 Solution

ID: past-exam-of-the-continuous-martingales-and-stochastic-calculus-course-of-the-university-of-oxford/2023/2/c/i/solution

Put

M_{t} = B_{t}^{3} - 3 t B_{t}

. It is an adapted process and is integrable at every finite time, because normal random variables have finite absolute moments of every order. Fix

0 \leq s < t

and let

D = B_{t} - B_{s}

. Then

D

F_{s}

, has mean zero, variance

t - s

, and third moment zero. Expanding the cube inside the conditional expectation gives

E [B_{t}^{3} ∣ F_{s}] = B_{s}^{3} + 3 B_{s} E D^{2} = B_{s}^{3} + 3 (t - s) B_{s} .

Also

E [B_{t} ∣ F_{s}] = B_{s}

. Therefore

E [M_{t} ∣ F_{s}] = B_{s}^{3} + 3 (t - s) B_{s} - 3 t B_{s} = B_{s}^{3} - 3 s B_{s} = M_{s} .

These are exactly the defining conditions for a martingale.

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