the minimization of a functional from stochastic differential equations - MathOverflow most recent 30 from http://mathoverflow.net2013-06-18T07:26:18Zhttp://mathoverflow.net/feeds/question/61472http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/61472/the-minimization-of-a-functional-from-stochastic-differential-equationsthe minimization of a functional from stochastic differential equationsJack2011-04-12T21:21:14Z2011-04-12T21:21:14Z
<p>Consider the following 1-dimensional Stochastic differential equation:</p>
<p>$dX_t=b(X_t)dt+\sigma(X_t)dB_t$</p>
<p>With some assumptions, one can have the stationary probability density $p(x)$:</p>
<p>$p(x)=\frac{C}{\sigma^2(x)}e^{\int_{-\infty}^{x}\frac{2b(y)}{\sigma^2(y)}dy}$,</p>
<p>where $C$ is the normalization constant, which can be done from the <a href="http://en.wikipedia.org/wiki/Fokker%E2%80%93Planck_equation" rel="nofollow">Fokker-Planck</a> equation. In the simple case where $\sigma(x)=1$ and $b(x)=-ax$ where $a>0$ is a constant, </p>
<p>$dX_t=-aX_tdt+dB_t$</p>
<p>one can find $a$ such that it minimizes the following functional</p>
<p>$F(f;q)=\int_{\bf R}[\sqrt{p(x)}-\sqrt{q(x)}]^2dx$</p>
<p>where $f(x)=-ax$, $p(x)$ is as the probability density defined above, and $q(x)=\frac{1}{\pi}e^{-x^2}$. Without need of calculus of variation, one can simply put $p(x)=q(x)$.</p>
<p><strong>Here is my question:</strong></p>
<blockquote>
<p><em>In the more general cases, say, $b(x)\leq 0$ and $\sigma(x)=1$, can one still be able to find a function $b(x)$ to minimize the functional $F(b;q)$?</em> Here $q(x)=\frac{1}{\pi(1+x^2)}$</p>
</blockquote>
<p><strong>Here is what I thought:</strong></p>
<p>One can still "force" $p(x)=q(x)$ to find $b(x)$. However, then one may get $b(x)=\frac{-x}{1+x^2}$, which may cause that $p(x)$ does not exist. Hence calculus of variations may be needed here. But I'm not sure if it will work or not since I am not able to calculate $\frac{d}{d\epsilon}|_{\epsilon=0}F(b+\epsilon h)$ here because of the ugly form of $p(x)$. Is any other technique needed?</p>