January 26, 2025
If B_t is Brownian motion let \overline{B}_t = \max_{0\le s\le t}B_s be the running max.
E[f(B_t) 1(\overline{B}_t > a)] = E[f(B_t) 1(B_t > a)] + E[f(2a - B_t) 1(B_t > a)]
Proof. Define the reflected Brownian motions B_t^a = B_t for t < \tau_a and B_t^a = 2a - B_t for t > \tau_a. We have \begin{aligned} E[f(B_t) 1(\overline{B}_t > a)] &= E[f(B_t) 1(\overline{B}_t > a,B_t > a)] + E[f(B_t) 1(\overline{B}_t > a,B_t < a)]\\ &= E[f(B_t) 1(B_t > a)] + E[f(B_t) 1(\overline{B}_t > a,B_t < a)] \\ &= E[f(B_t) 1(B_t > a)] + E[f(B^a_t) 1(\overline{B^a}_t > a,B^a_t < a)] \\ &= E[f(B_t) 1(B_t > a)] + E[f(2a - B_t) 1(\overline{B^a}_t > a,2a - B_t < a)] \\ &= E[f(B_t) 1(B_t > a)] + E[f(2a - B_t) 1(\overline{B^a}_t > a, B_t > a)] \\ &= E[f(B_t) 1(B_t > a)] + E[f(2a - B_t) 1(B_t > a)] \\ &= E[\bigl(f(B_t) + f(2a - B_t)\bigr) 1(B_t > a)] \\ \end{aligned}
Taking f(x) = 1 gives P(\overline{B}_t > a) = 2P(B_t > a).
Let \underline{B}_t = \min_{0 \le s \le t} B_s. (\underline{-B})_t = -\overline{B}_t.
\begin{aligned} E[f(B_t) 1(\underline{B}_t < a)] &= E[f(-B_t) 1(\underline{-B}_t < a)] \\ &= E[f(-B_t) 1(-\overline{B}_t < a)] \\ &= E[f(-B_t) 1(\overline{B}_t > -a)] \\ &= E[\bigl(f(-B_t) + f(-(-2a - B_t)\bigr) 1(B_t > -a)] \\ &= E[\bigl(f(-B_t) + f(2a + B_t)\bigr) 1(B_t > -a)] \\ &= E[\bigl(f(B_t) + f(2a - B_t)\bigr) 1(B_t < a)] \\ \end{aligned}
Taking f(x) = 1 gives P(\underline{B}_t < a) = 2P(B_t < a).
Let \tau_a = \inf\{t > 0\mid \underline{B}_t < a\} be the first time B_t hits level a < 0.
Note \tau_a > t if and only if \underline{B}_t < a.
Recall E[\tau] = \int_0^\infty P(\tau > t)\,dt if \tau\ge0.
P(\min\{\tau_a, T\} > t) = P(\tau_a > t, T > t) = P(\tau_a > t)1(T > t)
E[\min\{\tau_a, T\}] = \int_0^T P(\tau_a > t)\,dt.
Let X_t = \mu t + \sigma B_t.
Girsanov’s theorem states B_t - \alpha t is BM under P_\alpha where {dP_\alpha/dP|_t = \exp(\alpha B_t - \alpha^2t/2)}. Let g(x) = f(\mu t + \sigma x) and \alpha = \mu/\sigma.
\begin{aligned} E[f(X_t) 1(\underline{X}_t < a)] &= E[g(B_t) 1(\min_{0\le s\le t}\mu s + \sigma B_s < a)] \\ &= E[\exp(\alpha B_t - \alpha^2t/2) g(B_t - \alpha t) 1(\min_{0\le s\le t}\mu s + \sigma (B_s - \alpha s) < a)] \\ &= E[\exp((\mu/\sigma) B_t - (\mu/\sigma)^2t/2) g(B_t - (\mu/\sigma) t) 1(\min_{0\le s\le t}\sigma B_s < a)] \\ &= E[\exp((\mu/\sigma) B_t - (\mu/\sigma)^2t/2) g(B_t - (\mu/\sigma) t) 1(\min_{0\le s\le t}B_s < a/\sigma)] \\ &= E[h(B_t) 1(\underline{B}_t < a/\sigma)] \\ &= E[(h(B_t) + h(2a - B_t)) 1(B_t < a/\sigma)], \\ \end{aligned} where h(x) = \exp((\mu/\sigma) x - (\mu/\sigma)^2t/2) g(x - (\mu/\sigma) t).
If f(x) = 1 then g(x) = 1 and h(x) = \exp((\mu/\sigma) x - (\mu/\sigma)^2t/2).
If f(x) = 1(x < b) then g(x) = 1(\mu t + \sigma x < b) and {h(x) = \exp((\mu/\sigma) x - (\mu/\sigma)^2t/2) 1(\mu t + \sigma (x - (\mu/\sigma) t) < b)}
P(\underline{X}_t < a) = E[\bigl(\exp((\mu/\sigma) B_t - (\mu/\sigma)^2t/2) + \exp((\mu/\sigma) (2a - B_t) - (\mu/\sigma)^2t/2)\bigr)1(B_t < a/\sigma)].
E[\exp(\alpha B_t - \alpha^2t/2) 1(B_t < a)] = E[1(B_t + \operatorname{Cov}(\alpha B_t,B_t) < a)] = P(B_t < a - \alpha t).
E[\exp((\mu/\sigma) B_t - (\mu/\sigma)^2t/2) 1(B_t < a/\sigma)] = P(B_t < a/\sigma - (\mu/\sigma)t).
E[\exp((\mu/\sigma) (2a - B_t) - (\mu/\sigma)^2t/2) 1(B_t < a/\sigma)] =\exp(2a\mu/\sigma) P(B_t < a/\sigma + (\mu/\sigma)t).
E[\min\{\tau_a,T\}] = \int_0^T P(B_t < a/\sigma - (\mu/\sigma)t) + \exp(2a\mu/\sigma) P(B_t < a/\sigma + (\mu/\sigma)t)\,dt \\
Let Z be standard normal.
\int_0^T P(Z < a/\sqrt{t} + b\sqrt{t})\,dt
No closed form solution.
\begin{aligned} P(B_t < a, \underline{B_t} < \underline{a}) &= E[1(B_t < a) 1(\underline{B_t} < \underline{a})] \\ &= E[(1(B_t < a) + 1(2\underline{a} - B_t < a) 1(B_t < \underline{a})] \\ &= E[(1(B_t < a) 1(B_t < \underline{a})] + E[1(2\underline{a} - B_t < a) 1(B_t < \underline{a})] \\ &= P(B_t < \min\{a, \underline{a}\}) + P(B_t > 2\underline{a} - a, B_t < \underline{a}) \\ &= P(B_t < \min\{a, \underline{a}\}) + P(2\underline{a} - a < B_t < \underline{a}) \\ \end{aligned}
P(S_t - \underline{S}_t > a), S_t = s\exp(\mu t + \sigma B_t - \sigma^2 t/2).
P(X_t - \underline{X}_t > a), X_t = x + \mu t + \sigma B_t.