5

I am brushing up on my stochastic approximation. I am having a hard time with the following problem. I have the equation

$dX_t = \ln(1+ X_t^2)dt + X_tdB_t$

$X_0 = x$, with $x \in\mathbb R$

I know that this equation has a unique, strong solution. How can I prove this?

Kenta S
  • 18,181
Phil.D
  • 51
  • 3
    Sufficient conditions for existence and uniqueness are Lipschitz continuity of the drift and Brownian motion scaling. Have you seen a proof of this before? In this case, I would look at derivative of $\ln(1+X_t^2)$ since it is bounded on $\mathbb{R}$. Regardless, knowing that a strong solution exists and is unique (even though the proof is constructive using a technique like Picard–Lindelöf iteration) does not give you a nice form for the solution (which might be what the problem is asking for). – parsiad Jul 09 '16 at 13:31
  • I have not seen a proof like this before. Can you please show me how to implement this? – Phil.D Jul 10 '16 at 23:21
  • 1
    Page 12 in http://www.cmap.polytechnique.fr/~touzi/Fields-LN.pdf – parsiad Jul 11 '16 at 04:01
  • Thank you. The natural log is making my answer look really ugly. Should that be the case? – Phil.D Jul 11 '16 at 04:58
  • 1
    Which answer? Do you mean the derivative of $\ln(1+X_t^2)$? It's $2x/(x^2+1)$. The Lipschitz constant is bounded above since $\sup |2x/(x^2+1)|=1$. – parsiad Jul 11 '16 at 14:07
  • @par I do not understand why the Lipschitz constant is bounded above in the case of 2x/(x$^2$ + 1). Sorry to bother you about this, but would you mind showing me how you found that in context to this problem? Would really appreciate it :). – Phil.D Jul 11 '16 at 19:06
  • Let $f(x)=2x/(x^2+1)$. Note that $f(x)\rightarrow 0$ as $x\rightarrow \infty$, so that extrema of $f$ can only be attained at critical points of $f$. $f^\prime(x)=-2(x^2-1)/(x^2+1)^2$ has zeros at $x=\pm 1$. Note further that $|f(\pm 1)|=1$. A picture says a thousand words: http://www.wolframalpha.com/input/?i=2x%2F(x%5E2%2B1) – parsiad Jul 11 '16 at 21:15
  • 1
    @par Nice Socratic work here. (Re the supremum, note that $2|x|\leqslant x^2+1$ since $x^2+1-2|x|=(|x|-1)^2\geqslant0$, with the equality case coming freely.) – Did Jul 26 '16 at 11:36

0 Answers0