# Expectation of average, conditional on terminal value

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 $$\phi(z) = \frac{1}{\sqrt{2\pi}} e^{-\frac12 z^2} \enspace ?$$

• I'm afraid it's not this simple (but I am quite rusty with stocal so I can't remember the exact appraoch). The "problem" is we know the Brownian motion starts at $W_0 = 0$, and $W_T$ is also known. I believe the concept that you want to invoke is the Brownian bridge: en.wikipedia.org/wiki/Brownian_bridge Commented May 17 at 14:02
• @Rylan Yeah, it seemed to simple to me as well. Will read further. Commented May 17 at 14:10

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

• +1 Error functions are good. I'm actually trying to re-derive a similar problem and result from a paper I saw, which also contains erf. I'll wait for other potential answers/corrections before/if accepting your answer. Thanks. Commented May 17 at 15:48
• I accepted your answer. Just one question: shouldn't there also be a $\sigma$ multiplied with $bt/T$? Commented May 18 at 15:56
• I think you’re right about that — will edit this soon Commented May 18 at 16:13