Let $W_t$ be a Brownian motion and $M_t = \sup_{0 \leq s \leq t} W_s$. I want to show that $$ P( M_t > b, W_t < a) = P(W_t < a – 2b) \quad \forall a \leq b, b >0$$ but I'm stuck.
I have the following so far:
Let $T \equiv \inf \{t \ge 0: W_t > b \}$. This is an almost surely finite stopping time so that $\widetilde{W}_t \equiv W_{T+t} – b$ is a Brownian motion independent of $\mathcal{F}_T$ (assuming some right continuous filtration here) by the strong Markov property.
We note that the events $T < t$ and $M_t > b$ are equal almost everywhere. Thus the probability we seek can be calculated as: $$P(T < t, W_t < a) = P(T < t, \widetilde{W}_{t – T} < a – b) = E(E(f(T, \widetilde{W})|\mathcal{F}_T)) \\
= E( P(\widetilde{W}_{t-u} < a-b) \mathbf{1}(u < t) |_{u = T}) \quad \text{by independence} \quad \textbf{(EQ2)} \\
= E\left( P \left(Z < \frac{a-b}{\sqrt{t-T}} \right) \mathbf{1}(T <t)\right) \quad \textbf{(EQ1)} $$
for $f(u, Y) = \mathbf{1}(u < t)\mathbf{1}(Y_{t-u} < a -b)$, and $Z \sim N(0,1)$
Setting $ a = b$ in the above yields $P(T < t) = 2P(W_t > b)$ so that the quantity in EQ1 may be written as $$\frac{b}{2 \pi}\int_0^t \int_{-\infty}^{\frac{b-a}{\sqrt{t-u}}} \frac{\exp\left(-\frac{1}{2}(x^2 -b^2/u) \right)}{u^{3/2}} dx du $$
and I cannot simplify that to save my life. There might (hopefully) be an easier way, or some sort of easy way to recognize this integral is what we want it to be. Does anyone have any ideas?
If instead of the struckout argument, we simply say that due to the symmetry of the Gaussian $\widetilde{W}_{t-u}$, (EQ2) is equivalent to: $$E(P(\widetilde{W}_{t-u} > b-a) \mathbf{1}(u < t) |_{u = T}) = P(\widetilde{W}_{t-T} > b-a, T < t)\\ \text{applying the same independence argument in reverse} \\ = P(W_t > 2b-a, T< t) = P(W_t > 2b – a)$$ since $2b – a \geq b$ and thus $ W_t > 2b -a \implies T < t$
Best Answer
The question has been edited with a more correct solution than the one below. However, I will keep this as is since it captures enough of the intuition.
This is reflection principle. Intuitively, for all paths that cross $b$, and then go below $a$, you can flip the path $b$ onward, and get a path that goes above $2b - a$
As you define, let $T = \inf \{t \geq 0 W_t > b\} $, and note that $W_T$ is a well-defined construct. Note that $W_s$ and $W'_s = W_{t - s} - W_s$ are independent. So, we have that
$$P(M_t > b, W_t < a) = \int_0^t P(T = s, W'_{s} < a - b) ds $$
and we note that $P(T = s, W'_{s} < a - b) = P(T = s)P(W'_{s} < a - b) = P(T = s) P(W'_{s} > b - a) = P(T = s, W'_{s} > b - a)$
by the independence of the stopping time from the future, and the symmetry of gaussians. Note that
$\int_0^t P(T = s, W'_{s} > b - a) = \int_0^t P(T = s, W_t > 2b - a) = P(W_t > 2b - a)$ since $\{ T \in [0,t] \} \subset \{W_t > 2b - a\} $
The result now follows from the symmetry of $W_t$.