If one has

$dX_t = f(Y_t)\mu_1(X_t, t)dt + \sigma_1(X_t, t)dW_t$

$dY_t = \mu_2(Y_t, t) + \sigma_2(Y_t, t) dW_t$

driven by the same Brownian motion. Is it sensible to consider a conditional expectation of the type $Z_t = \mathbb{E}[X_t | Y_t = y_t]$ ?

Furthermore, is it sensible to consider $Z_t$ as satisfying an equation like:

$dZ_t = f(y_t) \mathbb{E}[(X_t, t) | Y_t = y_t]dt + \mathbb{E}[\sigma_1(X_t, t)dW_t | Y_t = y_t]$

What does conditioning do to the $dW_t$ term? If this was simply $\mathbb{E}$, then the entire diffusion term vanishes and I formally have a sort of ODE. The specific nature of this example I am willing to adjust if any of it is nonsense, but in general I am interested in understanding what I can say about

$$\mathbb{E} [X_t | Y_t ]$$ for $X_t, Y_t$ stochastic processes defined by an Ito SDE.

Here is an example that seems to work:

$$dX_t = Y_t X_t dt$$ $$dY_t = dW_t$$

This easily can be seen to have solution $X_t = \exp(\int_0^t Y_s ds)$ and $Y_t = W_t$. It is indeed the case that if I was to condition on the event $W_t = \omega_t$, then the solution of the equation

$$dZ = \omega_t Z_t dt$$

$Z_t = \exp(\int_0^t \omega_s ds)$ is indeed the conditional expectation of the directly calculated $X_t$.

  • $\begingroup$ Of course it is sensible to consider the conditional expectation $Z_t = \mathbb{E}[X_t | Y_t = y_t]\,.$ Your equation for $dZ_t$ is close to nonsense. What are the random variables there you are taking the conditional expectations of? $\endgroup$
    – Kurt G.
    Apr 3 at 7:46
  • $\begingroup$ Well what it really mean is taking the expectation of the integral equation for $X$, commuting $\int, \mathbb{E}$. Since any function $Y_t$ ought to be measurable w.r.t to the sigma algebra, I am inclined to think I can pull it out like constants. $\endgroup$ Apr 3 at 10:46
  • $\begingroup$ How about an attempt then? Choosing $\mu_1,\mu_2,\sigma_1,\sigma_2$ so that we have two explicit solutions should be of great help. No formal symbol pushing please. $\endgroup$
    – Kurt G.
    Apr 3 at 10:48
  • $\begingroup$ The easiest example I can think of is making $X_t$ an OU process with $Y_t$ dependent linear coefficient and making $Y_t$ Brownian motion, but this won't have any explicit solution. The main thing is I want to know what I can do with with conditional expectation vs just expectation for SDEs, since I know it to be true that were $Y_t \equiv y_t$ deterministic, I could take $\mathbb{E}$ and derive the formal ODE $\partial_t Z_t = f(y_t) \mathbb{E}[\mu_1(X_t,t)]$. $\endgroup$ Apr 3 at 11:02
  • $\begingroup$ I still have not freed you from the symbol pushing mode. 1) In the edited question, $Z_t = \exp(\int_0^t \omega_s\,ds)$ is the conditional expectation of $X_t = \exp(\int_0^t W_s\,ds)$ when you condition $X_t$ on the entire path of $W$ from zero to $t\,.$ In your notation you conditioned on $W_t$ only. 2) An SDE gives rise to a (not only formal) ODE for the expectation if and only if the drift is affine linear. That is because the expectation of the $dW$-term is zero. Think about how much of this carries over to the conidtional expectation. $\endgroup$
    – Kurt G.
    Apr 3 at 13:12


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