0
$\begingroup$

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 $2(1 - \Phi(a/\sqrt t))$.

$\endgroup$
  • 1
    $\begingroup$ I looked at your proof in detail and your mistake is not taking into account that $t$ can not take any values. Specifically, if $\tilde{W}_t^{T_a}\ge 2a-x$ and $t\le T_a$, then on the one hand $W_t\ge 2a-x$ and on the other hand $W_t<a$ which is not possible. $\endgroup$ – Ian Apr 27 '14 at 14:58
  • $\begingroup$ Similarly, if $M_t\ge a$, then necessarily $T_a\le t$. $\endgroup$ – Ian Apr 27 '14 at 15:05
1
$\begingroup$

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 $T$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.