Invariant measures for stochastic processes

Question

I have some doubts about the concept of invariant measure for a stochastic process. Let me introduce a definition.

Given $(\Omega, \mathcal{E}, \mathbb{P})$ a measure space, $H$ Hilbert space, let be $Y(t)$ the unique $H-$valued stochastic process which solves the following stochastic differential equation \begin{equation} \left\{ \begin{aligned} &dY(t) = b \bigl( Y(t) \bigr) dt + \sigma dW(t) \\ &dY(0) = y \end{aligned} \right. \end{equation} where $b \colon H \to H$ is Lipschitz continuous, $\sigma$ is a symmetric non-negative linear operator $H \to H$, $W(t)$ is a standard Brownian motion in $(\Omega, \mathcal{E}, \mathbb{P})$. We say that a measure $\mu \colon \mathcal{E} \to \mathbb{R}$ is invariant for the process $Y(t)$ if

\begin{equation} \int_H \mathbb{E} \left[ \varphi \bigl( Y(t,y) \bigr) \right] d \mu(y) = \int_H \varphi(x) d\mu(x) \qquad \forall \varphi \in B_b(H), \ \forall t \geq 0 \end{equation}

where $B_b(H)$ denotes the Banach space of all uniformly continuous and Borel bounded functions $\varphi \colon H \to \mathbb{R}$ with the usual norm.

Equivalently, $\mu$ is invariant if and only if \begin{equation} \mu (B) = \int_H \pi_t (y,B) \mu(dy) \qquad \forall B \in \mathcal{B}(H) \end{equation} where $\pi_t(y,\cdot)$ denotes the law of the random variable $Y(t,y)$, and where $\mathcal{B}(H)$ denotes the Borelian set of $H$.

Now, two questions. What is the intuitive meaning of such a measure $\mu$? Moreover, how to prove that $\mu$ has to satisfies the stationary Fokker-Planck equation?

Thanks in advance.

Answer 1

If you're just trying to understand this concept lets just focus on the case $H = \mathbb{R}$. Let's further assume that the test functions are $\phi \in C_0^2$

$\int_{\mathbb{R}} \mathbb{E}[\phi(Y(t,y))]d\mu(y) = \int_{\mathbb{R}} \phi(x) \mu(x)dx$

$\int_{\mathbb{R}} \left(\mathbb{E}[\phi(Y(t,y))]-\phi(y)\right)d\mu(y) = 0$

$\lim_{t \searrow 0} \int_{\mathbb{R}} \left(\frac{\mathbb{E}[\phi(Y(t,y))]-\phi(y)}{t} \right)d\mu(y) = 0$

We can commute the limit through the integral for $\phi \in C_0^2$ by dominated convergence theorem. Then inside the integral we get exactly the formula for the generator of ito process. You can figure out using Ito's Lemma that the generator $A$ is given by

$A\phi(x) = b(x) \phi'(x) + \frac{1}{2} \sigma^2 \phi''(x)$

So then you get

$ \int_{\mathbb{R}} A \phi(y) \mu(y) dy = 0$

Now if we let $A^*$ be the adjoint of $A$ we get

$\int_{\mathbb{R}} \phi(y) A^* \mu(y) dy = 0$

Since this is true for all $\phi(y)$ we have in fact $A^* \mu(y) = 0$

Which is exactly the stationary Fokker-Planck equation.

I think as far as intuition its best to look at the concept of a stationary distribution for Markov chains and then try to think by analogy. https://en.wikipedia.org/wiki/Markov_chain#Stationary_distribution_relation_to_eigenvectors_and_simplexes

Answer 2

About the meaning of a stationary measure $\mu$:

The solution $Y_t$ of your SDE depends on the start distribution $Y_0$. If $Y_0$ is distributed according to $\mu$, i.e. $\mathbb{P}[Y_0 \in A] = \mu(A)$ for all $A\in\mathcal{E}$, then the according solution $Y_t$ is always distributed according to $\mu$. This follows from:

$\mathbb{P}[Y_t \in A] =\mathbb{P}[Y_0 \in H , Y_t \in A] = \int_H \pi_t(y,A) \mu(dy) \overset{(*)}{=} \int_A \mu(dx) = \mu(A),$

where we have used the property of the stationary measure $\mu$ in $(*)$.