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Let $(W_t)_{0\leq t\leq 1}$ be a standard Wiener process on $[0,1]$, and let $\mathcal{F}_t$ be the natural filtration. Consider a BSDE $$ dX_t=f(t,X_t)dt+\sigma(t,X_t) dW_t $$ with terminal condition $X_1=x$, where $f(t,\cdot)$ and $\sigma(t, \cdot)$ are $\mathcal{F}_t$-adapted square integrable processes.

My question: is it possible for the BSDE to be well defined if $\sigma(t,X)=0$ for all $t\in [0,1]$ and all $X$? Also, it seems unlikely to me that I can treat such a case as an ODE since reversing time would screw with the progressive measurability of $f$. Am I wrong?

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  • $\begingroup$ You can treat it as an ODE. Since the resulting function is a deterministic function of time, progressive measurability amounts to the function itself being Borel measurable. $\endgroup$ – Michael Greinecker Jun 30 at 6:29
  • $\begingroup$ Sorry, I was imagining that the drift is not necessarily deterministic. I will edit my question accordingly. $\endgroup$ – tsm Jun 30 at 18:09
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I am not sure if I understand your question correctly. A typical Brownian BSDE has the form

$$dY_t = f(\omega, t, Y_t, Z_t)dt - Z_t dW_t$$

with terminal condition

$$Y_T = \xi \in \mathcal{F}^{W}_T$$

where $Y$ and $Z$ are two parts of the solution and required to be adapted to the Brownian filtration. If your question boils down to if there is a BSDE with deterministic terminal condition $\xi = x$ and (second part of the) solution process constant zero, the answer is yes. E.g., take

$$f(\omega, t, Y_t, Z_t) = W_{\frac{T}{2}}1_{(\frac{T}{2},\frac{3T}{4}]} - W_{\frac{T}{2}}1_{(\frac{3T}{4},T]}$$

and $x=0$.

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