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Parity game
From formulasearchengine
Revision as of 14:39, 2 January 2014 by en>Yoghurt(Make it clear that both players share the same token)
Let (X, Σ) be a measurable space and let f be a measurable function from X to itself. A measure μ on (X, Σ) is said to be invariant underf if, for every measurable set A in Σ,
In terms of the push forward, this states that f∗(μ) = μ.
The collection of measures (usually probability measures) on X that are invariant under f is sometimes denoted Mf(X). The collection of ergodic measures, Ef(X), is a subset of Mf(X). Moreover, any convex combination of two invariant measures is also invariant, so Mf(X) is a convex set; Ef(X) consists precisely of the extreme points of Mf(X).
In the case of a dynamical system (X, T, φ), where (X, Σ) is a measurable space as before, T is a monoid and φ : T × X → X is the flow map, a measure μ on (X, Σ) is said to be an invariant measure if it is an invariant measure for each map φt : X → X. Explicitly, μ is invariant if and only if
Put another way, μ is an invariant measure for a sequence of random variables (Zt)t≥0 (perhaps a Markov chain or the solution to a stochastic differential equation) if, whenever the initial condition Z0 is distributed according to μ, so is Zt for any later time t.
Consider the real lineR with its usual Borel σ-algebra; fix a ∈ R and consider the translation map Ta : R → R given by:
Then one-dimensional Lebesgue measureλ is an invariant measure for Ta.
More generally, on n-dimensional Euclidean spaceRn with its usual Borel σ-algebra, n-dimensional Lebesgue measure λn is an invariant measure for any isometry of Euclidean space, i.e. a map T : Rn → Rn that can be written as
The invariant measure in the first example is unique up to trivial renormalization with a constant factor. This does not have to be necessarily the case: Consider a set consisting of just two points and the identity map which leaves each point fixed. Then any probability measure is invariant. Note that S trivially has a decomposition into T-invariant components {A} and {B}.