MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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
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 $\kappa^2 t^2 X_0(1-X_0)$, with $t$ small. However since $X_n\in[0,1]$, $1$ is an upper bound for the variance.

share|cite|improve this question
Hi. 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 $dX = \sqrt{X(1-X)}\,dW$. – George Lowther Dec 24 '11 at 20:12
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 20:16
Also, the same question has been asked at math.stackexchange ( I'm not sure if this is the same user, but asking the same question at both sites is generally discouraged (unless some time has elapsed and it is decided that it would have been a better fit on the other site, which doesn't apply here). – George Lowther Dec 24 '11 at 20:25 I decided to vote to close the question on this site -- but, imo, it is a fine question for math.stackexchange. – George Lowther Dec 24 '11 at 20:26
as mentionned above, this is classical stuff. The usual reference for these kind of questions is 'Markov Processes: Characterization and Convergence' by Ethier and Kurtz. – Alekk Dec 24 '11 at 23:29
up vote 3 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 $$ X^\kappa(t) = X_0 + \int_0^t \sqrt{X^\kappa(s-)(1 - X^\kappa(s-))}\\,dW^\kappa(s). $$ 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.