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Markov Chain

Markov Chain

A first-Order Markov Chain

\[\begin{aligned} p(z^{(m+1)}\vert z^{(m)}, ...,z^{(1)} ) = p(z^{(m+1)}\vert z^{(m)})\end{aligned}\]

then given

  • initial variable: $p(z^{(0)})$
  • transition probabilities: $T_m(z^{(m)}, z^{(m+1)})$

we can obtain a specify Markov Chain

Homogeneous Markov Chain

Transition probabilities are the same for all $m$.

Stationary or Invariant Markov Chain

Each step in the chain leaves that distribution invariant.

For a homogeneous Markov chain with transition probabilities $T(z’,z)$, the distribution $p^*(z)$ is invariant:

\[\begin{aligned} p^*(z) = \sum_{z'} T(z',z)p^*(z') \end{aligned}\]

Note: a given Markov chain may have more than one invariant distribution. For instance: an identity transformation

Detailed Balance property

A sufficient but not necessary condition for ensuring that the required distribution $p(z)$ is invariant is to choose the transition probability to satisfy the property of detailed balance for the particular distribution $p^*(z)$

\[\begin{aligned} p^*(z')T(z',z) = p^*(z)T(z,z') \end{aligned}\]

$\text{Detailed Balance} \Longrightarrow \text{Stationary}$

\[\begin{aligned} \sum_{z'} p^*(z')T(z',z) &= \sum_{z'} p^*(z)T(z,z') \\ &= p^*(z)\sum_{z'} T(z,z') \\ &= p^*(z) \end{aligned}\]

Reversible Markov Chain

A Markov chain that respects detailed balance is said to be reversible.

Ergodicity Markov Chain

For a stationary Markov Chain, if for $m\rightarrow \infty$, $p(z^{(m)}$ converges to a required invariant distribution $p^*(z)$, irrespective of the choice of initial distribution $p(z^{(0)}$.

This property is called ergodicity.

And the invariant distribution is then called the equilibrium distribution.

Note: an ergodic Markov chain can have only one equilibrium distribution.

A homogeneous Markov chain will be ergodic, subject only to weak restrictions on the invariant distribution and the transition probabilities.

Base Transitions

In practice we often construct the transition probabilities from a set of ‘base’ transitions $B_1,…,B_K$. This can be achieved through a mixture distribution of the form

\[\begin{aligned} T(z',z) = \sum_{k=1}^K \alpha_k B_k(z',z) \\ \sum\alpha_k &= 1 \end{aligned}\]
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