# 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

$(L^* p_\phi)(t,y) =0$

1. 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
$\int \rho(y) dy =1$
2. 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$
3. Assuming that the formula derived above make sense more generally, compare the invariant measure of
$dX_t = -X_t + dW_t$
and
$dX_t = -sign(X_t) dt + \frac{1}{\sqrt{|X_t|}} dW_t$
4. 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

1. $I(t,\omega)=\int_0^t \sigma(s,\omega) dB(s,\omega)$is a stochastic integral then $I^2(t)-\int_0^t \sigma^2(s)ds$ is a martingale.
2. 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
$X_t=B_t^2\qquad Y_t=\int_0^t B_s ds \qquad Z_t=\int_0^t B_s^2 ds$