4th moment of a wiener process

Bazman · **Joined:** Mon, 29 Jan 2007 13:48:38 UTC **Posts:** 537

Hi there,

I know that:

$E[(X(t_j)-X(t_{(j-1)})]^4=\frac{3t^2}{n^2}$

I was even shown how to derive it in the dim and distant past but I have complete forgotten how. Can anyone enlighten me?

$X(t_j)-X(t_{j-1})$ is normally distributed with mean zero and variance t/n.

outermeasure · **Posted:** Wed, 14 Jul 2010 23:28:35 UTC

Bazman wrote:

Hi there,

I know that:

$E[(X(t_j)-X(t_{(j-1)})]^4=\frac{3t^2}{n^2}$

I was even shown how to derive it in the dim and distant past but I have complete forgotten how. Can anyone enlighten me?

$X(t_j)-X(t_{j-1})$ is normally distributed with mean zero and variance t/n.

Either evaluate the integral $\int_{-\infty}^\infty x^4e^{-x^2/2}\,\mathrm{d}x$ , or use the clever trick with $\exp(M_t-\frac{1}{2}\langle M\rangle_t)$ .

Bazman · **Joined:** Mon, 29 Jan 2007 13:48:38 UTC **Posts:** 537

Thanks OM,

I think that integral is beyond me?

Can you elaborate on the trick?

M_t is the moment. What does the t stand for its the moment at a given time step?

<M_t> is the quadratic variation right? (at a given time step)

where does this epxression even come from?

Baz

outermeasure · **Posted:** Thu, 15 Jul 2010 15:26:50 UTC

Bazman wrote:

Thanks OM,

I think that integral is beyond me?

Can you elaborate on the trick?

M_t is the moment. What does the t stand for its the moment at a given time step?

<M_t> is the quadratic variation right? (at a given time step)

where does this epxression even come from?

Baz

The integral is just an integration by parts and the definition of variance.

M_t is not the moment here. If M_t is a (local) martingale, then under suitable conditions (which the Brownian motion satisfies) the Doleans exponential exp(M_t-(1/2)<M>_t) is a martingale. So exp(W_t-t/2) is a martingale. But then it is just a matter of expanding and collecting the right terms that behaves like t^2.

Of course, you can also do it with Ito's formula (calculating dg(W)).

Bazman · **Joined:** Mon, 29 Jan 2007 13:48:38 UTC **Posts:** 537

Ah OK, its a little clearer.

But how can I expand the Doleans exponential? to get the fourth moment?

This just seems to be the same as applying ito to log X(t)?

If that is ocrrect then I follow you to this point, but how then do I build on this to get the 4th moment?

outermeasure · **Posted:** Fri, 16 Jul 2010 05:05:48 UTC

Bazman wrote:

Ah OK, its a little clearer.

But how can I expand the Doleans exponential? to get the fourth moment?

This just seems to be the same as applying ito to log X(t)?

If that is ocrrect then I follow you to this point, but how then do I build on this to get the 4th moment?

Just expand exp with the usual series exp(x)=1+x+x^2/2+x^3/6+x^4/24+...

Of course, to do things properly, you would want to scale time by c, expand, switch order of expectation and sum, and collect the terms with suitable powers of c (because the martingale property tells you $\mathbb{E}(\mathcal{E}(W_{ct}))=1$ ).

Bazman · **Joined:** Mon, 29 Jan 2007 13:48:38 UTC **Posts:** 537

Right had a bash at he integration by parts

$y=x^4 y'=4x^3 x=-\frac{-e^{\frac{x^2}{2}}}{x} x'=e^{\frac{-x^2}{2}}$

$-x^3 e^{\frac{-x^2}{2}} + \int4x^2 e^{\frac{x^2}{2}}$

applying integration by parts to the last section

$y=4x^2 y'=8x x=\frac{e^{-\frac{-x^2}{2}}}{x} x'=e^{\frac{-x^2}{2}}$

$-4x e^{\frac{-x^2}{2}} + \int 8 e^{\frac{-x^2}{2}}$

applying integration by substitution to the remaining integral with $u=\frac{-x^2}{2}$ gives

$-4x e^{\frac{-x^2}{2}} + \int 8 \frac{e^{\frac{-x^2}{2}}}{x}$

I think that the last part should go to zero at x goes to plu sand minus infinity but not sure how to prove it?

also all the parts above have a $e^{\frac{-x^2}{2}}$ term and so all tend to zero so far as I can see.

