Let $B_t$ be a standard Brownian Motion, $B_0=0$, and $T>0$. Given $0 \leq t \leq T$ we have the nice result that
$$ \mathbb{E}[ B_t | B_T=x ] = tx/T$$
(of course we can't condition on 0-probability events, I'm using this notation throughout as short-hand for the "proper" case where we look at functional form of expectation conditional on random variables)
It follows that the time average of the process is just $x/2$:
$$ \mathbb{E} \frac{1}{T} \int_{s=0}^T B_s ds = \frac{1}{T} \int_{s=0}^T sx/T = x/2 $$
i.e. the expected value of the Gaussian Bridge tethered at $(0, 0)$ and $(T, x)$ is just the mean of the two end points.
What about if we also condition on the min/max that the process achieves over the interval $[0, T]$? I.e. what is
$$f(t) = \mathbb{E}[B_t|B_T=x, \min_{0 \leq s \leq T} B_s = x_{lo}, \max_{0 \leq s \leq T} B_s = x_{hi}] $$
Initially, I thought (wrongly I believe) that it will in fact be, neatly, the arithmetic average of initial value (0), $x$ and the extremes, $x_{lo}$ and $x_{hi}$. My reasoning was thus:
- Suppose extremes are attained at (random) times $T_{lo}, T_{hi}$
- By same logic as above, $\mathbb{E} B_s | x, x_{lo}, x_{hi}, T_{lo}, T_{hi}$ will be a piecewise linear function, through the points $(0, 0), (T, x), (T_{lo}, x_{lo}), (T_{hi}, x_{hi})$
- Conditional on these variables, the time-average value of $B_s$ is $\frac{0+x+x_{lo}+x_{hi}}{4}$
- Trivially taking expectation over $T_{lo}$ and $T_{hi}$ yields the result
But it seems $g(t) = \mathbb{E} B_t | 0 \leq t \leq T_{lo}, B_{T_{lo}} = x_{lo}$ is no longer a linear function (and indeed the stopped process is not a Gaussian process anymore).
Suggestions or solutions welcome! I should add, this is not homework, but a problem I'm pursuing for personal curiosity.