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Let $x_t$ be a zero mean, time homogeneous Markovian process (chiefly look at the case where the value is in $1$ dimension) over time $t$ starting from $x_0=0$. Is it necessary that, in continuous time setting, the variance of $x_t$ does not decrease over $t$?

What are the examples of $x_t$ where the variance at $t$ decreases over some interval of $t$? The following are my successful ( 1) ) and failed ( 2) ) attempts in constructing the examples.

1) In discrete time and discrete state, the followig is a very simple example where the variance periodically oscillates over time.

$$x_{t+1} = \eta(1-|x_t|),\, x_0=0;\, \eta\in\{-1,1\},\mbox{ with probability of } \frac{1}{2} \mbox{ on each value of }\eta.$$

2) In continuous time, but discontinuous path setting. I first looked at the following jump diffusion process.

$$dx_t = -\alpha x_t dt+dz_t+ y\eta dN_t,\, x_0 = 0,$$ where $\alpha\gg 0$, $z_t$ is the standard brownian motion with mean $0$ and standard deviation $t$, $N_t$ is the Poisson process with frequency $0<\lambda\ll 1$, $\eta$ takes on values $-1$ or $1$ with $0.5$ probability each, $z_{t_1}$, $N_{t_2}$ and $\eta$ are independent of each other at arbitrary $t_1$ and $t_2$, and constant $y\gg 1$.

It does not seem a correct example. One can solve this equation and one will find the variance of this process is the sum of the variance from $dz_t$ and that from $dN_t$ due their independence. We may have to make the jumps negatively correlated to $z_t$.

Another setup I thought of is to shift $x_t$ beyond a barrier directly back to the $x=0$ line. So the process resides on the topology of two cylinders touched along a longitude. However, it seems to me, even this set up with $x_t$ being either a standard Browniam motion or mean reverting one still has its variance increasing over time.

Therefore, I am still without a valid example in this setup.

3) What are the examples for continuous path? As Martin Hairer shows in his solution below, this can be achieved in higher-than-1 dimensional value space. This circumvents the difficulty of having to revisit the same point going away from $0$ and come closer to $0$. Now if we tackle the difficulty of revisiting the same value, and restrict the process to $1$ dimension. What is the answer? I suspect the satisfactory process does not exist. Can anyone prove this if the variance has to increase over $t$?

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Simple counterexample: $X_t = 0$ for all $t$. –  cardinal Dec 11 '13 at 2:38
@cardinal: I meant "increase" in the weak sense of "not decrease". That is a good point nevertheless. I now made it clear in the question. Take a look. –  Hansen Dec 11 '13 at 2:47
@StephanSturm: You are right. For martingale, it would be an application of Jensen's inequality to show the non-decreasing over time. However, mean-reverting process and jump process are not martingale. Any ideas on those? –  Hansen Dec 11 '13 at 5:37
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4 Answers

The following is an example: take a Markov chain with states $0,\pm 1,\pm 1/2,\pm 1/3,..$ with jump rates $\lambda_{0,1}=\lambda_{0,-1}=\lambda_{1/k,1/(k+1)}=\lambda_{-1/k,-1/(k+1)}=1$ and rates $0$ otherwise, and start it at state $0$. The initial variance is $0$, then increases, but as $t\to\infty$ the variance decreases.

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@ofer zeitouni: This Markov chain is in continuous time, and the jump rates are the entries of the transition matrix? –  Hansen Dec 11 '13 at 6:04
The rate matrix. The MC is in continuous time and the jump rates are as I wrote (so the rate matrix has row-sum 0). BTW, one can modify the chain to be irreducible if needed by replacing the two lines I have put at 0 by many loops at $0$ of longer and longer lengths so that at large time, whp you are in a very long loop of length $L_t$ with $L_t\to 0$ as $t\to\infty$ –  ofer zeitouni Dec 11 '13 at 6:13
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Here is a counterexample with continuous paths in dimension 2. Let $B$ be a standard Brownian motion and set $$ x_t = {B_t^3\over 1+B_t^4}\;,\quad y_t = {B_t\over 1+B_t^4}\;. $$ It clearly has mean zero. Furthermore, the pair $(x,y)$ weakly converges to $0$ as $t \to \infty$, but the variance is non-zero at any finite time. Finally, one can check that the map $t \mapsto (t^3/(1+t^4), t/(1+t^4))$ is injective, so the pair $(x_t,y_t)$ is Markov.

The example is slightly strange because there is no specification on what the process should do when started outside the range of the curve. Furthermore, it is clearly not Feller. However, both of these problems could be cured at the expense of not being as explicit...

