Suppose $B_t$ and $W_t$ are two independent Brownian motions and $\tau$ is the first hitting time of $B_t$ to some $a >0$. Compute the distribution of $W_{\tau}$.
We can try the characteristic function method. Consider $E[e^{iu W_{\tau}}] = E[E[e^{iuW_{\tau}}|F_{B}]]$. Now we would like to use the Independence Lemma to get that this is $E[E[e^{iuW_t}]|_{\tau=t}] = E[e^{\frac{-tu^2}{2}} |_{\tau=t}] = E[e^{\frac{-\tau u^2}{2}}] = e^{-|u|a}$ so that $W_{\tau}$ has the same characteristic function as the Cauchy distribution and the problem would be solved. But this is not quite the Independence Lemma.
The only Independence Lemma that I know says that $E[f(X,Y)|F_Z] = E[f(x,Y)]|_{X=x}$ when $Y$ is independent of $F_Z$ and $X$ is measurable wrt $F_Z$ and $f$ is a Borel function. Does anyone know if there is some other more complicated Independence Lemma that would yield the above computation?
Thank you for your help.
Your computation is valid. Probably the easiest way to show it is to note that, for every $t$, $$ E(\mathrm e^{\mathrm iuW_\tau}\mid\tau=t)=E(\mathrm e^{\mathrm iuW_t}\mid\tau=t)=E(\mathrm e^{\mathrm iuW_t}), $$ where the second identity holds thanks to the independence of $W_t$ and $\tau$. One knows that, for every $t$, $E(\mathrm e^{\mathrm iuW_t})=\mathrm e^{-u^2t/2}$, hence $$ E(\mathrm e^{\mathrm iuW_\tau}\mid\tau)=\mathrm e^{-u^2\tau/2}, $$ which implies that $$ E(\mathrm e^{\mathrm iuW_\tau})=E(\mathrm e^{-u^2\tau/2}), $$ and you are done since you are able to compute the RHS.