Let $(W_t)_{t \in \mathbb{R}_{+}}$ be a Wiener process and $M_t=\max_{0 \leq s \leq t}X_s$, $R_t=2M_t-W_t$. Then the distribution of $(M, R)$ is given as
$\mathbb{P}\{M_t \in dx, R_t \in dy\} = dxdy2h_{t}(y), 0 \leq x \leq y$ and
$\mathbb{P}\{M_{s+t} \in dx, R_{s+t} \in dy | M_s = a, R_s = b\} \\ = I(a, dx)dyk_t(b-a,y-x)+dxdy2h_t(b+y-2a), 0<a\leq b, a\leq x \leq y$
(here $x \rightarrow I(a, x)$ is an indicator function) where
$h_t(a)=\frac{ae^{-\frac{a^2}{2t}}}{\sqrt{2\pi t^3}}, k_t(x,y)=\frac{1}{\sqrt{2\pi t}}(e^{-\frac{(x-y)^2}{2t}}-e^{-\frac{(x+y)^2}{2t}})$.
The question is to calculate the distribution in terms of $R$, that is, to prove
$\mathbb{P}\{R_t \in dx\} = dx2xh_t(x)$ and
$\mathbb{P}\{R_{s+t} \in dy|R_s=x\} = dy\frac{y}{x}k_t(x,y)$
While the first equation is trivial, I am stuck with the second equation above. I'm not even sure how to calculate it (how to combine those relationships so that I can get the formula for $\mathbb{P}\{R_{s+t} \in dy|R_s=x\}$). Any help would be greatly appreciated.
Here's how you would combine those two relationships. You're given the conditional density of $M_{s+t},R_{s+t}\mid M_s,R_s$ so integrate it with respect to the first argument to get the density for $R_{s+t}\mid M_s,R_s$. To get this into what you need, use the law of total expectation with respect to a conditional density to integrate out the $M_s$, i.e., integrate the last result wrt the density of $M_s\mid R_s$. You can compute that using the density version of bayes rule from the joint $M_s,R_s$, given to you, and the marginal $R_s$, which you calculated in the first part.