The sixth moment of Brownian motion · Problem 180

s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.

Idea. Itô's formula turns the $k$ -th moment into an integral of the $(k - 2)$ -th moment, so the moments can be computed two at a time.

Step 1: the recursion. Taking expectations in Itô's formula (the $d W$ term has mean 0), $E [W_{t}^{k}] = \frac{k ( k - 1 )}{2} \int_{0}^{t} E [W_{s}^{k - 2}] d s .$

Step 2: climb.

$E [W_{t}^{2}] = \int_{0}^{t} 1 d s = t$ .
$E [W_{t}^{4}] = 6 \int_{0}^{t} s d s = 3 t^{2}$ .
$E [W_{t}^{6}] = 15 \int_{0}^{t} 3 s^{2} d s = 15 t^{3}$ .

Step 3. At $t = 1$ , $E [W_{1}^{6}] = 15$ .

Check. $W_{1}$ is standard normal, and the even moments of a standard normal are $E [Z^{2 m}] = 1 \cdot 3 \cdot 5 \dots (2 m - 1)$ . For $2 m = 6$ this is $1 \cdot 3 \cdot 5 = 15$ .