# Ornstein-Uhlenbeck type process with thresholding

(Edited) I met a univariate Ornstein-Uhlenbeck type process but with self soft-thresholding: $$dX(t) = - c\ \mbox{sgn}(X(t))\big[|X(t)|-c_1 t^{\mu}\big]_+ dt + \sigma dB(t), \quad X(0)=0,$$ where $B(t)$ is the standard Brownian motion, and $c,c_1,\sigma>0$ are constants, and $[\cdot]_+$ is a soft-thresholding operator with $[a]_+=a$ if $a> 0$ and $[a]_+=0$ if $a\leq 0$.

My Question: does the stationary distribution exist for the SDE above, for different $\mu\in[0,1]$? I suspect that there is a regime switching phenomenon at around $\mu=1/2$...

Any comment will be much appreciated!

• I think $X(t)$ will be oscillating between $(k+o(1))\sqrt t$ and $(-k+o(1))\sqrt t$ as $t\to\infty$, for some real $k>0$. So, it seems that no stationary distribution exists. Also, in view of the law of the iterated logarithm for the Brownian motion, with probability $1$ the paths of $X$ will be substantially different from those of the Brownian motion. Also, there seems to be no reason to believe that there exists a closed-form solution to this SDE. – Iosif Pinelis May 20 '18 at 21:32
• Thank you. Suppose I consider the following process instead:$$dX(t) = - c\ \mbox{sgn}(X(t))\big[|X(t)|-c_1 t^{\gamma}\big]_+ dt + \sigma dB(t), \quad X(0)=0,$$ where $c_1>0$, I wonder if anything can be said about the behavior of $X(t)$ with respect to different value of $0 < \gamma\leq 1$? – Nick May 21 '18 at 1:49
• I think, again in view of the law of the iterated logarithm, it will depend on whether $\gamma\le1/2$. If so, I think $X(t)$ will be oscillating between $(k+o(1))t^\gamma$ and $(−k+o(1))t^\gamma$ as $t\to\infty$. If $\gamma>1/2$, then I think $X$ will be similar to $\sigma B$. – Iosif Pinelis May 23 '18 at 20:44
• Thank you! I ran some simulations. For $\gamma>1/2$, the behavior of $X$ does look like $\sigma B$. For $\gamma<1/2$, $X$ explodes. However, rather than oscillating between positive and negative values, $X$ explodes toward one of the directions. Some simulation paths explode toward $+\infty$, some toward $-\infty$, but they do not oscillate. – Nick May 24 '18 at 3:11
• I think you should recheck your simulation at least for $\gamma\le1/2$. If $X$ takes a very large positive (say) value, then there will be a strong negative trend. – Iosif Pinelis May 24 '18 at 11:05