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Tricky! Putting the value of the process in 2 dimension avoids having to revisit the same point before coming back to the origin. The injective map is nice, too, to keep the transformed process Markovian. It cleverly circumvents the difficult point. I would very much like to accept this marvelous solution, but since only one solution can be "accepted", I would want to wait for the one dimensional case (or cylindrical symmetric case for higher dimension) which tackles the difficulty of revisiting-same-points head on. Thank you, very much, Martin. –  Hansen Dec 11 '13 at 19:02
I did up-vote your solution though. –  Hansen Dec 11 '13 at 19:52
My hunch would be that if you insist on continuous paths, mean 0, and dimension 1, then the statement might be correct... –  Martin Hairer Dec 12 '13 at 14:38
Do you have any idea on how to proceed to prove it? Yuri Bakhtin had a special case of diffusion with symmetry proved as shown in his answer. –  Hansen Dec 14 '13 at 1:14
I think that a variant of Yuri's proof works in $1D$ for any strong Markov process with continuous sample paths that is symmetric under $x \mapsto -x$. Strong Markov (rather than just Markov) is essential since one can build a counterexample that first runs away from the origin on dyadics and then comes back on irrationals. Not sure whether the symmetry assumption is also needed... –  Martin Hairer Dec 14 '13 at 8:48
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The first part of this answer was written when I overlooked the zero mean condition that takes this question to another level of difficulty.

If you want an example where things are computable precisely try $dX=-Xdt+XdW(t)$ with initial condition, say, $X(0)=1$ (you can shift the entire example by 1 if you insist on $X(0)=0$). The solution of this equation is a specific case of geometric Brownian motion. See http://en.wikipedia.org/wiki/Geometric_Brownian_motion for some basic facts and variance in particular.

It's clear why it behaves this way with this choice of coefficients: all solutions approach 0 fast, so initial state $X(0)=1$ and terminal state $X(\infty)=0$ are deterministic, but there is some randomness in between. The result is that variance grows for a while and then starts decaying to 0.

Update upon seeing Martin's answer: In 2d you can have very simple examples since you can avoid the Markovian difficulties by letting every point to be passed no more than once. Take any curve $\gamma$ starting at $0$ and such that it does not pass any point twice and does not have common points with $-\gamma$ besides $0$. Then let your process go along $\gamma$ or $-\gamma$ with probabilities $1/2$ and $1/2$. By tweaking $\gamma$ you can have any given behavior of the variance.

Update 2: If we restrict ourselves to 1d symmetric diffusions, i.e., $dX=b(X)dt+\sigma(X)dW$ with $b(-x)=-b(x)$ and $\sigma(-x)=\sigma(x)$, then, I think, I have an answer for $b,\sigma\in C^1$ and $\sigma(x)>c$ for all $x$ and some $c>0$.

Under these assumptions $Y=X^2$ satisfies an SDE with coefficients allowing for strong solutions and for the comparison principle. To compare the distributions of $Y$ at times $t$ and $s$ with $s<t$, start two strong solutions $Y_0$ and $Y_1$ of the aforementioned SDE driven by the same Wiener process but with different initial conditions. We choose the initial condition for $Y_0$ to be nonrandom and concentrated at $0$, and we let the distribution of the initial condition of $Y_1$ to coincide with that of $Y(t-s)$. It is a.s.-positive and thus at time $0$ dominates the initial condition $0$ with probability $1$. By comparison principle, at all times, $Y_1\ge Y_0$ with probability $1$. At time $s$ though the distribution of $Y_1$ coincides with that of $Y(t)$ and the distribution of $Y_0$ coincides with that of $Y(s)$. Conclusion: $Y(t)$ stochastically dominates $Y(s)$ and, therefore, has greater expectation. So $X(t)$ has greater variance than $X(s)$.

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The first sentence asks for the process to have zero mean. –  Nate Eldredge Dec 11 '13 at 7:19
Hi Yuri, there is no way by a simple shift to bring it back to mean $0$, as the mean of your process is time dependent. But you can easily fix it by starting a process at $0$, making a jump to $\pm 1$, and then start your geometric BM. This is the continuous version of my example, and it is time inhomogeneous (since you will never revisit $0$). However, this still has discontinuous paths (at $x=0$). –  ofer zeitouni Dec 11 '13 at 7:25
Oh, I missed the mean zero requirement. Of course, Nate and Ofer, you are right. –  Yuri Bakhtin Dec 11 '13 at 7:30
@YuriBakhtin: I agree with ofer zeritouni that the fixing of $0$ starting point by adding a jump at $t=0$, would make the process time inhomogeneous. I thought of that jump in my attempt for case 2) but had to make the jump time homogenous which destroyed the variance decreasing property. Any other ideas? –  Hansen Dec 11 '13 at 19:08
@YuriBakhtin: I agree with your edit regarding the 2d construction inspired by Martin Hairer's answer. Please see my comment below his answer and my modification of question in case 3). I would like to tackle the difficulty of the process forced to revisit the same points. So I restrict the process to 1 dimension especially for case 3) of continuous path. Any ideas? –  Hansen Dec 11 '13 at 20:13
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Here is the long version of my comment.

