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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?

ALEXANDER
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    What is $n$? You seem to be dealing with 1-dimensional BM, so where does $n$ come from? – Kavi Rama Murthy Mar 27 '18 at 09:49
  • In case $n$ is the dimension of the Brownian motion $(B_t)$, this is direct from expanding the square on the RHS of the identity $$|B_t-B_s|^4=(X_1^2+\cdots+X_n^2)^2$$ where $(X_k)$ is i.i.d. $N(0,|t-s|)$ and using the moments $E(Z^2)=1$ and $E(Z^4)=3$ of every standard $N(0,1)$ random variable $Z$. – Did Mar 28 '18 at 07:21

2 Answers2

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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.

yama
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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.

Olórin
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  • "you seem to use a one-dimensional brownian motion so that your n would be equal to 1, which I assume from now" Don't. – Did Mar 28 '18 at 07:21
  • Going from $1$ to $n$ is obvious, btw. Oh, and usually, outside the affine line, you write $|\cdot|$ and not $|\cdot|$. – Olórin Mar 28 '18 at 07:41
  • ?? This "obvious" thing is what the OP is asking about (and yes, their notational choices are faulty, which does not mean that answers should not address what they are really asking). – Did Mar 28 '18 at 07:58
  • I say that showing it for any $n$ is obvious when you know it for $n = 1$ . And that the latter boils down to integrating by parts and knowing $\int_{\mathbf{R}} e^{-x^2 / 2} dx$. – Olórin Mar 28 '18 at 08:23
  • OK, so you do not read comments made at you. MSE pathology at its worst. No big deal. – Did Mar 28 '18 at 08:32