I have some issues with this problem:
"Consider the Stochastic Differential Equation: $$dX_t=b(X_t)dt+\sigma(X_t)dW_t$$ $$X_0=c \in \mathbb{R}$$ Consider the stopping time $\tau$: $$\tau = \inf \hspace{1mm}( t : X_t \notin (a,b))$$ The stopping time $\tau$ is almost surely finite for each open interval $(a,b) \subset \mathbb{R}$ . Consider the function $f$ such that $$f \in C^2(\mathbb{R})$$ $$b(x)f'+\frac{1}{2}\sigma^2(x)f''=0$$ Compute $p:= \mathbb{P}(X_{\tau}=b)$
in the case when $X_t= x+bt + \sigma W_t$,
with $b , \sigma \in \mathbb{R}$"
In order to do this exercise, I already proved that the following equations are true, as suggested as a Hint:
$$(1) \hspace{5mm} f(X_{\min (t,\tau)})= f(x)+ \int_0^{\min (t,\tau)} f'(X_u)\sigma(X_u)dW_u \hspace{4mm} \forall t>0$$
$$(2) \hspace{5mm} df= f'\sigma dW_t$$
$$(3) \hspace{5mm} \mathbb{E}(f(X_{\tau}))=f(x)$$
$$(4) \hspace{5mm} p=\frac{f(x)-f(a)}{f(b)-f(a)}$$
Even if I proved all this stuff, I don't know how to compute $p$ in the case when
$$X_t= x+bt + \sigma W_t$$
Can someone help me please? Thank you
We have that $dX_t=bdt+\sigma dW_t,\,X_0=x$. I suppose $x \in (a,b)$. To find $f$ we must find a solution to the second order ODE described in OP. Consider $$f(y)-f(z)=\int_{z}^ye^{-\frac{2b}{\sigma^2}s}ds=\frac{\sigma^2}{2b}(e^{-\frac{2b}{\sigma^2}z}-e^{-\frac{2b}{\sigma^2}y})$$ Indeed $$\frac{d}{dy}f(y)=e^{-\frac{2b}{\sigma^2}y},\,\frac{d^2}{dy^2}f(y)=-\frac{2b}{\sigma^2}e^{-\frac{2b}{\sigma^2}y}$$ So $$\frac{f(x)-f(a)}{f(b)-f(a)}=\frac{e^{-\frac{2b}{\sigma^2}a}-e^{-\frac{2b}{\sigma^2}x}}{e^{-\frac{2b}{\sigma^2}a}-e^{-\frac{2b}{\sigma^2}b}}$$