Expectation of average, conditional on terminal value

Question

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 1

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