Invariant measures

[math] \newcommand{\R}{\mathbb{R}} \newcommand{\A}{\mathcal{A}} \newcommand{\B}{\mathcal{B}} \newcommand{\N}{\mathbb{N}} \newcommand{\C}{\mathbb{C}} \newcommand{\Rbar}{\overline{\mathbb{R}}} \newcommand{\Bbar}{\overline{\mathcal{B}}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\E}{\mathbb{E}} \newcommand{\p}{\mathbb{P}} \newcommand{\one}{\mathds{1}} \newcommand{\0}{\mathcal{O}} \newcommand{\mat}{\textnormal{Mat}} \newcommand{\sign}{\textnormal{sign}} \newcommand{\CP}{\mathcal{P}} \newcommand{\CT}{\mathcal{T}} \newcommand{\CY}{\mathcal{Y}} \newcommand{\F}{\mathcal{F}} \newcommand{\mathds}{\mathbb}[/math]

Definition (Invariant measure)

Let [math]\mu[/math] be a positive measure on [math]E[/math] such that [math]\mu(x) \lt \infty[/math] for all [math]x\in E[/math] and [math]\mu\not\equiv 0[/math]. We say that [math]\mu[/math] is invariant for the transition matrix [math]Q[/math] if for all [math]y\in E[/math] we get

[[math]] \mu(y)=\sum_{x\in E}\mu(x)Q(x,y). [[/math]]

With matrix notation, this means

[[math]] \mu Q=\mu. [[/math]]

Since for all [math]n\in\N[/math], [math]Q_n=Q^n[/math], we have [math]\mu Q_n\mu=\mu[/math].

Interpretation

Assume that [math]\mu(E) \lt \infty[/math], which is always the case when [math]E[/math] is finite. We can assume that [math]\mu(E)=1[/math]. Then for all measurable maps [math]f:E\to\R_+[/math], we get that

[[math]] \E_\mu[f(X_1)]=\sum_{x\in E}\mu(x)\sum_{y\in E}Q(x,y)f(y)=\sum_{y\in E}f(y)\sum_{x\in E}\mu(x)Q(x,y)=\sum_{y\in E}\mu(y)f(y). [[/math]]

Hence under [math]\p_\mu[/math], we get that [math]X_1[/math] [math]{law}\atop{=}[/math] [math]X_0\sim \mu[/math]. Using the fact that [math]\mu Q_n=\mu[/math], we show that under [math]\p_\mu[/math], we get that [math]X_n\sim \mu[/math]. For all measurable maps [math]F:\Omega\to\R_+[/math] we have

[[math]] \E_\mu[F\circ \Theta_\one]=\E_\mu[\E_{X_1}[F]]=\sum_{x\in E}\mu(x)\E_x[F]=\E_\mu[F], [[/math]]

which implies that under [math]\p_\mu[/math], we get that [math](X_{1+n})_{n\geq 0}[/math] has the same law as [math](X_n)_{n\geq 0}[/math].

Example

For the r.v. on [math]\mathbb{Z}^d[/math], we get [math]Q(x,y)=(y,x)[/math]. One then immediately checks that the counting measure on [math]\mathbb{Z}^d[/math] is invariant.

Definition (Reversible measure)

Let [math]\mu[/math] be positive measure on [math]E[/math] such that [math]\mu(x) \lt \infty[/math] for all [math]x\in E[/math]. [math]\mu[/math] is said to be reversible if for all [math]x,y\in E[/math] we get that

[[math]] \mu(x)Q(x,y)=\mu(y)Q(y,x). [[/math]]

Proposition

A reversible measure is invariant.

Show Proof

If [math]\mu[/math] is reversible, we get that

[[math]] \sum_{x\in E}\mu(x)Q(x,y)=\sum_{x\in E}\mu(y)Q(y,x)=\mu(y). [[/math]]

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There exists invariant measures which are not reversible, for example the counting measure is not reversible if the [math]\mathcal{G}[/math] is not symmetric, i.e. [math]\mathcal{G}(x)=\mathcal{G}(-x)[/math].

