Consider the one dimensional SDE
\[dX_t = f(X_t) dt + g(X_t) dW_t\]
which we assume has a unique global in time solution. For simplicity let us assume that there is a positive constant \(c\) so that \( 1/c < g(x)<c\) for all \(x\) and that \(f\) and \(g\) are smooth.
A stationary measure for the problem is a probability measure \(\mu\) so that if the initial distribution \(X_0\) is distributed according to \(\mu\) and independent of the Brownian Motion \(W\) then \(X_t\) will be distributed as \(\mu\) for any \(t \geq 0\).
If the functions \(f\) and \(g\) are “nice” then the distribution at time \(t\) has a density with respect to Lebesgue measure (“dx”). Which is to say there is a function \(p_x(t,y)\) do that for any \(\phi\)
\[\mathbf E_x \phi(X_t) = \int_{-\infty}^\infty p_x(t,y)\phi(y) dy\]
and \(p_\phi(t,y)\) solves the following equation
\[\frac{\partial p_\phi}{\partial t}(t,y) = (L^* p_\phi)(t,y)\]
with \( p_\phi(0,y) = \phi(y)\) where \(\phi(z)\) is the density with respect to Lebesgue of the initial density. (The pdf of \(X_0\) .)
\(L^*\) is the formal adjoint of the generator \(L\) of \(X_t\) and is defined by
\[(L^*\phi)(y) = – \frac{\partial\ }{\partial y}( f \phi)(y) + \frac12 \frac{\partial^2\ }{\partial y^2}( g^2 \phi)(y) \]
Since we want \(p_x(t,y)\) not to change when it is evolved forward with the above equation we want \( \frac{\partial p}{\partial t}=0\) or in other words
\[(L^* p_\phi)(t,y) =0\]
- Let \(F\) be such that \(-F’ = f/g^2\). Show that \[ \rho(y)=\frac{K}{g^2(y)}\exp\Big( – 2F(y) \Big)\] is an invariant density where \(K\) is a normalization constant which ensures that
\[\int \rho(y) dy =1\] - Find the stationary measure for each of the following SDEs:
\[dX_t = (X_t – X^3_t) dt + \sqrt{2} dW_t\]\[dX_t = – F'(X_t) dt + \sqrt{2} dW_t\] - Assuming that the formula derived above make sense more generally, compare the invariant measure of
\[ dX_t = -X_t + dW_t\]
and
\[ dX_t = -sign(X_t) dt + \frac{1}{\sqrt{|X_t|}} dW_t\] - Again, proceding fromally assuming everything is well defined and makes sense find the stationary density of \[dX_t = – 2\frac{sign(X_t)}{|X_t|} dt + \sqrt{2} dW_t\]
