Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

Tanaka's SDE [see wikipedia article] $\text{d}X_t = \operatorname{sgn}( X_t ) \; \text{d}B_t$ with $X_0 = 0$, where $\operatorname{sgn}(x) = 1$ if $ x\ge 0$ and $\operatorname{sgn}(x) = -1$ if $x<0$, is known to not admit strong solutions.

I take this to mean that given a Brownian motion $\hat{B}_t$ upfront, it might be impossible to build $X_t$ adapted to the filtration generated by $\hat{B}_t$, which I take to mean that Euler-Maruyama would not converge to the solution of this SDE?

If I am correct, how does one simulate processes satisfying SDE that do not admit strong solutions?

But most likely I am wrong, and misunderstand the concept of strong/weak solutions. Does existence/non-existence of strong solutions bear any consequences for numerical simulation ?


share|cite|improve this question
up vote 8 down vote accepted

You are correct in saying that Tanaka's SDE $$ \begin{align} dX&={\rm sgn}(X)\,dB,\\ X_0&=0 \end{align} $$ does not admit strong solutions, so that $X$ cannot be constructed as a function of $B$. As mentioned by TheBridge, the Brownian motion $B$ can be written as $\int{\rm sgn}(X)\,dX$, which is a function of $\vert X\vert$ (it is the martingale part of $\vert X\vert$) and, consequently, ${\rm sgn}(X_t)$ is independent of $B$.

However, the Euler-Maruyama method still works here. You do not obtain approximate solutions converging to a function of $B$, but you do get convergence in distribution. It is not too hard to show how this happens. Suppose that we are constructing solutions over an interval $[0,T]$. Then, for a positive integer $N$ you would construct an approximation $X^{(N)}$, say, by setting $$ \begin{align} &X^{(N)}_0=0,\\ &X^{(N)}_\frac{k+1}N=X^{(N)}_{\frac kN}+{\rm sgn}\left(X^{(N)}_{\frac kN}\right)\,\left(B_\frac{k+1}N-B_\frac kN\right) \end{align} $$ for $k=0,1,\ldots,N-1$. This defines the discrete approximation at times $k/N$. As the term ${\rm sgn}(X^{(N)}_\frac kN)$ is independent of $B_\frac{k+1}N-B_\frac kN$, the process $X$ has normally distributed independent increments. Actually, I think it is a bit easier to extend the definition of $X^{(N)}$ to all intermediate times $\frac kN\le t\le\frac{k+1}N$ by $$ X^{(N)}_t=X^{(N)}_{\frac kN}+{\rm sgn}\left(X^{(N)}_{\frac kN}\right)\,\left(B_t-B_\frac kN\right), $$ in which case $X^{(N)}$ is a Brownian motion. This can be expressed conveniently as a stochastic integral $$ X^{(N)}_t=\int_0^t{\rm sgn}(X_{\lfloor sN\rfloor/N})\,dB_s $$ where $\lfloor\cdot\rfloor$ is the floor function. This does not converge, in probability, to a limit as $N\to\infty$. However, we do have $$ B_t=\int_0^t{\rm sgn}(X^{(N)}_{\lfloor sN\rfloor/N})\,dX^{(N)}_s. $$ So, if $\tilde X$ is any standard Brownian motion (defined on any probability space), we have equality of distributions $$ \left(B,X^{(N)}\right)\sim\left(\int{\rm sgn}(\tilde X_{\lfloor sN\rfloor/N})\,d\tilde X_s,\tilde X\right). $$ We do have convergence ${\rm sgn}(\tilde X_{\lfloor sN\rfloor/N})\to{\rm sgn}(\tilde X_s)$ outside of the (zero measure) set of times where $\tilde X_s=0$. Bounded convergence of stochastic integrals means that we get convergence in distribution $$ \left(B,X^{(N)}\right)\xrightarrow{\rm d}\left(\tilde B,\tilde X\right) $$ where $\tilde B=\int{\rm sgn}(\tilde X)\,d\tilde X$. Finally, $\tilde X$ does indeed satisfy Tanaka's SDE with respect to the driving Brownian motion $\tilde B$, but is not a function of $\tilde B$.

So, on an individual sample path for $B$, we do not have convergence of the paths of $X^{(N)}$ to a function of $B$, but do have convergence in distribution to some process which actually has additional randomness not contained in the paths of $B$. What is happening is that, when $X^{(N)}$ becomes small, ${\rm sgn}(X^{(N)})$ takes the values $1$ and $-1$ with roughly equal probability. Only a small change in $B$ (or small change in the time when $B$ is sampled) will reverse the sign of ${\rm sgn}(X^{(N)}_s)$. So, as $N$ becomes large, ${\rm sgn}(X^{(N)}_s)$ becomes independent of $B$ in the limit. You do get pathwise convergence of $\vert X^{(N)}\vert$. In fact $$ \vert X^{(N)}_t\vert\to B_t-\min_{s\le t}B_s $$ but ${\rm sgn}(X^{(N)})$ moves around at random, and only converges in distribution.

share|cite|improve this answer
Thank you very much for a very clear explanation! – Sasha Aug 15 '11 at 5:13

Here is a proof, supposing $X_t$ admit a strong solution for a given brownian motion $B_t$. Then we have : $ dB_t=sgn(X_t)^2dB_t=sgn(X_t)dX_t$ but then by Tanaka's formula we have : $|X_t|=\int_0^tsgn(X_s)dX_s+2L^X_t$.

So $B_t=|X_t|-2L^X_t$ is $|X_t|$-adapted (local time of $X_t$ at 0 is $|X_t|$ adapted). On the other hand $X_t$ is $B_t$-adapted by hypothesys which exhibits the desired contradiction.


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.