Home » Stochastic Calculus » PDEs and SDEs

# Category Archives: PDEs and SDEs

## One dimensional stationary measure

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 \[dX_t = – 2\frac{sign(X_t)}{|X_t|} dt + \sqrt{2} dW_t\]

## Associated PDE

Show that if

- \[I(t,\omega)=\int_0^t \sigma(s,\omega) dB(s,\omega)\]is a stochastic integral then \[I^2(t)-\int_0^t \sigma^2(s)ds\] is a martingale.
- What equation must \(u(t,x)\) satisfy so that

\[ t \mapsto u(t,B(t))e^{\int_0^t V(B(s))ds} \]

is a martingale? Here \(V\) is a bounded function. Hint: Set \(Y(t)=\int_0^t V(B(s))ds\) and apply It\^0’s formula to \(Z(t,B(t),Y(t))=u(t,B(t))\exp(Y(t))\).

## Making the Cube of Brownian Motion a Martingale

Let \(B_t\) be a standard one dimensional Brownian

Motion. Find the function \(F:\mathbf{R}^5 \rightarrow \mathbf R\) so that

\begin{align*}

B_t^3 – F\Big(t,B_t,B_t^2,\int_0^t B_s ds, \int_0^t B_s^2 ds\Big)

\end{align*}

is a Martingale.

Hint: It might be useful to introduce the processes

\[X_t=B_t^2\qquad Y_t=\int_0^t B_s ds \qquad Z_t=\int_0^t B_s^2 ds\]