# BSDE without volatility

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?

• 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. – Michael Greinecker Jun 30 at 6:29
• Sorry, I was imagining that the drift is not necessarily deterministic. I will edit my question accordingly. – tsm Jun 30 at 18:09

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 $$x=0$$.