Deriving the Fourth Moment of Brownian Motion

# Variance of Brownian Motion


# Variance of Brownian Motion












Can someone point me into the right direction to calculate this one: $E(B^4_t)=3t^2$

I had tried using the following property with no luck:

$E(B^4_t)=E(B^2_tB^2_t)=E(\int B^2 dt )E(\int B^2 dt )=[E(\int B^2 dt )]^2=[\int E(B^2) dt]^2=[\int t dt]^2$

Any other suggestion will be appreciated. Thanks!

## Answer by Daneel Olivaw (score 4, accepted)

https://quant.stackexchange.com/a/39043

Apply Itô's Lemma to $W_t^4$: $$ \text{d}(W_t^4)=4W_t^3\text{d}W_t+6W_t^2\text{d}t$$

Integrate: $$ W_t^4=4\int_0^tW_s^3\text{d}W_s+6\int_0^tW_s^2\text{d}s$$

The first term is an Itô integral, which is by construction a martingale, with expectation $0$ hence: $$E[W_t^4]=6\int_0^tE[W_s^2]\text{d}s=6\int_0^ts\text{d}s=3t^2$$

## Answer by David C (score 6)

https://quant.stackexchange.com/a/28413

At time $t$, Brownian Motion $B_t$ is simply a normal random variable $N(0,t)$.

The Moment Generating Function for a normal $N(\mu,\sigma^2)$ random variable is as follows: $$M(x) = exp(\mu x + \frac{1}{2}\sigma^2 x^2)$$ Furthermore, the fourth moment is given as the fourth derivative of this equation: $$M''''(x) = exp(\mu x + \frac{1}{2}\sigma^2 x^2)\Big( (\mu + \sigma^2x)^4 + 6\sigma^2(\mu + \sigma^2 x)^2 + 3\sigma^4 \Big)$$ So the expectation of $B_t^4$ is just the fourth moment, evaluated at $x=0$ (with parameters $\mu = 0$, $\sigma^2 = t$): $$E(B_t^4) = M''''(0) = 3\sigma^4 = 3t^2 $$

## Answer by user16651 (score 6)

https://quant.stackexchange.com/a/28421

Other Way

$$\mathbb{E}\left[ \,{{e}^{iuB_t}} \right]=\exp \left( iu\,\mathbb{E}\left[ B_t \right]+\frac{1}{2}{{(\,iu\,)}^{2}}\operatorname{Var}(B_t\,) \right)={\exp \left( -\frac{1}{2}{{u}^{2}}t \right)}$$ We know $$\mathbb{E}\left[{{e}^{iuB_t}} \right]=E\left[1+iuB_t-\frac{1}{2\,!}{{u}^{2}}{{B_t}^{2}}-\frac{1}{3\,!}i{{u}^{3}}{{B_t}^{3}}+\frac{1}{4\,!}{{u}^{4}}{{B_t}^{4}}+\cdots \right]$$ therefore

$${\exp \left( -\frac{1}{2}{{u}^{2}}t \right)}=1+iu\mathbb{E}\left[ B_t \right]-\color{green}{\frac{1}{2!}{{u}^{2}}\mathbb{E}\left[ {{B_t}^{2}} \right]}-\frac{1}{3!}i{{u}^{3}}E\left[ {{B_t}^{3}}\right]+\color{red}{\frac{1}{4!}{{u}^{4}}\mathbb{E}\left[ {{B_t}^{4}} \right]}+\cdots \tag1$$ On the other hand $$\exp \left( -\frac{1}{2}{{u}^{2}}t \right)=1-\color{green}{\frac{1}{2}\,{{u}^{2}}t}+\color{red}{\frac{1}{2!}\left( \frac{1}{4}{{u}^{4}}{{t}^{2}} \right)}-\frac{1}{3 !}\left( \frac{1}{8}{{u}^{6}}{{t}^{3}} \right)+\frac{1}{4 !}\left( \frac{1}{16}{{u}^{8}}{{t}^{4}} \right)-\cdots\tag2 $$ $(1)$ and $(2)$ $$\frac{1}{4!}{{u}^{4}}\mathbb{E}\left[ {{B_t}^{4}} \right]=\frac{1}{2 !}\left( \frac{1}{4}{{u}^{4}}{{t}^{2}} \right)$$ thus $$\mathbb{E}\left[ {{B_t}^{4}} \right]=3t^2$$ Generally, we have

> $$\left\{ \begin{align} & E\left[ {{B}^{2n+1}}(t) \right]=0\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ & \quad E\left[ {{B}^{2n}}(t) \right]=\frac{(2n)!}{{{2}^{n}}n\,!}\,{{t}^{n}} \\ \end{align} \right.$$