Question

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 1

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.)