Finding a stochastic differential equation as limit of a discrete stochastic process - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T03:27:16Z http://mathoverflow.net/feeds/question/84231 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/84231/finding-a-stochastic-differential-equation-as-limit-of-a-discrete-stochastic-proc Finding a stochastic differential equation as limit of a discrete stochastic process gm 2011-12-24T19:35:33Z 2012-08-02T18:22:00Z <p>I stumbled upon a problem that seems simple but I cannot tackle it. Let $X_n$ be a discrete process defined by the following algorithm.</p> <p>Choose $X_0\in[0,1]$, set $\kappa>0$ small enough and<br/> $X_{n+1}=X_n+\kappa(I_n-X_n)$<br/> with $I_n=1$ with probability $X_n$ and $I_n=0$ with probability $1-X_n$.</p> <p>In other words the $X_n$ decreases with probability $1-X_n$ by $\kappa X_n$ and increases with probability $X_n$ by $\kappa(I_n-X_n)$ so $E[X_{n+1}]=X_n$.</p> <p>The point is that $\kappa$ can be arbitrarily small so we can take its limit to $0$ while decreasing linearly the time step. This naturally should give an SDE (in this case I would expect it to be non-linear). So my question is how can one find this SDE or the PDE that gives the probability density.</p> <p>I should add that for short times it looks like a random walk (which is expected I guess) with the variance being proportional to $\kappa^2 t^2 X_0(1-X_0)$, with $t$ small. However since $X_n\in[0,1]$, $1$ is an upper bound for the variance.</p> http://mathoverflow.net/questions/84231/finding-a-stochastic-differential-equation-as-limit-of-a-discrete-stochastic-proc/102690#102690 Answer by Jason Swanson for Finding a stochastic differential equation as limit of a discrete stochastic process Jason Swanson 2012-07-19T17:46:13Z 2012-07-19T18:07:52Z <p>Here is a sketch of how you might approach this. Let $$\xi_j = \frac{\kappa(I_{j-1} - X_{j-1})} {\sqrt{X_{j-1}(1 - X_{j-1})}}.$$ The two "hard" results that must be proven are: (1) For each $t$, $$\lim_{\kappa\to0} E\big[\max\{|\xi_j|: 1 \le j \le \lfloor\kappa^{-2}t\rfloor\}\big] = 0,$$ and (2) for each $t$, $$\sum_{j=1}^{\lfloor\kappa^{-2}t\rfloor} \xi_j^2 \to t,$$ in probability as $\kappa\to0$. The rest of the proof would then be the following "soft" argument based on general theory.</p> <p>First, let $W^\kappa(t)=\sum_{j=1}^{\lfloor\kappa^{-2}t\rfloor} \xi_j$. Using the two results above, one can use the martingale central limit theorem (Theorem 7.1.4 in <a href="http://www.amazon.com/Markov-Processes-Characterization-Convergence-Probability/dp/047176986X" rel="nofollow">Ethier &amp; Kurtz</a>) to prove that $W^\kappa\Rightarrow W$, where $W$ is a standard Brownian motion.</p> <p>Next, we take the difference equation which defines the sequence ${X_n}$ and rewrite it as an integral equation. More specifically, if we define $X^\kappa(t)=X_{\lfloor\kappa^{-2}t\rfloor}$, then we may write $$X^\kappa(t) = X_0 + \int_0^t \sqrt{X^\kappa(s-)(1 - X^\kappa(s-))}\,dW^\kappa(s).$$ There is nothing deep here, just a change of notation, really.</p> <p>Finally, we use Theorem 5.4 in <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.aop/1176990334" rel="nofollow">Kurtz &amp; Protter</a> to prove that $(X_0,X^\kappa,W^\kappa) \Rightarrow(X_0,X,W)$, where $X$ is the unique strong solution to $dX=\sqrt{X(1-X)}\,dW$, $X(0)=X_0$.</p> <p>A watered-down version of Theorem 5.4 in Kurtz &amp; Protter is available as Theorem 2.3 in <a href="http://math.swansonsite.com/instructional/skor_lemmas.pdf" rel="nofollow">these lecture notes</a>. This version is sufficient for your purposes, and it may be easier to digest. Also, to use this theorem, you must show that, for every version of $(X_0,W)$, the limiting SDE has a unique strong solution for all time. This follows, for example, from Proposition 5.2.13, Theorem 5.5.4, and Corollary 5.3.23 in <a href="http://www.amazon.com/Brownian-Stochastic-Calculus-Graduate-Mathematics/dp/0387976558" rel="nofollow">Karatzas &amp; Shreve</a>.</p>