Sufficient condition for a measure to be invariant. 1. Given two stochastic processes on a probability ...

Feb 21, 2018 · An invariant measure (or maybe stationary measure) is sometimes a vector $\pi$ that satisfies $\pi P = \pi$, but not necessarily $\sum_i \pi_i = 1$. (This makes a difference for infinite Markov chains, where we can't necessarily divide by $\sum_i \pi_i$ to normalize.)

When dealing with manifolds, volume forms are often conflated with measures. Here, the volume form on the upper half plane is $|y|^{-2}dxdy$, so the measure of a Borel set is its integral with respect to that volume form.

Let $\lambda$ be a translation-invariant measure on the Borel sets that puts positive and finite measure on the right-open unit interval $[0,1)$ then $\lambda$ is a positive multiple of Lebesgue measure. Here is an outline of the proof: Every Borel measure is …

Oct 21, 2019 · Since invariant measures have many properties that hold almost everywhere (such as Poincaré recurrence, ergodicity and many more), it is extremely useful to find an absolutely continuous invariant measure. Let me explain this in more details using a fundamental example.

Dec 20, 2018 · I am looking for the proof of the theorem in Markov chain theory which roughly states that a recurrent Markov chain admit an essentially unique invariant measure (See the theorem at the end for the precise statement). The theorem is stated here (Theroem 10) but no proof or a reference is provided. Definitions Transition Probability Kernel

Jul 23, 2015 · $\begingroup$ Do you need to actually find the measure (if yes, analytically or numerically?) or to prove its existence? $\endgroup$ – demitau Commented Jul 23, 2015 at 8:40

Sep 7, 2015 · $\begingroup$ By Perron-Frobenius, the largest eigenvalue is $1$ and it has only one eigenvector, namely the invariant measure. So in the diagonalizable case your result follows pretty much immediately ($\rho$ is the next largest eigenvalue).

Preferably in a book that I can just quote as in "it is a well-known fact that the Lebesgue measure is the unique rotation-invariant measure on the sphere [1]" :) $\endgroup$ – Yoni Rozenshein Commented Apr 7, 2015 at 13:15

Is there a proof for the rotation invariance of the Lebesgue measure that doesn't use already that one has $$ \lambda(A^{-1}(B)) = \vert \operatorname{det} A \vert ^{-1} \lambda (B) \qquad \text{for all } A \in \operatorname{GL}(\mathbb R^d), \ B \in \mathcal B^d?$$ For example one can show easily that $\lambda$ is invariant under translation ...