Example

[Random walk on a graph]

The measure [math]\mu(x)=\vert A_x\vert[/math] is reversible. Indeed, if [math]\{x,y\}\in A[/math], we get

[[math]] \mu(x)Q(x,y)=\vert A_x\vert\frac{1}{\vert A_x\vert}=1=\mu(y)Q(y,x). [[/math]]

Example

[Ehrenfest's model] This is the Markov chain on [math]\{1,...,k\}[/math] with transition matrix

[[math]] \begin{cases}Q(j,j+1)=\frac{k-j}{k},& 0\leq j\leq k-1\\ Q(j,j-1)=\frac{j}{k},&1\leq j\leq k\end{cases} [[/math]]

A measure [math]\mu[/math] is reversible if and only if for [math]0\leq j\leq k-1[/math] we have

[[math]] \mu(j)\frac{k-j}{k}=\mu(j+1)\frac{j+1}{k}. [[/math]]

One can check that [math]\mu(j)=\binom{k}{j}[/math] is a solution.

Theorem

Let [math]x\in E[/math] be recurrent. The formula

[[math]] \mu(y)=\E_x\left[\sum_{k=0}^{H_x-1}\one_{\{ X_k=y\}}\right] [[/math]]

defines an invariant measure. Moreover, [math]\mu(y) \gt 0[/math] if and only if [math]y[/math] is in the same recurrence class as [math]x[/math].

Show Proof

Let us first note that if [math]y[/math] is not in the same recurrence class as [math]x[/math], then

[[math]] \E_x[N_y]=U(x,y)=0, [[/math]]

which implies that [math]\mu(y)=0[/math]. For [math]y\in E[/math], we get that

[[math]] \begin{multline*} \mu(y)=\E_x\left[\sum_{k=1}^{H_x}\one_{\{ X_k=y\}}\right]=\sum_{z\in E}\E_x\left[\sum_{k=1}^{H_x}\one_{\{ X_{k-1}=z,X_k=y\}}\right]=\sum_{z\in E}\sum_{k\geq 1}\E_x\left[\one_{\{k\leq H_x,X_{k-1}=z\}}\one_{\{ X_k=y\}}\right]\\=\sum_{z\in E}\sum_{k\geq 1}\E_x\left[\one_{\{k\leq H_x,X_{k-1}=z\}}\right]=\sum_{z\in E}\E_x\left[\sum_{k=1}^{H_x}\one_{X_{k-1}=z}\right]Q(z,y)=\sum_{z\in E}\mu(z)Q(z,y). \end{multline*} [[/math]]

We showed that [math]\mu Q=\mu[/math]. Thus, it follows that [math]\mu Q_n=\mu[/math] for all [math]n\geq 0[/math]. In particular, we get

[[math]] \mu(x)=1=\sum_{z\in E}\mu(z)Q_n(z,x). [[/math]]

Let [math]y[/math] be in the same recurrence class as [math]x[/math]. Then there exists some [math]n\geq 0[/math] such that [math]Q_n(y,x) \gt 0[/math], which implies that

[[math]] \mu(y) \lt \infty. [[/math]]

We can also find [math]m\geq 0[/math] such that [math]Q_m(x,y) \gt 0[/math] and

[[math]] \mu(y)=\sum_{z\in E}\mu(z)Q_m(z,y)\geq Q_m(x,x) \gt 0. [[/math]]

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If there exists several recurrence classes [math]R_i[/math] with [math]i\in I[/math] for some index set [math]I[/math] and if we set

[[math]] \mu_i(y)=\E_{x_i}\left[\sum_{k=0}^{H_{x_i}-1}\one_{\{ X_k=y\}}\right], [[/math]]

we obtain a invariant measure with disjoint supports.

Theorem

Let us assume that the Markov chain is irreducible and recurrent. Then the invariant measure is unique up to a multiplicative constant.

