I am given that for $b,a,\sigma >0$ and $x \in (-a,b)$ and $\nu \in \mathbb{R}$, I have the following stochastic differential equation:
$$ dZ_t = \nu \,dt + \sigma\, dW_t$$ $$ Z(0) = x$$
and $\tau = \inf\{t>0:Z(t)=b \text{ or } Z(t) = -a\}$
and $u(x) = b\mathbb{P}[Z(\tau)=b\mid Z_0 = x] -a\mathbb{P}[Z(\tau)=-a\mid Z_0 = x]$, I want to find
$\mathbb{E}[u(Z(\tau))\mid Z_0 = x] - u(x)$ and the differential eq that $u(x)$ satisfies.
Attempt at solution: We know that for a starting point $x$, $u(x) = b\mathbb{P}[Z(\tau)=b\mid Z_0 = x] -a\mathbb{P}[Z(\tau)=-a\mid Z_0 = x]$ so then $u(Z(\tau)) = b\mathbb{P}[Z(\tau)=b\mid Z_0 = Z(\tau)] -a\mathbb{P}[Z(\tau)=-a\mid Z_0 = Z(\tau)]$ but $\mathbb{P}[Z(\tau)=b\mid Z_0 = Z(\tau)] = 1/2 = \mathbb{P}[Z(\tau)=-a\mid Z_0 = Z(\tau)]$ So then $u(Z(\tau)) = \frac{1}{2}b + \frac{1}{2}a$ But this seems wrong since it doesnt depend on time? Also how would I apply Ito to $u(x)$? It is a function of a probability of $Z(\tau)$. How would I write it as an explicit function since I cant take partials of probability?
Any help would be greatly appreciated! Thanks!
After "so then", you seem to use the identity that if $v(x)=P[U=b\mid V=x]$ for every $x$ then $v(U)=P[U=b\mid V=U]$. This is wrong, and you could have been suspicious about this, due to the fact that in the setting of the exercise, the event $U=V$ is impossible.
Instead, note that if $U$ takes values in $\{u_k\}_k$, then $\sum\limits_{k}u_kP[U=u_k\mid V=x]=E[U\mid V=x]$, by definition.