Reflected process - Brownian motion

Question

I am still new to stochastic processes and I tried to do this exercise. I don't know how to go on.

Define the maximum process \begin{align*} M_t = \max_{0 \leqslant s \leqslant t} W_s, \end{align*} and hitting times \begin{align*} T_a = \inf\{s : W_s = a\}. \end{align*} Give an expression for $\mathbb P(T_a \leqslant t)$ in terms of the maximum process $M_t$.

For any fixed $T$ and a standard Brownian motion $W_t$, the reflected process defined by \begin{align*} \tilde W_t^T = \begin{cases} W_t, & t \leqslant T \\ 2W_T-W_t, & t > T, \end{cases} \end{align*} is itself a standard Brownian motion.

Prove that for any $x \leqslant a$, \begin{align*} \mathbb P(M_t \geqslant a, W_t \leqslant x) = \mathbb P(\tilde W_t^{T_a} \geqslant 2a-x) = 1 - \Phi\left(\frac{2a-x}{\sqrt t}\right). \end{align*} Using symmetry, conclude that \begin{align*} \mathbb P(T_a \leqslant t) = 2(1 - \Phi(a/\sqrt t)). \end{align*} For the first part, we have that \begin{align*} \mathbb P(T_a \leqslant t) &= \mathbb P(\inf\{s : W_s = a\} \leqslant t) = \mathbb P\left(\max_{0 \leqslant s \leqslant t} W_s \geqslant a\right) = \mathbb P(M_t \geqslant a). \end{align*} Assume that $\tilde W_t^{T_a} \geqslant 2a-x$.

If $t \leqslant T_a$, it is $\tilde W_t^{T_a} = W_t$.

Suppose that $M_t < a$. Then \begin{align*} 2a-x \leqslant W_t \leqslant \max_{0 \leqslant s \leqslant t} W_s = M_t < a, \end{align*} which implies $x > a$. So we must have $M_t \geqslant a$.

But how can I prove in this case, that $W_t \leqslant x$?

If $t > T_a$, we have \begin{align*} M_t = \max_{0 \leqslant s \leqslant t} W_s \geqslant \max_{0 \leqslant s \leqslant T_a} W_s \geqslant W_{T_a} = a. \end{align*} Moreover, \begin{align*} W_t &= 2W_{T_a} + W_t - 2W_{T_a} = 2a - (2W_{T_a} - W_t) = 2a - \tilde W_t^{T_a} \leqslant 2a - (2a-x) = x. \end{align*} Conversely, assume that $M_t \geqslant a \geqslant x$ and $W_t \leqslant x \leqslant a$.

If $t \leqslant T_a$, I do not know how to show that $W_t \geqslant 2a-x$.

If $t > T_a$, we get that \begin{align*} \tilde W_t^{T_a} &= 2W_{T_a} - W_t = 2a - W_t \geqslant 2a - x. \end{align*}

We get the second equality as follows: \begin{align*} \mathbb P(\tilde W_t^{T_a} \geqslant 2a-x) &= 1 - \mathbb P(\tilde W_t^{T_a} \leqslant 2a-x). \end{align*} Since $\tilde W_t^{T_a}$ is itself a standard Brownian motion with mean 0 and variance $t$, we get \begin{align*} \mathbb P(\tilde W_t^{T_a} \geqslant 2a-x) = 1 - \Phi\left(\frac{2a-x}{\sqrt t}\right). \end{align*} Now, we see that \begin{align*} \mathbb P(T_a \leqslant t) &= \mathbb P(M_t \geqslant a) = \mathbb P(M_t \geqslant a, W_t \leqslant x) + \mathbb P(M_t \geqslant a, W_t > x) \\ &= 1 - \Phi\left(\frac{2a-x}{\sqrt t}\right) + \mathbb P(M_t \geqslant a, W_t > x) \end{align*}

Here, I do not see how to obtain $2(1 - \Phi(a/\sqrt t))$.

Answer 1

Let us start with your first problem. You want to calculate

$$\mathbb P(M_t \geqslant a, W_t \leqslant x).$$

As you noted,

\begin{align*} \mathbb P(M_t \geqslant a, W_t \leqslant x)=\mathbb P(T_a \leqslant t, W_t \leqslant x)&=\mathbb P(T_a \leqslant t, W_t -2W_{T_a}\leqslant x-2a)\\ &=\mathbb P(T_a \leqslant t, \tilde{W}_t^{T_a}\geqslant 2a-x), \end{align*}

where the last line follows because $t\ge T_a$. Lastly,

$$\mathbb P(T_a \leqslant t, \tilde{W}_t^{T_a}\geqslant 2a-x)=\mathbb P(\tilde{W}_t^{T_a}\geqslant 2a-x)-\mathbb P(T_a > t, \tilde{W}_t^{T_a}\geqslant 2a-x),$$

but if $T_a>t$, $\{\tilde{W}_t^{T_a}\ge 2a-x\}=\{W_t\ge 2a-x\}\subset\{W_t\ge a\}\subset\{T_a\le t\}$, which means that $\mathbb P(T_a > t, \tilde{W}_t^{T_a}\geqslant 2a-x)=0$.

For your second question, take $x=a$. The second term will simplify since $\{W_t>a\}\subset\{M_t\ge a\}$.

Note for the first part that we need the reflected Brownian motion to be a Brownian motion for any $T$ hitting time (you wrote for any fixed $T$).