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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.

Choose $X_0\in[0,1]$, set $\kappa>0$ small enough and
$X_{n+1}=X_n+\kappa(I_n-X_n)$
with $I_n=1$ with probability $X_n$ and $I_n=0$ with probability $1-X_n$.

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 $E[X_{n+1}]=X_n$.

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.

I should add that for short times it looks like a random walk (which is expected I guess) with the variance being proportional to $\kappa t X_0(1-X_0)$, with $t$ small. However since $X_n\in[0,1]$, $1$ is an upper bound for the variance.

Edit: It is not $\kappa t X_0(1-X_0)$, it is $\kappa^2 t^2 X_0(1-X_0)$

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Of course the PDE that gives the probability density is the Fokker–Planck equation. –  tst Dec 24 '11 at 13:37
    
This question has also been asked at math.overflow (mathoverflow.net/questions/84231). I voted to close the question there, but I think it suits this site fine. –  George Lowther Dec 24 '11 at 20:29
    
I'll also add my comment which I posted on mathoverflow here. "I think this is quite standard stuff, discretely approximating an SDE by a binary tree. Taking a time step size of $\kappa^2$ will converge weakly to a solution to the SDE $dX=\sqrt{X(1−X)}\,dW$. So, I'm not really sure that this is research level, and is just a special case of the general method of discretely approximating SDEs. I think Roger's and Williams book (vol 2) should cover this, although maybe not in this precise form." –  George Lowther Dec 24 '11 at 20:31
    
...by which I mean, the weak compactness method for SDEs described in Rogers and Williams together with uniqueness for the SDE $dX=\sqrt{X(1-X)}\,dW$ should work here. –  George Lowther Dec 24 '11 at 20:34
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1 Answer 1

up vote 1 down vote accepted

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.

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 Ethier & Kurtz) to prove that $W^\kappa\Rightarrow W$, where $W$ is a standard Brownian motion.

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 $$ (1) \qquad X^\kappa(t) = X_0 + \int_0^t \sqrt{X^\kappa(s-)(1 - X^\kappa(s-))}\,dW^\kappa(s). $$ There is nothing deep here, just a change of notation, really.

Finally, we use Theorem 5.4 in Kurtz & Protter to prove that $(X_0,X^\kappa,W^\kappa) \Rightarrow(X_0,X,W)$, where $X$ is the unique strong solution to $dX=\sqrt{X(1-X)}\,dW$, $X(0)=X_0$.

A watered-down version of Theorem 5.4 in Kurtz & Protter is available as Theorem 2.3 in these lecture notes. 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 Karatzas & Shreve.

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Thank you very much for your feed back, it was extremely helpful. I surely need quite a bit of time to work this out. However I have a question on intuition. What is the intuition behind the definition of $\xi_J$? Is this something standard that I miss? –  tst Jul 26 '12 at 20:04
    
Once you identify the limiting SDE, then you know you want Equation (1) to hold. The definition of $\xi_j$ is chosen precisely so that Equation (1) works out. –  Jason Swanson Jul 27 '12 at 10:44
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