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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 second moment. So $X(t)$ has greater variance than $X(s)$.

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 $X(s)$.

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 variance.

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 second moment. So $X(t)$ has greater variance than $X(s)$.

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.

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 variance.