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Consider the following SDE with $X_0 = 1$, $$ dX_t = X_t\operatorname{sign}(X_t) \, dt + X_t \, dW_t, $$ where $\operatorname{sign}(x) = \mathbb{1}_{\{x \ge 0\}} -\mathbb{1}_{\{x < 0\}}$. How am I supposed to solve this SDE?

Van Tom
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  • What methods do you know for proving the existence/non-existence of a strong solution? You must have seen some previous examples, easier ones. Also, which book are you consulting for this material? (Just want to see the notation, background etc.) – Sarvesh Ravichandran Iyer Feb 08 '21 at 03:58
  • @TeresaLisbon Well, I can show that this SDE satisfies the Lipschitz condition and linear growth condition, therefore, there exists a unique strong solution. But then how do I find it? – Van Tom Feb 08 '21 at 06:04
  • Using some heuristics, I think $X_t = e^{W_t}$ works. This seems to satisfy the equation. Note that $sign(X_t) = 1$ a.s. . I think any non-negative solution of the equation $dX_t = X_t dt +X_t dW_t$ will solve your equation because the sign term might as well not be there. If you can verify this, I will put it as an answer. – Sarvesh Ravichandran Iyer Feb 08 '21 at 06:31
  • Normally, the definition of the sign function must be $\operatorname{sign}(x) = \mathbb{1}{{x > 0}} -\mathbb{1}{{x < 0}}$. Are you sure we have the sign function in the equation? – NN2 Feb 08 '21 at 08:42
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    @TeresaLisbon I think $X_t = e^{t/2+W_t}$ satisfies the equation. But I derived this $X_t$ by simply ignoring the sign function. So the problem is how do I prove that $X_t$ will never hit zero? – Van Tom Feb 08 '21 at 09:29
  • @NN2 Yes, you are correct. I put the sign function wrongly. I'll edit it. Thanks! – Van Tom Feb 08 '21 at 09:31
  • @VanTom Of course $X_t$ will never hit zero , because it's the exponential of something real, so always stays positive. That's why I felt that the sign was a barrier to the entire equation. – Sarvesh Ravichandran Iyer Feb 08 '21 at 09:39
  • Also I'd made an error in the earlier comment : I was meaning to say $X_t = e^{W_t + \frac t2}$ but forgot to add that part in. Having said that, the sign argument still holds. – Sarvesh Ravichandran Iyer Feb 08 '21 at 09:59
  • @TeresaLisbon Yeah, I got that. If you put it as an answer, I'll accept it. – Van Tom Feb 08 '21 at 10:03
  • @VanTom Did it. You seem to have just gotten started in stochastic calculus. I will follow some of your questions, since I may be able to help you out somewhat in this respect. I will not answer all of them, but I will visit them for sure. Thanks for your time. – Sarvesh Ravichandran Iyer Feb 08 '21 at 10:14
  • @TeresaLisbon That is awesome, and yes, I am a newbie in stochastic calculus. Thank you so much! – Van Tom Feb 08 '21 at 10:52

1 Answers1

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We assume that $X_t = f(t,W_t)$ is an Ito process, so that we can assume the Ito formula to find a candidate for the solution of the equation, by matching the $dt$ and $dW_t$ drifts. Write $f \equiv f(t,x)$. We proceed to do some rough work in the next part.

Note that by the Ito formula, we have : $$ dX_t = \left(\frac{\partial f}{\partial t} + \frac{1}{2}\frac{\partial f}{\partial x^2}\right) dt + \frac{\partial f}{\partial x} dW_t $$

Therefore, we must find $f$, such that $\frac{\partial f}{\partial x} = f$ and $\frac{\partial f}{\partial t} + \frac{1}{2}\frac{\partial f}{\partial x^2} = f \mbox{sign}\{f\}$, along with $f(1,0) = 1$. Indeed, from the first condition we obtain that $f(t,x) = g(t)e^{x}$ for some $t$, then by the second condition we obtain that $g(t) = e^{\frac t2}$ fits, therefore we get $X_t = e^{\frac t2 + W_t}$ to fit in the boundary condition.

Note that $\mbox{sign}(X_t) = 1$ a.s. since $X_t =e^{R_t}$ for a real-valued random variable. Hence that particular term never contributes to the equation.

Of course, this is rough work since we did not rigorously solve the ODEs above, for example. But then we have a guess : one can easily verify now that the $X_t$ obtained from the rough work is indeed a strong solution using the Ito formula.

This solution is unique, since the coefficients satisfy the Lipschitz and linear growth conditions.

Sarvesh Ravichandran Iyer
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  • Since you mentioned that $X_t=e^{t/2+W_t}$ is the unique strong solution. How can I find a weak solution to this SDE, i.e. the tuple $(X, B)$, where $B$ is certain Brownian Motion? – Van Tom Feb 08 '21 at 10:57
  • There's a result of Yamada and Watanabe in $1971$ that guarantees weak uniqueness in the presence of strong uniqueness. So any weak solution is also in law equal to $X_t$. This is then the only weak solution. See section 5.3.D of Karatzas and Shreve, "Brownian Motion and Stochastic Calculus". I thought about the Tanaka equation for some time, but the law doesn't change. – Sarvesh Ravichandran Iyer Feb 08 '21 at 11:41
  • @TeresaLisbon: How do you prove that $ sign(X_t) = 1$ a.s? It seems to me that you supposed that you could ignore $sign(X_t)$ in order to get $X_t = e^{R_t}$ and after that you said because $X_t = e^{R_t}$ then $sign(X_t) = 1$ a.s. But what if because of the sign function, then $X_t \ne e^{R_t}$ from the beginning? – NN2 Feb 08 '21 at 22:45
  • @NN2 From the rigorous proof, $X_t = e^{\frac t2 + W_t}$ is a non-negative process which satisfies the given differential equation, as you can check from Ito's rule. There's no other process thanks to strong uniqueness. Now, it so happens that this process is non-negative. You are right , it could have been that $X_t$ changed sign and therefore the sign term could come into play. Somehow , that was not the case here. – Sarvesh Ravichandran Iyer Feb 09 '21 at 03:05
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    @TeresaLisbon: Ok, I see. Thank you. – NN2 Feb 09 '21 at 11:28
  • @NN2 I spent some time deliberating over the same issue as you, so I can understand that you might have been a little confused. I attempted to clarify this in the answer ,maybe I did not do enough. I will edit your thought in, thank you very much. – Sarvesh Ravichandran Iyer Feb 09 '21 at 15:08