Approximation of the law of a stochastic process - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T21:09:02Z http://mathoverflow.net/feeds/question/29934 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/29934/approximation-of-the-law-of-a-stochastic-process Approximation of the law of a stochastic process AlGoRiS 2010-06-29T16:00:08Z 2010-09-15T11:22:16Z <p>Hello Dear fellows,</p> <p>I thank you in advance for your help and ideas.</p> <p>I have just read an article and want you to help me understand the rational behind a part of it.</p> <p>We have two processes $v_t$ and $V_t$ </p> <p>such that:</p> <p>$dv_{t}=\omega v_{t} dW_{t}$ $W_{t}$ is Brownian and $V_{t}=\sqrt{\frac{1}{T} \int_0^T {v_{t}}^{2}\,dt}$ </p> <p>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.</p> <p>The next statement is the one that i did not fully undestand the way they authors did have it :</p> <p>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 $\rho (z,z_{1})=\frac{\sqrt{3}}{2}$ </p> <p>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.</p> <p>Does someone have an idea how to prove the law equality above?</p> <p>Many thanks</p> http://mathoverflow.net/questions/29934/approximation-of-the-law-of-a-stochastic-process/34491#34491 Answer by zhoraster for Approximation of the law of a stochastic process zhoraster 2010-08-04T09:58:09Z 2010-08-04T09:58:09Z <p>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. </p> <p>(In fact, by considering expectations of greater powers of $V_T$, one can see that $V_T$ cannot be log-Gaussian with any parameters.)</p>