What is the Ito integral of a Brownian motion raised to a power?

[Euge]

Here's this week's problem!

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Problem. Let $B_s$ be a Brownian motion. Compute the Ito integral \(\displaystyle \int_a^b B_s^n\, dB_s,\) where $n$ is a positive integer.

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[Euge]

No one solved this week's problem. You can find my solution below.

Spoiler

Let $A_n = \int_a^b B_s^n\, dB_s$. By the fundamental theorem of Ito calculus,

\(\displaystyle A_n = \frac{B_b^{n+1} - B_a^n}{n + 1} - \frac{n\sigma^2}{2}A_{n-1}\)

for all $n \ge 1$. Thus

\(\displaystyle A_1 = \frac{B_b^2 - B_a^2}{2} - \frac{\sigma^2(b - a)}{2}\)

and for $n \ge 2$ (using induction)

\(\displaystyle A_n = \frac{B_b^{n+1} - B_a^{n+1}}{n+1}+\sum_{k = 2}^n (-1)^{n-k+1}n!\frac{B_b^k - B_a^k}{k!}\left(\frac{\sigma^2}{2}\right)^{n-k+1} + (-1)^n n! \left(\frac{\sigma^2}{2}\right)^n (b - a).\)