I wanted to follow the answer in this question, there are two steps in particular that I wanted to follow. Here let $B_{t}$ be a Brownian motion starting from a point $x$ contained in an interval $[a,b]$.
Let $\tau := \inf\{B_{t} \notin [a,b]\}$ which we write as an independent sum $\sigma$ and $\tau-\sigma$ where $\sigma := \inf\{B_{t} \notin [x-z,x+z]\}$ and let $$ u(x)=E_x[\mathrm e^{-\theta^2\tau/2}]. $$ where the subscript $x$ denotes the starting point of the Brownian motion.
Since $\sigma$ and $\tau-\sigma$ are independent, we can write
$$u(x)= E_x[\mathrm e^{-\theta^{2}\sigma/2}])E_{x}[\mathrm e^{-\theta^2(\tau-\sigma)/2}]$$ but then using the using the Markov property at time $\sigma$ it is essentially claimed that
$$ E_x[\mathrm e^{-\theta^2(\tau-\sigma)/2}]=\,\tfrac12(u(x+z)+u(x-z)). $$
I wanted to understand why, the strong Markov property in my book is written as $E_{x}[Y_{\sigma} \circ \theta_{\sigma}|\mathcal{F}_{\sigma}]=E_{B_{\sigma}}[Y_{\sigma}]$
