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Conditional Expectation of Integrated Brownian Motion

Article Quant Q&A · Author: solid
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Summary

The document derives the conditional expectation of the time integral of a standard Brownian motion over an interval, given the information available at the interval’s start. The result is the interval length multiplied by the Brownian value at that start.

It gives two main routes: use the martingale property to evaluate the conditional expectation inside the integral, or separate the process into its starting value and an independent future increment with conditional mean zero. An integration-by-parts argument provides another derivation. The explanations establish an expectation identity, not the distribution or pathwise value of the integral; the result relies on standard Brownian motion and its usual filtration assumptions.

Key ideas

  • The conditional mean of a future Brownian value, given current information, equals the current value.
  • The integral can be evaluated in conditional expectation by integrating that conditional mean over time.
  • Future Brownian increments are independent of the filtration at the starting time and have mean zero.
  • Integration by parts offers an alternative derivation using martingale properties.

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# Show that $E [ \int_s^t W_u \, du \,|\, \mathcal{F}_s ] = (t - s) W_s$


# Show that $E [ \int_s^t W_u \, du \,|\, \mathcal{F}_s ] = (t - s) W_s$












Show that

$$E \left[ \int_s^t W_u \, du \,|\, \mathcal{F}_s \right] = (t - s) W_s$$

where $W_u$ is a standard Brownian motion and $\mathcal{F}_s$ is the filtration up to time $s $.

## Answer by Kurt G. (score 9, accepted)

https://quant.stackexchange.com/a/80689

I prefer to use integration by parts and write $$ \int_s^tW_{\color{red}u}\,d\color{red}u=tW_t-sW_s-\int_s^tu\,dW_u\,. $$ Then, by the martingale property of $W_t$ and the $dW_u$-integral, $$ \mathbb E\left[\int_s^tW_u\,du\, \Bigg|\,{\cal F}_s\right]=tW_s-sW_s-0\,. $$

## Answer by KaiSqDist (score 5)

https://quant.stackexchange.com/a/80683

Given that the expectation of a Wiener process is a martingale (the value of the Wiener process in the next period $t$ has the same value as the current $s$) and if you plot the value of the Wiener process against time, your integral evaluates to give us the "area of a rectangle", which is $(t-s)W_s$

Please let me know if you need more details.

## Answer by Andrea (score 4)

https://quant.stackexchange.com/a/81532

$\mathbb{E}$ and $\int$ commute so

$E \left[ \int_s^t W_u \, du \,|\, \mathcal{F}_s \right] = \int_s^t \mathbb{E} [ W_u \,|\, \mathcal{F}_s ] \, du = \int_s^t W_s \, du = (t - s)W_s$

## Answer by solid (score 2)

https://quant.stackexchange.com/a/81525

$$ \mathbb{E}\left[\int_s^t W_u \, du \, \middle| \, \mathcal{F}_s \right] = \mathbb{E}\left[\int_s^t (W_u - W_s + W_s) \, du \, \middle| \, \mathcal{F}_s \right] = \underbrace{\mathbb{E}\left[\int_s^t (W_u - W_s) \, du \, \middle| \, \mathcal{F}_s \right]}_{1)} + \underbrace{\mathbb{E}\left[\int_s^t W_s \, du \, \middle| \, \mathcal{F}_s \right]}_{2)} = (t - s) W_s $$

1) $$ \mathbb{E}\left[\int_s^t (W_u - W_s) \, du \, \middle| \, \mathcal{F}_s \right] = \int_s^t \mathbb{E}(W_u - W_s) \, du = 0, \quad \text{since } W_u - W_s \text{ is independent from } \mathcal{F}_s. $$ 2) $$ \mathbb{E}\left[\int_s^t W_s \, du \, \middle| \, \mathcal{F}_s \right] = W_s \int_s^t du = W_s (t - s). $$

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