Show Proof

Let [math]\mu[/math] be an invariant measure. We can show by induction that for [math]p\geq 0[/math] and for all [math]x,y\in E[/math], we get

[[math]] \mu(y)\geq \mu(x)\E_x\left[\sum_{k=0}^{p\land (H_x-1)}\one_{\{ X_k=y\}}\right].(\star) [[/math]]

First, if [math]x=y[/math], this is obvious. Let us thus suppose that [math]x\not=y[/math]. If [math]p=0[/math], the inequality is immediate. Let us assume [math](\star)[/math] holds for [math]p[/math]. Then

[[math]] \begin{align*} \mu(y)&=\sum_{z\in E}\mu(z)Q(z,y)\geq \mu(x)\sum_{z\in E}\E_x\left[\sum_{k=0}^{p\land (H_x-1)}\one_{\{ X_k=z\}}\right]Q(z,y)\\ &=\mu(x)\sum_{z\in E}\sum_{k=0}^p\E_x[\one_{\{ X_k=z,k\leq H_x-1\}}]Q(z,y)\\ &=\mu(x)\sum_{z\in E}\sum_{k=0}^p \E_x[\one_{\{ X_k=z,k\leq H_x-1\}}]\one_{\{ X_{k-1}=y\}}\\ &=\mu(x)\E_x\left[\sum_{k=0}^{p\land (H_x-1)}\one_{\{ X_{k-1}=y\}}\right]=\mu(x)\E_x\left[\sum_{k=1}^{(p+1)\land (H_x)}\one_{\{ X_k=y\}}\right]. \end{align*} [[/math]]

This establishes the result for [math]p+1[/math]. Now, if we let [math]p\to\infty[/math] in [math](\star)[/math], we get

[[math]] \mu(y)\geq \mu(x)\E_x\left[\sum_{k=0}^{H_x-1}\one_{\{X_k=y\}}\right]. [[/math]]

Let us fix [math]x\in E[/math]. The measure [math]\nu_x(y)=\E_x\left[\sum_{k=0}^{H_x-1}\one_{\{ X_k=y\}}\right][/math] is invariant and we have [math]\mu(y)\geq \mu(x)\nu_x(y)[/math]. Hence for all [math]n\geq 1[/math] we get

[[math]] \mu(x)=\sum_{z\in E}\mu(z)Q_n(z,x)\geq \sum_{x\in E}\mu(x)\nu_x(z)Q_n(z,x)=\mu(x)\nu_x(x)=\mu(x). [[/math]]

Therefore we get that [math]\mu(z)=\mu(x)=\nu_x(z)[/math] for all [math]z[/math] such that [math]Q_n(z,x) \gt 0[/math]. Since the chain is irreducible, there exists some [math]n\geq 0[/math] such that [math]Q_n(z,x) \gt 0[/math]. This implies finally that

[[math]] \mu(z)=\mu(x)\nu_x(z). [[/math]]

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Corollary

Let us assume the chain is irreducible and recurrent. Then one of the following hold.

• There exists an invariant probability measure [math]\mu[/math] and for [math]x\in E[/math] we have
  [[math]] \E_x[H_x]=\frac{1}{\mu(x)}. [[/math]]
• All invariant measures have infinite total mass and for [math]x\in E[/math] we get
  [[math]] \E_x[H_x]=\infty. [[/math]]

In the first case, the basis is said to be positive recurrent and in the second case it is said to be negative recurrent.

If [math]E[/math] is finite, then only the first case can occur.

Show Proof

[Proof of Corollary] We know that in this situation all invariant measures are proportional. Hence they all have finite mass or infinite mass. For case [math](i)[/math], let [math]\mu[/math] be the invariant probability measure and let [math]x\in E[/math]. Moreover, let

[[math]] \nu_x(y)=\E_x\left[\sum_{k=0}^{H_x-1}\one_{\{ X_k=y\}}\right]. [[/math]]

Then for some [math]C \gt 0[/math] we get [math]\mu=C\nu[/math]. We can determine [math]C[/math] by

[[math]] 1=\mu(E)=C\nu_x(E), [[/math]]

which implies that [math]C=\frac{1}{\nu_x(E)}[/math] and thus

[[math]] \mu(x)=\frac{\nu_x(x)}{\nu_x(E)}=\frac{1}{\nu_x(E)}. [[/math]]

But on the other hand we have

[[math]] \nu_x(E)=\sum_{y\in E}\E_x\left[\sum_{k=0}^{H_x-1}\one_{\{ X_k=y\}}\right]=\E_x\left[\sum_{k=0}^{H_x-1}\left(\sum_{y\in E}\one_{\{ X_k=y\}}\right)\right]=\E_x[H_x]. [[/math]]

In case [math](ii)[/math], [math]\nu_x[/math] is infinite and thus

[[math]] \E_x[H_x]=\nu_x(E)=\infty. [[/math]]

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General references

Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].