Let $B$ be a brownian motion and $B^*_t:=\sup_{s\leq t} B_s$. I want to show that for $y>0, x\leq y$, $$\Bbb{P}(B_t\leq x, B^*_t\leq y)=F\left(\frac{x}{\sqrt t}\right)-F\left(\frac{x-2y}{\sqrt t}\right)$$ where $F$ is the distribution function of a normal random variable $N\sim \mathcal{N}(0,1)$.
I somehow don't see how to procedure, I thought about using the strong markov property for $W_t:=B_{t+\rho}-B_{\rho}$ where $\rho$ is a suitable stopping time, because we have done something similar in class. There we took $\rho:=\inf\{t>0: B_t\in (y,\infty\}$ to be the first entry time, which is an $\Bbb{F}^{B_+}$ stopping time, but I think this does not work in our case here. Therefore I wanted to ask if someone could help me futher.
What we have proven in class is that for all $y>0$ and $x\leq y$ $$\Bbb{P}(B_t\leq x, B^*_t\geq y)=\Bbb{P}(B_t\geq 2y-x)$$
$$\begin{align} \Bbb{P}(B_t\leq x, B^*_t\leq y)&=\Bbb{P}(B_t\leq x)-\color{blue}{\Bbb{P}(B_t\leq x, B^*_t\geq y)}\\ &=\color{red}{\Bbb{P}(B_t\leq x)}-\color{blue}{\Bbb{P}(B_t\geq 2y-x)} \hspace{1cm}\text{(as you learnt in class)}\\ &=\color{red}{F\left(\frac{x}{\sqrt t}\right)}-\color{blue}{F\left(\frac{x-2y}{\sqrt t}\right)} \end{align}$$