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Conditional Expected Average of a Geometric Brownian Motion

Article Quant Q&A · Author: Frido
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Summary

The document considers the expected time integral of a stock price under a zero-drift geometric Brownian motion, conditional on its terminal price. It corrects an attempted approach that treats the intermediate price as a simple function of the terminal value and an independent normal draw. Instead, the answer uses the conditional distribution of Brownian motion given its endpoint, represented by a Brownian bridge. This preserves the dependence between the path at an intermediate time and the known terminal value.

Applying the normal moment-generating function gives a conditional expectation for the stock price at each time. Fubini’s theorem then reduces the expected integral to a one-dimensional time integral, whose expression is provided in the answer and may involve error functions when evaluated explicitly. The response is framed as a tentative derivation, with its author inviting corrections. It does not discuss parameter estimation, market calibration, or applications such as pricing Asian options, and the result depends on the stated model assumptions.

Key ideas

  • Conditioning on the terminal stock price changes the distribution of the path at intermediate times.
  • A Brownian bridge gives the conditional distribution of Brownian motion between its starting point and known endpoint.
  • The normal moment-generating function simplifies the conditional expectation at each time.
  • Fubini’s theorem converts the expected time integral into an integral of conditional expectations.
  • The resulting time integral can have a closed form involving error functions.

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StatisticsDerivatives pricing
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# Expectation of average, conditional on terminal value


# Expectation of average, conditional on terminal value












Silly question, but for some reason I'm a bit uncertain about this this trivial example perhaps:

I have the following simple BS model $$ S_T = S_t \exp \left\{ -\frac12 \sigma^2 (T-t) + \sigma (W_T - W_t) \right \} $$

I'd like to compute the following conditional expectation: $$ E_0 \left[ \left. \int_0^T S_t dt\, \right| S_T \right] $$

Can't I just write $$ E_0 \left[ \left. \int_0^T S_t dt\, \right| S_T \right] = \int_0^T \int_{-\infty}^{\infty} S_T \exp \left\{ \frac12 \sigma^2 (T-t) - z\sigma \sqrt{T-t}\right \} \phi(z) \, dz \, dt $$ with $$ \phi(z) = \frac{1}{\sqrt{2\pi}} e^{-\frac12 z^2} \enspace ? $$

## Answer by Rylan (score 2, accepted)

https://quant.stackexchange.com/a/79406

My attempt at an answer (as I've said in the comments, I'm quite rusty... any corrections are appreciated)

We have $S_0 > 0$, and since we know $S_T$ we know $\sigma \sqrt{T-t}W_T = b$ for some $b \in \mathbb{R}$.

Using the general case of Brownian bridge here, we have:

$$S_t | S_T = S_0e^{-\frac{\sigma^2t}{2} + \frac{bt}{T} + \sigma \sqrt{\frac{(T-t)t}{T}}z}$$

And we want $$E\Big(\int_0^T S_0e^{-\frac{\sigma^2t}{2} + \frac{bt}{T} + \sigma \sqrt{\frac{(T-t)t}{T}}z}dt\Big)$$

Using Fubini, this gives us

$$\int_0^TE\Big( S_0e^{-\frac{\sigma^2t}{2} + \frac{bt}{T} + \sigma \sqrt{\frac{(T-t)t}{T}}z}\Big)dt$$

We use the MGF of the normal to simplify the term in the expectation and we get $$E\Big( S_0e^{-\frac{\sigma^2t}{2} + \frac{bt}{T} + \sigma \sqrt{\frac{(T-t)t}{T}}z}\Big) = S_0e^{-\frac{\sigma^2t}{2} + \frac{bt}{T} + \frac{\sigma^2(T-t)t}{2T}}$$

meaning that the total term is

$$S_0\int_0^T e^{-\frac{\sigma^2t}{2} + \frac{bt}{T} + \frac{\sigma^2(T-t)t}{2T}}dt$$

which Wolfram gives me a nasty looking result for which contains error functions

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.