Splitting Integral of Brownian motion over stopping times Let $B_s$ be a Brownian Motion and denote be $E^y$ the law of BM started at $B_0 = y \in \mathbb{R}$. The stopping time $\tau_{[a,b]}$ denotes the exit time from the intervall $[a,b]$.
Now let $a\leq b\leq x \leq c$ be real one dimensional numbers and consider a function $f:[a,c]\rightarrow \mathbb{R}$ (such that the integrals below make sense).
Are the following equalities true? I splitted the integration intervall in the first step and used the Markov property in the second step.
$$E^x(\int_0^{\tau_{[a,c]}} f(B_s) ds)= E^x(\int_0^{\tau_{[b,c]}}f(B_s)ds)+E^x(\int_{\tau_{b}}^{\tau_{[a,c]}} f(B_s)ds ~|~ \tau_b < \tau_c)=E^x(\int_0^{\tau_{[b,c]}}f(B_s)ds)+E^b(\int_{0}^{\tau_{[a,c]}} f(B_s)ds)$$
Thanks in advance
 A: You have a little mistake in your second equality. By bringing in the expectation with respect to the conditional probability $\Bbb P (\cdot \; | \tau_b < \tau_c)$ you have forgotten to multiply with the inverse of reweighting factor $1/ \Bbb P ( \tau_b < \tau_c)$, namely $ \Bbb P ( \tau_b < \tau_c)$. My calculation goes as follows.
Note that $\tau_{[a,c]} = \tau_a \wedge \tau_c := \min (\tau_a ,\tau_c)$. We have
$$\Bbb E^x[ \int_0^{\tau_{[a,c]}} f(B_s) \text d s ] = \Bbb E^x[ 1_{\{\tau_b < \tau_c\}}\int_0^{\tau_{[a,c]}} f(B_s) \text d s ] + \Bbb E^x[ 1_{\{\tau_b > \tau_c\}}\int_0^{\tau_{[a,c]}} f(B_s) \text d s ]$$
The case $\tau_b < \tau_c$ implies that $\tau_b \leq \tau_c \wedge \tau_a $. Thus by splitting the Riemann integral the above equals
$$\Bbb E^x[ 1_{\{\tau_b < \tau_c\}}\int_0^{\tau_b} f(B_s) \text d s ]+ \Bbb E^x[ 1_{\{\tau_b < \tau_c\}}\int_{\tau_b}^{\tau_c \wedge \tau_a} f(B_s) \text d s ] +\Bbb E^x[ 1_{\{\tau_b > \tau_c\}}\int_0^{\tau_{[a,c]}} f(B_s) \text d s ]$$
Treating the middle term by using the markov property and using that $B_{\tau_b} = b$ we get
$$\Bbb E^x[ 1_{\{\tau_b < \tau_c\}}\int_{\tau_b}^{\tau_c \wedge \tau_a} f(B_s) \text d s ] = \Bbb E^x[ 1_{\{\tau_b < \tau_c\}} \Bbb E^x[\int_{\tau_b}^{\tau_c \wedge \tau_a} f(B_s) \text d s | \mathcal F_{\tau_b}] ] \\
=\Bbb E^x[ 1_{\{\tau_b < \tau_c\}} \Bbb E^{B_{\tau_b}}[\int_{0}^{\tau_c \wedge \tau_a} f(B_s) \text d s ] ] = \Bbb E^x[ 1_{\{\tau_b < \tau_c\}} \Bbb E^{b}[\int_{0}^{\tau_c \wedge \tau_a} f(B_s) \text d s ] ]\\
= \Bbb P^x (\tau_b < \tau_c) E^{b}[\int_{0}^{\tau_c \wedge \tau_a} f(B_s) \text d s ] $$
Now we observe that the case $\tau_b > \tau_c$ implies $\tau_a\wedge \tau_c = \tau_c = \tau_b \wedge\tau_c$ and the case $\tau_b < \tau_c$ implies $\tau_b = \tau_b \wedge \tau_c$. Putting this together yields
$$\Bbb E^x[ \int_0^{\tau_{[a,c]}} f(B_s) \text d s ] \\
= \Bbb P^x (\tau_b < \tau_c) E^{b}[\int_{0}^{\tau_c \wedge \tau_a} f(B_s) \text d s ] + \Bbb E^x[ 1_{\{\tau_b < \tau_c\}}\int_0^{\tau_b\wedge \tau_c} f(B_s) \text d s ] + \Bbb E^x[ 1_{\{\tau_b > \tau_c\}}\int_0^{\tau_b\wedge \tau_c} f(B_s) \text d s ] \\
= \Bbb P^x (\tau_b < \tau_c) E^{b}[\int_{0}^{\tau_{[a,c]}} f(B_s) \text d s ] + E^x[ \int_0^{\tau_{[b,c]}} f(B_s) \text d s ]$$
