Approximation of the law of a stochastic process

2010-06-29T16:00:08Z

Hello Dear fellows,

I thank you in advance for your help and ideas.

I have just read an article and want you to help me understand the rational behind a part of it.

We have two processes $v_t$ and $V_t$

such that:

$dv_{t}=\omega v_{t} dW_{t}$ $W_{t}$ is Brownian and $V_{t}=\sqrt{\frac{1}{T} \int_0^T {v_{t}}^{2}\,dt}$

We know that $v_{t}=(law)v_{0}\exp(\omega z \sqrt{T}-\frac{\omega^{2}T}{2})$ $z$ is a gaussian with mean 0 and std 1.

The next statement is the one that i did not fully undestand the way they authors did have it :

They state $V_{T}=(law)\exp(\omega z_{1} \sqrt{\frac{T}{3}}-\frac{\omega^{2}T}{6})$ ou $z_{1}$ is a gaussian with mean 0 and std 1 and $\rho (z,z_{1})=\frac{\sqrt{3}}{2}$

I have tried to approximate $V_{T}$ by $\frac{1}{T} \int_0^T v_{t}\,dt$ (We know that $\frac{1}{T} \int_0^T v_{t}\,dt \leq V_{T}$) But I did not obtain any pertinent result this way.

Does someone have an idea how to prove the law equality above?

Many thanks

Answer by zhoraster for Approximation of the law of a stochastic process

2010-08-04T09:58:09Z

This is wrong. Let for simplicity $\omega = 1$. Write $$ E[V_T^2] = \frac 1T \int _0^T E[\exp(2W_t-t)]dt = 1/T\int_0^Te^{t} dt = (e^{T}-1)/T. $$ This is inconsistent with their claim.

(In fact, by considering expectations of greater powers of $V_T$, one can see that $V_T$ cannot be log-Gaussian with any parameters.)

Approximation of the law of a stochastic process - MathOverflow

Approximation of the law of a stochastic process

Answer by zhoraster for Approximation of the law of a stochastic process