I would like to consider the "conditional distribution" of the Brownian sample paths conditional on certain sample path functionals, in a similar way that one considers the Brownian bridge. For example, consider the functional $\phi: C[0,1] \rightarrow \mathbb{R}$ defined by, say

$$ \phi( W(\cdot) ) = \left( \int_0^1 W dW \right)^2, $$ or $$ \phi( W(\cdot) ) = \int_0^1 W^2_t dt, $$ etc., what is the process obtained by conditioning $W$ on $\phi$? Are there any results of this type?

| cite | improve this question | | | | |
  • 4
    $\begingroup$ The first one is easy. $\int_0^1 WdW = W_1^2-1$ so of we condition on the square of that to be a fixed positive number, the resultant distribution is easily obtained in terms of the Brownian bridge. $\endgroup$ – Shalop May 12 '19 at 2:16

I don't know of any literature on the subject of conditioning Brownian motion in this way, however, I believe that when the functional $\phi$ is nice enough (in the sense of Malliavin calculus), it is possible to write the conditioned process as the solution of some SDE: $dX_t = dB_t+G(t,X_{\cdot})dt$. Here $G$ is some predictable function from $\Bbb R_+ \times C_0[0,1] \to \Bbb R$ which depends on $\phi$ and also on the value we condition on.

I do not know the precise/rigorous conditions on $\phi$ which make this possible and I do not intend to pursue that here. Instead, I will give some intuitionist, "physics-level" derivation of how to compute the drift $G$ from the functional $\phi$.

So, suppose we want to condition on $\phi(W)=C$. Assume that the law of $\phi(W)$ has a continuous density $p_{\phi}(x)$ with respect to Lebesgue measure which is strictly positive near $C$ (there are conditions to ensure this using Malliavin Calculus, e.g. $C$ should be a regular value of $\phi$, and the Malliavin matrix of $\phi$ should be invertible and satisfy some regularity conditions in some neighborhood of $\phi^{-1}(\{C\})$). Then define $$F(\phi,C) := \frac1{p_{\phi}(C)}\partial_t|_{t=0} \big(\partial_x|_{x=C}\Bbb E[W_t\cdot 1_{\{\phi(W) < x\}}]\big),\tag{1}$$ which (as may be verified using regular conditioning) will just be the drift felt by the conditional process at $t=0$. Furthermore, if $t \ge 0$ and $f,g \in C[0,1]$ then we define $\psi(t,f,g) := \phi\big(f(t^{-1}\cdot)*g((1-t)^{-1}\cdot)\big)$, where $f(t^{-1}\cdot)\in C[0,t]$ is the map $u \mapsto f(t^{-1}u)$ (and similarly for $g$); and moreover "$*$" denotes concatenation of paths: more specifically, if $\gamma \in C_0[0,t]$ and $\mu \in C_0[0,1-t]$ then $\gamma * \mu\;(u) = \gamma(u)$ for $u \le t$ and $\gamma * \mu\; (u) = \gamma(t) + \mu(u-t)$ for $u \in [t,1]$. Finally, we are ready to define the drift function $$G(t,\omega) := \frac{F\big(\;\psi(t,\omega(t\cdot),(1-t)^{1/2}\cdot)\;, C\big)}{1-t}.\tag{2}$$ The way to interpret this last expression is to note that for fixed $\omega \in C_0[0,1]$ and $t \ge 0$, the map $g \mapsto \psi\bigg(t\;,u\mapsto \omega(tu)\;, v\mapsto (1-t)^{1/2}g(v)\bigg)$ is a functional on $C[0,1]$, hence we can plug it into $F(\cdot,C)$.

Now, with the way we have defined $G$, we claim that the solution of $dX_t = dB_t + G(t,X_{\cdot})dt$ is essentially Brownian motion conditioned on $\phi=C$. The intuitive reason for this (which is easy to visualize) is that at each infinitesimal step, $X_t$ feels a drift in the direction which allows the remainder of the path to be distributed according to a Brownian motion conditional on the constraint that if we concatenate this Brownian path with the history of $X$ so far, then applying $\phi$ to the resultant path gives $C$. Then one may check (perhaps tediously) that mathematically formulating this whole procedure gives the above SDE. Unfortunately it remains to be shown that the solution $X$ actually exists and that its law is the same as that obtained by other conditioning procedures, like regular conditioning / disintegration. However I did some computations which suggest that this is actually true, and in fact the resulting law even appears to be weakly continuous in $C$ if $\phi$ is nice.

