Does the variance of a continuous time, time homogeneous, Markov process starting from one point necessarily not decrease? 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 $t$?
 A: 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. 
A: 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 $X(s)$.
A: 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...
A: 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 $t$.