Where am I going wrong?

Bazman · **Joined:** Mon, 29 Jan 2007 13:48:38 UTC **Posts:** 537

outermeasure wrote:

Bazman wrote:

Ah OK, its a little clearer.

But how can I expand the Doleans exponential? to get the fourth moment?

This just seems to be the same as applying ito to log X(t)?

If that is ocrrect then I follow you to this point, but how then do I build on this to get the 4th moment?

Just expand exp with the usual series exp(x)=1+x+x^2/2+x^3/6+x^4/24+...

Of course, to do things properly, you would want to scale time by c, expand, switch order of expectation and sum, and collect the terms with suitable powers of c (because the martingale property tells you $\mathbb{E}(\mathcal{E}(W_{ct}))=1$ ).

Correct me If I am wrong but in the above expansion each term divided by the relevant factorial correspends to the given moment. So the fourth moment $\frac{x^4}{24}$ . However you still have to calculate the expectaionof x^4 so I think you wil end up using Ito's lemma anyway right?

With this in mind:

$d(x_t^4)=4x_t^3dx +\frac{1}{2}x_t^2dt$

integrating
$x_t^4-x_0^4= \int_0^t 4x_t^3dx_t + 6X_t^2-0$

Taking expectations

x_t^4= + 6X_t^2

given that x_t^2

is equal to the variance t/n

this gives

x_t^4= + 6t^2/n

This appears to be in the right direction but is still wrong.

Can you clarify my first point as well as where I am going wrong in my application of ito's lemma.

outermeasure · **Posted:** Sun, 18 Jul 2010 16:32:07 UTC

Something wrong with your integration by parts.

Let $u=x^3,\,\mathrm{d}v=xe^{-x^2/2}\,\mathrm{d}x$ , then $\mathrm{d}u=3x^2,\,v=e^{-x^2/2}$ , so

$\displaystyle \begin{aligned} \int_{-\infty}^\infty x^4 \dfrac{1}{\sqrt{2\pi}}e^{-x^2/2}\,\mathrm{d}x &=\int_{-\infty}^\infty 3x^2\dfrac{1}{\sqrt{2\pi}}e^{-x^2/2}\,\mathrm{d}x\\ &=3 \int_{-\infty}^\infty\dfrac{1}{\sqrt{2\pi}}e^{-x^2/2}\,\mathrm{d}x\\ &=3 \end{aligned}$
So $\mathbb{E}\left[X_{\frac{j}{n}t}-X_{\frac{j-1}{n}t}\right]^4=3\left(\mathop{\mathrm{Var}}\left[X_{\frac{j}{n}t}-X_{\frac{j-1}{n}t}\right]\right)^2=3\left(\frac{t}{n}\right)^2$

From Ito's formula, you get $X(t)^4-6\int_0^tX(s)^2\,\mathrm{d}s$ is a martingale. Also, remember X(t)^2-t

is a martingale (actually we only need $\mathbb{E}X_t^2=t$ ). So $\mathbb{E}[X(t_j)-X(t_{j-1})]^4=\mathbb{E}(X_{t/n}^4)=6\int_0^{t/n} s\,\mathrm{d}s=3\frac{t^2}{n^2}$ .

Bazman · **Joined:** Mon, 29 Jan 2007 13:48:38 UTC **Posts:** 537

thanks om!!

