X_t = mu t + sigma W_t with X_0 = 0. For mu < 0 find P(sup_t X_t ≥ a) for a > 0, and for mu > 0 find E[tau_a] where tau_a is the first passage to a. State the martingale you used.

X_t = mu t + sigma W_t with X_0 = 0. For mu < 0 find P(sup_t X_t ≥ a) for a > 0, and for mu > 0 find E[tau_a] where tau_a is the first passage to a. State the martingale you used.

Approach: Find the exponent theta making exp(theta X_t) a martingale, then apply optional stopping at the passage time. For the positive drift case use the martingale X_t - mu t instead.

P(sup_t X_t ≥ a) = exp(2 mu a / sigma^2) for mu < 0, and E[tau_a] = a/mu for mu > 0. The process exp(theta X_t) has drift (theta mu + theta^2 sigma^2/2) exp(theta X_t) dt by Ito, so it is a martingale exactly when theta = -2 mu / sigma^2, which is positive when the drift is negative. Optional stopping at the first passage time tau_a on the bounded region below a gives 1 = E[exp(theta X_{tau})] = exp(theta a) P(tau_a < ∞) + 0, because on the non-hitting paths X tends to minus infinity and the exponential vanishes, so P(tau_a < ∞) = exp(-theta a) = exp(2 mu a / sigma^2). For mu > 0 the level is reached with probability 1, and applying optional stopping to the martingale X_t - mu t gives a - mu E[tau_a] = 0, so E[tau_a] = a / mu.

Follow-up: With mu < 0, what is the law of sup_t X_t, and what does it give for the probability of a 20% drawdown on a strategy with mu = 0.08 and sigma = 0.15?

Key concepts: exponential martingale, optional stopping, first passage, drift.