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Let $W_t$ be standard Brownian motion, and define $$ X_t := \int_0^t W_s ~\textrm{d}s. $$ The marginal distributions of $X_t$ are easy to write down (see here), but it doesn't seem possible to express $X_t$ as a function $f(t,W_t)$ of time and the process $W_t$ itself.

Is it indeed impossible? And could somebody refer me to a proof?
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    $\begingroup$ I think you could show that the conditional distribution of $X_1$ given $W_1$ is not constant. $\endgroup$ – Nate Eldredge Nov 5 '15 at 20:42
  • $\begingroup$ Oh right, that sounds totally doable. Thanks! $\endgroup$ – RP_ Nov 5 '15 at 20:46
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Intuitive argument: by symmetry we should have $P(X_1 \ge 0 \mid W_1 = 0) = \frac{1}{2}$. However, $P(f(1, W_1) \ge 0 \mid W_1 = 0)$ is either $0$ or $1$ depending on whether $f(1,0) \ge 0$.

Of course this is not really a proof because I conditioned on an event of probability 0. So let's try to use the same idea in an actual proof.

Consider the conditional probability $Y = P(X_1 \ge 0 \mid W_1)$.

Suppose $X_1 = f(1, W_1)$ where $f$ is Borel; then we have $Y = 1_B(W_1)$ a.s., where $B = \{x : f(1,x) \ge 0\}$. In particular we have $$P(Y \in \{0,1\}) = 1. \tag{*}$$

But since $X_1, W_1$ are both linear functionals of the Gaussian process $\{W_t\}$, they are jointly Gaussian with mean 0. Of course $W_1$ has variance 1, and as shown in your link, $X_1$ has variance $1/3$. Also, using Fubini's theorem we have $$\begin{align*} E[W_1 X_1] &= E\left[ W_1 \int_0^1 W_t\,dt\right] \\ &= E\left[\int_0^1 W_1 W_t\,dt\right] \\ &= \int_0^1 E[W_1 W_t]\,dt \\ &= \int_0^1 t\,dt = \frac{1}{2}.\end{align*}$$ So if we set $Z = W_1 - 2 X_1$, then $Z$ is a normal random variable with mean 0 and variance $1/3$ which is independent of $W_1$. We can write the event $\{X_1 \ge 0\}$ as $\{Z \le W_1\}$. So by independence we have $$Y = P(X_1 \ge 0 \mid W_1) = P(Z \le W_1 \mid W_1) = \Phi(\sqrt{3}W_1) \quad \text{a.s.}$$ where $\Phi$ is the standard normal cdf. In particular $0 < Y < 1$ almost surely, contradicting (*).

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  • $\begingroup$ Very nice, thanks! One question: I already knew $X_1,W_1$ to be Gaussians with zero mean, but you seem to use the general argument "since $X_1,W_1$ are both linear functionals of the Gaussian process $\{W_t\}$, they are jointly Gaussian with mean $0$" for this. I don't see exactly why this should be true in this generality. Are you referring to some kind of general result here, or am I missing something obvious? (If by "linear functional" you had meant "finite linear combination of certain $W_{t_i}$", then I'd have been on board with it, but clearly this doesn't apply to $X_1$.) $\endgroup$ – RP_ Nov 6 '15 at 13:21
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    $\begingroup$ @René: Yes, I'm referring to something more general; "linear functional" here can mean something like "continuous linear functional on the Banach space $C([a,b])$". If you want to prove it more directly, write $X_1$ as the limit of Riemann sums $X_1^n$, each of which is a finite linear combination of $W_{t_i}$. We have $(X_1^n, W_1) \to (X_1, W_1)$ almost surely, so $(X_1, W_1)$ is a limit of jointly Gaussian vectors, so by an argument like this it too is Gaussian $\endgroup$ – Nate Eldredge Nov 6 '15 at 14:33

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