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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 ?

Thanks.

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up vote 9 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.

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Thank you very much for a very clear explanation! –  Sasha Aug 15 '11 at 5:13
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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.

Regards

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