Expected value of the Brownian motion to the power of $n$

Question

Let $B$ be a Brownian motion. How can $\mathbb{E}\left[B_t^n\right]$ and $\mathbb{E}\left\lvert[B_t\right\rvert^n]$ with $n\in\mathbb{N}$ and $t\geq 0$ be computed?

Answer 1

Let $s<t$. From $B_t-B_s$ being independent of $\mathcal{F}_s$ and $B_t-B_s \sim \mathcal{N}\left(0, t-s\right)$ follows:

\begin{align*} \mathbb{E}\left[\left(B_t - B_s\right)^n\vert \mathcal{F}_s\right] = \mathbb{E}\left[\left(B_t - B_s\right)^n\right] &= \left(t-s\right)^\frac{n}{2}\left(n-1\right)!! \cdot \begin{cases} 0 &\text{if}\ n\ \text{is odd}\\ 1 &\text{if}\ n\ \text{is even} \end{cases}\\ \mathbb{E}\left[\left\lvert B_t - B_s\right\rvert^n\vert \mathcal{F}_s\right] = \mathbb{E}\left[\left\lvert B_t - B_s\right\rvert^n\right] &= \left(t-s\right)^\frac{n}{2}\left(n-1\right)!! \cdot \begin{cases} \sqrt{\frac{2}{\pi}} &\text{if}\ n\ \text{is odd}\\ 1 &\text{if}\ n\ \text{is even} \end{cases} \end{align*}

Because of $B_t = B_t - B_0 \sim \mathcal{N}\left(0, t\right)$ we also have:

\begin{align*} \mathbb{E}\left[B_t^n\right] &= t^\frac{n}{2}\left(n-1\right)!! \cdot \begin{cases} 0 &\text{if}\ n\ \text{is odd}\\ 1 &\text{if}\ n\ \text{is even} \end{cases}\\ \mathbb{E}\left[\left\lvert B_t\right\rvert^n\right] &= t^\frac{n}{2}\left(n-1\right)!! \cdot \begin{cases} \sqrt{\frac{2}{\pi}} &\text{if}\ n\ \text{is odd}\\ 1 &\text{if}\ n\ \text{is even} \end{cases} \end{align*}

Here, $n!!$ denotes the double factorial of $n$, i.e. the product of all natural numbers from 1 to $n$ that have the same parity as $n$. The double factorial should not be confused with the factorial function iterated twice, which is written as $\left(n!\right)!$.

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For $n\leq 4$ we have: \begin{align*} &\mathbb{E}\left[\left(B_t - B_s\right)^1\vert \mathcal{F}_s\right] = \mathbb{E}\left[\left(B_t - B_s\right)^1\right] = 0 &&\mathbb{E}\left[B_t^1\right] = 0\\ &\mathbb{E}\left[\left(B_t - B_s\right)^2\vert \mathcal{F}_s\right] = \mathbb{E}\left[\left(B_t - B_s\right)^2\right] = t-s &&\mathbb{E}\left[B_t^2\right] = t\\ &\mathbb{E}\left[\left(B_t - B_s\right)^3\vert \mathcal{F}_s\right] = \mathbb{E}\left[\left(B_t - B_s\right)^3\right] = 0 &&\mathbb{E}\left[B_t^3\right] = 0\\ &\mathbb{E}\left[\left(B_t - B_s\right)^4\vert \mathcal{F}_s\right] = \mathbb{E}\left[\left(B_t - B_s\right)^4\right] = 3\left(t-s\right)^2 &&\mathbb{E}\left[B_t^4\right] = 3t^2\\\\ &\mathbb{E}\left[\left\lvert B_t - B_s\right\rvert^1\vert \mathcal{F}_s\right] = \mathbb{E}\left[\left\lvert B_t - B_s\right\rvert^1\right] = \left(t-s\right)^{\frac{1}{2}}\sqrt{\frac{2}{\pi}} &&\mathbb{E}\left[\left\lvert B_t\right\rvert^1\right] = t^{\frac{1}{2}}\sqrt{\frac{2}{\pi}}\\ &\mathbb{E}\left[\left\lvert B_t - B_s\right\rvert^2\vert \mathcal{F}_s\right] = \mathbb{E}\left[\left\lvert B_t - B_s\right\rvert^2\right] = t-s &&\mathbb{E}\left[\left\lvert B_t\right\rvert^2\right] = t\\ &\mathbb{E}\left[\left\lvert B_t - B_s\right\rvert^3\vert \mathcal{F}_s\right] = \mathbb{E}\left[\left\lvert B_t - B_s\right\rvert^3\right] = 2\left(t-s\right)^{\frac{3}{2}}\sqrt{\frac{2}{\pi}} &&\mathbb{E}\left[\left\lvert B_t\right\rvert^3\right] = 2t^{\frac{3}{2}}\sqrt{\frac{2}{\pi}}\\ &\mathbb{E}\left[\left\lvert B_t - B_s\right\rvert^4\vert \mathcal{F}_s\right] = \mathbb{E}\left[\left\lvert B_t - B_s\right\rvert^4\right] = 3\left(t-s\right)^2 &&\mathbb{E}\left[\left\lvert B_t\right\rvert^4\right] = 3t^2 \end{align*}