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Let $W_t$ be a standard Brownian motion. Let $T$ be the terminal date, $X_T=x$, and $$ dX_t=f_tdt+B_tdW_t $$ where $f_t$ and $B_t$ (yet to be determined) have to be adapted to the filtration generated by $W$.

Assume $x$ is a constant. One possible solution is that $f_t=B_t=0$ so that $X_t=x, \forall t$. Is it possible to have other solutions where $f$ or $B$ are not always 0?.

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The are surely many ways to do this. One classical example of this kind of process is the Brownian Bridge from $0$ to $x$, given by the SDE

$$ dX_t = \frac{x - X_t}{1-t}dt + dW_t. $$

This is solved by $X_t = tx + (1-t)\int_0^t{\frac{dW_s}{1-s}}$. As $t$ approaches $1$, $X_t$ approaches $x$ almost surely.

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(Cross-listed on Math Stackexchange) So I think I figured out an answer to my question. Here is an example where $f$ and $B$ are not necessarily zero.

Let $Y_t$ be some Ito process with $dY_t=\tilde f_tdt+\tilde B_t dW_t$ for some adapted processes $\tilde f, \tilde B$. Define $X_t=(1-e^{T-t})Y_t+x$. We have that $X_T=x$ and \begin{align} dX_t=&e^{T-t}Y_tdt+(1-e^{T-t})dY_t\\[8pt] =&[e^{T-t}Y_t+(1-e^{T-t})\tilde f_t]dt+(1-e^{T-t})\tilde B_tdW_t\\[8pt] =&f_tdt+B_t dW_t \end{align} where $f_t=e^{T-t}Y_t+(1-e^{T-t})\tilde f_t$ and $B_t=(1-e^{T-t})\tilde B_t$. As $Y, \tilde f, \tilde B$ are adapted processes, I am pretty sure that $X$ is a well-defined BSDE.

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