First, let us show that if $X$ is any $1D$ strong Markov process with $X_0 = 0$ which is symmetric under sign inversion (i.e the Markov semigroup commutes with the composition operator with $x \mapsto -x$) and has continuous trajectories, then its variance cannot decrease. The proof very closely follows Yuri's argument. Let $\delta>0$ be fixed and construct a process $Y$ with the same law as $X$ as follows. $Y$ runs independently from $X$ until the stopping time $$ \tau = \inf\{t > 0\,:\, |Y_t| \ge |X_{t+\delta}|\}\;. $$ For $t > \tau$, we then set $Y_t = X_{t+\delta}$. Since trajectories are continuous, one has $|Y_\tau| = |X_{\tau + \delta}|$ and $|Y_t| \le |X_{t+\delta}|$ for every $t$. By the strong Markov property, $Y$ has the same law as $X$ and so we have shown that $|X_{t}|$ is stochastically dominated by $|X_{t+\delta}|$ for every $t, \delta > 0$, which is more than we need.

Another counterexample: The above construction relied on the strong Markov property to conclude that $Y$ and $X$ are equal in law. If one drops this, the conclusion fails in general even in the $1D$ symmetric case. Consider the following (slightly crazy) example. Let $\mu$ be the random measure on the dyadics in $[0,1]$ which is such that $\mu(\{k/2^n\})$ is exponentially distributed with mean $3^{-n}$ whenever $k$ contains no factor $2$. Furthermore, these exponential random variables are all independent. Let also $Z$ be the solution to the SDE $$ dZ = -10 Z\,dt + Z\,dW\;,\qquad Z_0 = 1\;. $$ (The only properties we need are that the law of $Z$ has a nice density w.r.t. Lebesgue, only charges positive numbers, has continuous trajectories, and converges to $0$ in mean square as $t \to \infty$.) Finally, take a random coin toss $\sigma \in \{\pm1\}$ independent of all of the above. We then set $X_t = \sigma \bar X_t$, where $\bar X_t$ is built as follows. Write $\tau = \mu([0,1])$ for the total mass of $\mu$. For $t \le \tau$, we then set $$ \bar X_t = \inf\{y\ge 0\,:\, \mu((0,y]) \ge t\}\;. $$ Loosely speaking, $\bar X$ runs through all dyadics in increasing order and, when it is located at $k 2^{-n}$ it jumps to "the next" dyadic with rate $3^{-n}$. For $t \ge \tau$, we then set $\bar X_t = Z_{t-\tau}$. Obviously, the variance of $X$ is not monotone. It is a nice exercise to show that $X$ has continuous trajectories (easy) and is Markov but not strong Markov (harder). Hint: Dyadics have measure $0$ under the law of $Z_t$ for any fixed $t$.

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Excuse my obtuseness, but could you please explicate "by the strong Markov property, Y has the same law as X"? –  Hansen Dec 15 '13 at 23:00
I have some confusions with your proof. First you state "Let $\delta>0$ be fixed and construct a process $Y_t$ with the same law as $X_{t+\delta}$ as follows.", then you claim "$Y$ has the same law as $X$", which I interpret as $Y(t)$ has the same law as $X(t)$ which I suppose has a different law from that of $X(t+\delta)$. I am also having difficulty understanding how the laws of $X(t)$ and $Y(t)$ (or Y(t+\delta) ) are the same given $Y(t)<X(t+\delta),\,\forall t<\tau$ and $Y(t)$ is independent from $X(t)$ for $t<\tau$. Could you please explain? –  Hansen Dec 16 '13 at 16:20
Sorry, I got myself confused with the indices... Now it should be correct. –  Martin Hairer Dec 17 '13 at 8:58
I still do not understand your proof. There seems to be something wrong, if I may say. Consider $X$ of the kind where $|X_t|>at$, where $a$ is a positive constant. Since $Y(t)$ is to be chosen arbitrarily initially so long as it is independent of $X(t)$, consider $Y(t)\equiv 0$. Then the stopping time $\tau=\infty$. By construction, $0=|Y(t)|<|X(t)|, \forall t>0$. The laws of $X$ and $Y$ are different. Could you explain the contradiction, or tell me what I have missed? –  Hansen Dec 18 '13 at 17:05
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