It is interesting to note that if we apply these heuristics to $\phi(\omega) = \omega(1)$ then we recover the SDE of Brownian bridge: $dX_t = dB_t + \frac{C-X_t}{1-t}dt$. For another example, lets say that $\phi(\omega) = p(\omega(1))$ where $p$ is a quadratic polynomial with two real roots $r,s$. Then we set $f(a,b) = \frac{ae^{-a^2/2}+be^{-b^2/2}}{e^{-a^2/2}+e^{-b^2/2}}$. Then, one way to understand the conditional process (with $\phi(W)=0$) is in terms of a Brownian bridge (see my comment), but another way is as the solution of the SDE $dX_t = dB_t +\frac{f(r-X_t,s-X_t)}{1-t}dt$. With some modifiction, this SDE formulation seems to make sense even for problems such as conditioning $W$ to stay positive (e.g., take $\phi = \inf$ and let $C=0$).

Another remark is that it seems from this description that the conditioned process is necessarily absolutely continuous wrt Brownian motion, but actually this is not true because the drift term $G$ can have small blow-ups along the way which can make it true that $\int_0^t G(s,X_{\cdot})^2ds = +\infty$ (think about $\phi(\omega) = \omega(1/2)$, which gives a Brownian bridge up to time $1/2$, but then just evolves as Brownian motion). A final remark is that in practice it might be difficult to compute $F$ and $G$ (the quantities appearing in $(1)$ and $(2)$). To compute $F$, in principle one only needs to know how to compute $p_{\phi}$ and $\Bbb E[W_t|\phi(W)]$, and in certain cases (like Brownian bridge) this is quite explicit, however in other cases it might be very hard. Computing $G$ can be even harder, however in certain cases it can be computed easily from $F$, even if we cannot explicitly compute $F$ itself (this is the case for your second example where $\phi(f) = \int f^2$, because this functional satisfies a nice additivity property with respect to concatenation of paths).

| cite | improve this answer | | | | |
  • $\begingroup$ One more remark: there is a discrete-time version of this procedure. Specifically if $X$ is a markov chain on some discrete state space $S$, and if $f:S^N \to \Bbb R$ is some functional, then we can condition on $f(X_1,...,X_N)=C$ and the result is an (inhomogeneous) markov chain $(\bar X_1,...,\bar X_N)$ whose (time-dependent) transition densities can be easily derived from those of $X$ (it will be like a "time-varying Doob h-transform"). $\endgroup$ – Shalop May 14 '19 at 22:57
  • $\begingroup$ Another remark on the "computations" mentioned above: Basically if $U \subset C_0[0,1]$ is open, then we can (similarly to above) define $F(U) = \partial_t|_{t=0} \Bbb E[W_t|W\in U]$, and analogously we can also define $G(U;t,\omega)$ using the same procedure as above. Then we claim that the solution of the SDE $dX_t = dB_t +G(U;t,X_{\cdot})dt$ has the same law as that of $W$ conditioned to stay in $U$. This may be proved by noting that if $Y$ denotes $W$ conditioned to stay in $U$,then $Y_t - \int_0^t G(U;s,Y_{\cdot})ds$ is a martingale whose quadratic variation... $\endgroup$ – Shalop May 15 '19 at 0:47
  • $\begingroup$ ...equals $t$, and thus the latter is a standard BM (note by Girsanov that this also gives the seemingly nontrivial identity $\Bbb E[1_U|\mathcal F_t] = \Bbb P(W \in U) e^{\int_0^t G(U;s,W_{\cdot})ds -\frac12\int_0^t G(U;s,W_{\cdot})^2ds}$). Once you have this characterization of the conditioned process for open sets, you may apply it to the open sets $U_{\epsilon}:=\phi^{-1}(C-\epsilon, C+\epsilon)$; then as $\epsilon \to 0$ it is easily shown that $F(U_{\epsilon}) \to F(\phi,C)$ and similarly for $G$. Then one may argue that the respective laws of the SDE's also converge as $\epsilon \to 0$. $\endgroup$ – Shalop May 15 '19 at 19:49
  • $\begingroup$ did you come up with this? If so, wouldn't it deserve a paper? $\endgroup$ – Tom Sep 8 '19 at 13:33
  • $\begingroup$ @Tom Yes I came up with this, but I’m not sure if it’s publishable because this kind of thing seems classical in spirit. I imagine that there might already be existing results from ~30 years ago on this type of subject, though I’m not sure. $\endgroup$ – Shalop Sep 8 '19 at 16:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.