More specifically, I have some diffusion $Y_t$ and a filtration $\mathcal{F}_t$. $Y_t$ is not $\mathcal{F}_t$-measurable, but $Y_t$ is not independent of $\mathcal{F_t}$. My question is, if we define $$ Z_t = \mathbb{E}\left[ Y_t \mid \mathcal{F_t} \right] $$ then what kind of conditions can guarantee that $ \;dZ_t = \mathbb{E}\left[ dY_t \mid \mathcal{F_t} \right] \;\;?$

In other words we need a condition that will allow us to say something along the lines of $$ \mathbb{E}\left[ \int_0^t dY_u \mid \mathcal{F_t} \right] = \int_0^t \mathbb{E}\left[ dY_u \mid \mathcal{F_u} \right] \;.$$ I have tried to rewrite the above with the limit definition of the stochastic integral and go from there, but I didn't manage to get anywhere with it.

Clarification: Some commenters suggested that I provide a more concrete example. I must admit that the above question is kind of vague and not well defined.

Let's suppose that $Y_t$ is the solution to the SDE: $$ dY_t = Y_t\, B(t,X_t,W_t)\, dW_t , \; Y_0=1$$ where $W_t$ is some Wiener process which is adapted to its natural filtration $\mathcal{F}_t^W$, and $X_t$ is some other diffusion adapted to a filtration $\mathcal{G}_t \supseteq \mathcal{F}_t^W$. Now let's define $Z_t$ just as before, $$ Z_t = \mathbb{E}\left[ Y_t \mid \mathcal{F_t}^W \right] \;.$$

My question is whether it is possible to say that $$Z_t = 1 + \int_0^t \mathbb{E}\left[ Y_u\, B(u,X_u,W_u) \mid \mathcal{F_u}^W \right] \, dW_u \;,$$ or to know what conditions would need to be met so that the above holds true.

Attempt I have tried approaching this problem via the definition of the stochastic integral. Let's suppose that we define $u_j^{(n)} = t \times \frac{j}{2^n}$. Then we can say that $$ \mathbb{E}\left[ Y_t \mid \mathcal{F_t}^W \right] - Y_0 = \mathbb{E}\left[ \int_0^t Y_u \,B(u,X_u,W_u)], dW_u \mid \mathcal{F_t}^W \right] = \mathbb{E}\left[ \lim_{n\rightarrow \infty} \sum_{j=0}^{2^n -1} Y_{u_j^{(n)}} B_{u_j^{(n)}} (W_{u_{j+1}^{(n)}} -W_{u_j^{(n)}}) \mid \mathcal{F_t}^W \right] \;.$$ If I was able to pass the limit outside of the expected value I'd be done. I can't seem to find an argument that allows me to do that, however. I have tried to find some term that dominates each of the sums, but to no avail. If anybody could help, it would be greatly appreciated!

  • $\begingroup$ Why do you need that? Is there any application in mind? $\endgroup$ – Gordon Oct 28 '16 at 20:04
  • $\begingroup$ Can you clarify what you understand by $dY_t$? Usually such notation has a specific meaning - as such, I am very unsure about the meaning of the term $\mathbb{E}[dY_t \, | \, \mathcal{F}_t]$ and it's integration and therefore also the rest of your question. $\endgroup$ – Furrer Oct 28 '16 at 20:29
  • $\begingroup$ In the LHS you already have $\displaystyle \mathbb{E}\left[\int_0^t dY_u\bigg|\mathcal{F}_t\right] = \mathbb{E}\left[Y_t - Y_0|\mathcal{F}_t\right] $ $\endgroup$ – BGM Oct 29 '16 at 3:20
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    $\begingroup$ @Gordon, the application in mind is related to filtering. $\endgroup$ – Hotdog2000 Oct 31 '16 at 21:12
  • $\begingroup$ I also had a similar problem... $\endgroup$ – AIM_BLB Nov 8 '16 at 20:32

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