqrt{\frac{1}{6}\pi t^3}}e^{-\frac{z^2}{\frac{1}{6} t^3}} dz. \end{align*} That is, the conditional distribution of $X_t$ given $W_t=x$ is normal with the density function \begin{align*} \frac{1}{\sqrt{\frac{1}{6}\pi t^3}}e^{-\frac{(y-\frac{1}{2}tx )^2}{\frac{1}{6} t^3}}. \end{align*}

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