Proving $E^{x}[|B_s-B_t|^4]=n(n+2)|t-s|^2$

Question

Prove that:

$$E^{x}\left[ \left| B_{t}-B_{s}\right| ^{4}\right] =n\left( n+2\right) \left| t-s\right| ^{2},$$ where $B$ is a brownian motion.

Now there are two methods I have tried, but I am clearly missing something vital as my attempts are incorrect. Would someone please let me know where I am going wrong?

Attempt 1:

Useful equations:

$$E[B_t^2]=t,\ E[B_t^3]=0,\ E[B_t^4]=3t^2.$$

First idea, as the assumption that B being a brownian motion with mean(0) and variance(t or s)

$$(B_t-B_s)^4=-4 B_s^3 B_t+6 B_s^2 B_t^2-4 B_s B_t^3+B_s^4+B_t^4.$$

Now substituting the useful equations using I don't get the right answer, hence either my calculations are wrong or my assumptions,(understanding) is wrong.

Attempt 2:

Useful equations 2

$$E^x\left[(B_t-x)^2\right]=nt,\\ E^x\left[((B_t-x)(B_s-x))^2\right]=n·\min(s,t),\\ E^x\left[(B_t-B_s)^2\right]=n(t-s).$$

Mini question when stating min(s,t) I understand that it takes the min of (s,t) but if t and s represent time and t>s does that if both start at t=s=0 that $B_t$ being a vector has more elements in it then $B_s$?

$$\left((B_t-x)-(B_s-x)\right)^4.$$

Now expanding this out and using Useful equations 2 I still not get the right answer, hence I am doing something wrong with the calculations or my attempt in itself is flaud.

Would someone please take me through the steps?

Answer 1

It is an exercise of a book "Stochastic Differential Equations" by Øksendal, isn't it? I have also considered about it a couple of days. You seem to confuse 1-dimensional solution and n-dimensional solution.

In n-dimensional case, $ |B_{t}-B_{S}| $ is $$ |B_{t}-B_{S}|=\sqrt{\left(B_{t}^{(1)}-B_{s}^{(1)}\right)^{2}+\left(B_{t}^{(2)}-B_{s}^{(2)}\right)^{2}+\ldots+\left(B_{t}^{(n)}-B_{s}^{(n)}\right)^{2}} $$

where $ B_{t}^{(i)} $ and $ B_{s}^{(i)} $ are i-th elements of $ B_{t} $ and $ B_{s} $, respectively. Accordingly, $$ |B_{t}-B_{s}|^{4}=\left\{\left(B_{t}^{(1)}-B_{s}^{(1)}\right)^{2}+\left(B_{t}^{(2)}-B_{s}^{(2)}\right)^{2}+\ldots+\left(B_{t}^{(n)}-B_{s}^{(n)}\right)^{2}\right\}^{2} $$

Since elements of i and j are independent, the expectation with an initial location, x, is $$ E^{x}\left[|B_{t}-B_{s}|^{4}\right]=\sum_{i}^{n}E^{x}\left[\left(B_{t}^{(i)}-B_{s}^{(i)}\right)^{4}\right]+\sum_{i}^{n}E^{x}\left[\left(B_{t}^{(i)}-B_{s}^{(i)}\right)^{2}\right]\sum_{j\neq i}^{n}E^{x}\left[\left(B_{t}^{(j)}-B_{s}^{(j)}\right)^{2}\right] $$ $$ =n3(t-s)^{2}+n(t-s)(n-1)(t-s) $$ $$ =n(n+2)(t-s)^{2} $$

I am not a mathematician originally, therefore, this may include mistakes. It could be also helpful that some hints for exercises of Øksendal book are provided by http://www.quantsummaries.com/oksendal.pdf.

Answer 2

First you don't write gaussian brownian motion as brownian motion is a gaussian process. (See Oksendal's "Stochastic Differential Equations", chapter 2 section 2.) (Thx for having edited your question.)

Second, you seem to use a one-dimensional brownian motion so that your $n$ would be equal to $1$, which I assume from now.

Last, by definition of $B$ you know that $B_t - B_s$ (with $t \geq s$) has the law of a centered normal distribution with variance $t-s$. Up to a change of variables, calculating the expectation of $(B_t - B_s)^4$ amounts to calculate a sum of the form $\int_{\mathbf{R}} x^4 e^{-x^2 / 2} dx$. Integrating by parts four times and knowing the (well-known) value of $\int_{\mathbf{R}} e^{-x^2 / 2} dx$ will allow you to conclude.