Bazman · **Joined:** Mon, 29 Jan 2007 13:48:38 UTC **Posts:** 537

outermeasure wrote:

Something wrong with your integration by parts.

Let $u=x^3,\,\mathrm{d}v=xe^{-x^2/2}\,\mathrm{d}x$ , then $\mathrm{d}u=3x^2,\,v=e^{-x^2/2}$ , so

$\displaystyle \begin{aligned} \int_{-\infty}^\infty x^4 \dfrac{1}{\sqrt{2\pi}}e^{-x^2/2}\,\mathrm{d}x &=\int_{-\infty}^\infty 3x^2\dfrac{1}{\sqrt{2\pi}}e^{-x^2/2}\,\mathrm{d}x\\ &=3 \int_{-\infty}^\infty\dfrac{1}{\sqrt{2\pi}}e^{-x^2/2}\,\mathrm{d}x\\ &=3 \end{aligned}$
OM in the first line you have added in the $\dfrac{1}{\sqrt{2\pi}}$ which obvioulsy from the fact that dx is normal (and should have been included all along) its not derived in any way correct?

In the 2nd line you have applied intrgeation by parts again using the same dv term but now with u=x right?

If the answer to this integration is just 3 why do we then multiply by Var^2

So $\mathbb{E}\left[X_{\frac{j}{n}t}-X_{\frac{j-1}{n}t}\right]^4=3\left(\mathop{\mathrm{Var}}\left[X_{\frac{j}{n}t}-X_{\frac{j-1}{n}t}\right]\right)^2=3\left(\frac{t}{n}\right)^2$

From Ito's formula, you get $X(t)^4-6\int_0^tX(s)^2\,\mathrm{d}s$ is a martingale. Also, remember X(t)^2-t

is a martingale (actually we only need $\mathbb{E}X_t^2=t$ ). So $\mathbb{E}[X(t_j)-X(t_{j-1})]^4=\mathbb{E}(X_{t/n}^4)=6\int_0^{t/n} s\,\mathrm{d}s=3\frac{t^2}{n^2}$ .

outermeasure · **Posted:** Mon, 19 Jul 2010 15:24:57 UTC

Bazman wrote:

outermeasure wrote:

Something wrong with your integration by parts.

Let $u=x^3,\,\mathrm{d}v=xe^{-x^2/2}\,\mathrm{d}x$ , then $\mathrm{d}u=3x^2,\,v=e^{-x^2/2}$ , so

$\displaystyle \begin{aligned} \int_{-\infty}^\infty x^4 \dfrac{1}{\sqrt{2\pi}}e^{-x^2/2}\,\mathrm{d}x &=\int_{-\infty}^\infty 3x^2\dfrac{1}{\sqrt{2\pi}}e^{-x^2/2}\,\mathrm{d}x\\ &=3 \int_{-\infty}^\infty\dfrac{1}{\sqrt{2\pi}}e^{-x^2/2}\,\mathrm{d}x\\ &=3 \end{aligned}$

OM in the first line you have added in the $\dfrac{1}{\sqrt{2\pi}}$ which obvioulsy from the fact that dx is normal (and should have been included all along) its not derived in any way correct?

In the 2nd line you have applied intrgeation by parts again using the same dv term but now with u=x right?

If the answer to this integration is just 3 why do we then multiply by Var^2

Yes, integration by parts again in the middle equality (or you may recall the variance of a standard normal to jump to the answer). I could put the $\sqrt{2\pi}$ factor is at the end, without changing anything. It is there to emphasise I am calculating $\int(-)\,\mathrm{d}\mathbb{P}$ .

The reason for multiplying by Var^2 is that we are not calculating the fourth moment of a N(0,1), but a N(0,t/n). The Var^2 is precisely the scaling factor of X^4.

Bazman · **Joined:** Mon, 29 Jan 2007 13:48:38 UTC **Posts:** 537

thanks again OM!!

You really did spread a little happiness today :-)

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