M_t = ∫_0^t sigma(s) dW_s with sigma(s) = 0.2 sqrt(s). Show M is a time changed Brownian motion, then compute P(sup_{s ≤ 4} M_s ≥ 1) to four decimal places.

M_t = ∫_0^t sigma(s) dW_s with sigma(s) = 0.2 sqrt(s). Show M is a time changed Brownian motion, then compute P(sup_{s ≤ 4} M_s ≥ 1) to four decimal places.

Approach: Compute the quadratic variation clock by the Ito isometry, note that the running maximum only depends on the total clock elapsed, then apply the reflection principle at that clock value.

0.0771. The Ito isometry gives the quadratic variation clock [M]_t = ∫_0^t sigma(s)^2 ds = ∫_0^t 0.04 s ds = 0.02 t^2, so [M]_4 = 0.32, and the time change M_t = B_{[M]_t} holds with the clock deterministic, which makes B independent of it. The running maximum over [0,4] is therefore the running maximum of a standard Brownian motion over the clock interval [0, 0.32]. The reflection principle gives P(sup ≥ 1) = 2(1 - N(1/sqrt(0.32))) = 2(1 - N(1.7678)) = 2(0.03855) = 0.0771. With deterministic volatility the whole problem reduces to one number, the accumulated variance, and a path with front loaded volatility reaching the same total 0.32 would give the same answer for the terminal law and the same answer here.

Follow-up: If instead sigma is stochastic and correlated with W, why does this reduction fail and what inequality still holds for the maximum?

Key concepts: time change, Ito isometry, reflection principle, deterministic volatility.