A Difference between a limiting distribution and a Stationary distribution.

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user366312

Gold Member
66
2
Summary
Kindly provide me examples of a limiting distribution, and a stationary distribution.
I am totally confused with these two terms.
Consider a Markov chain ##(X_n)_n## on ##S=\{1, 2\}## with initial distribution ##α## and the transition matrix

##P =
\begin{bmatrix}
2/3 & 1/3 \\
2/3 & 1/3 \\
\end{bmatrix}##

1. Limiting distribution = ?
2. Stationary distribution = ?
My Solution:

##\underline {\text{Limiting Distribution}}##

##
P^2 = \begin{bmatrix}
2/3 & 1/3 \\
2/3 & 1/3 \\
\end{bmatrix}

\begin{bmatrix}
2/3 & 1/3 \\
2/3 & 1/3 \\
\end{bmatrix} =

\begin{bmatrix}
2/3 & 1/3 \\
2/3 & 1/3 \\
\end{bmatrix}
##

So, the limiting distribution of ##P## is ##P## itself.

____

##\underline {\text{Stationary Distribution}}##

Let the stationary distribution ##\pi = \begin{bmatrix} p & 1-p \end{bmatrix}##.

So,

##\pi P = \pi ##

##\Rightarrow \pi (P-1) = 0##

##\Rightarrow \begin{bmatrix} p & 1-p \end{bmatrix} \left(\begin{bmatrix}
2/3 & 1/3 \\
2/3 & 1/3 \\
\end{bmatrix} - \begin{bmatrix}
1 & 0 \\
0 & 1 \\
\end{bmatrix}\right) = 0##

##\Rightarrow \begin{bmatrix} p & 1-p \end{bmatrix} \begin{bmatrix}
-1/3 & 1/3 \\
2/3 & -2/3 \\
\end{bmatrix} = 0##

##\Rightarrow \begin{bmatrix} \frac{-p}{3}+\frac{2}{3}+\frac{-2p}{3} & \frac{p}{3} + \frac{-2}{3} + \frac{2p}{3} \end{bmatrix} = 0##

##\Rightarrow p = 2/3##

So,

##\pi = \begin{bmatrix} \frac{2}{3} & \frac{1}{3} \end{bmatrix}##

___



  • Are the terms Limiting distribution and Stationary distribution properly perceived here?
 
Last edited:

StoneTemplePython

Science Advisor
Gold Member
1,069
508
Summary: Kindly provide me examples of a limiting distribution, and a stationary distribution.
I am totally confused with these two terms.

All finite state time homogenous markov chains have (at least one) stationary distribution ##\mathbf \pi^T##.

If the chain is irreducible then ##\mathbf \pi^T## is unique, and if in addition the chain is aperiodic then

##\lim_{k \to \infty } P^k \to \mathbf{1 \pi}^T##

hence for any valid starting distribution vector ##\mathbf v##

##\lim_{k \to \infty } \mathbf v^T P^k \to\mathbf v^T \big(\mathbf{1 \pi}^T\big) = \big(\mathbf v^T \mathbf 1 \big) \mathbf \pi^T = \mathbf \pi^T##
since ##\mathbf v^T \mathbf 1 = 1##

hence the stable distribution is the limiting distribution.

Limits need not exist for periodic chains, and discussions of limits get muddled with reducible chains. These concepts are better illustrated with larger matrices.
- - - - --
process-wise for this example, ##P^2 = P## so ##P## is idempotent and already the limitting form. You can check that ##P## is rank one and maybe written as ##P=\mathbf{1\pi}^T##

For the second part you'd be smarter to run gaussian elimination on ##\big(P-I\big)^T## and find the sole non-zero vector in the nullspace (then rescale as needed).
 

user366312

Gold Member
66
2
All finite state time homogenous markov chains have (at least one) stationary distribution ##\mathbf \pi^T##.

If the chain is irreducible then ##\mathbf \pi^T## is unique, and if in addition the chain is aperiodic then

##\lim_{k \to \infty } P^k \to \mathbf{1 \pi}^T##

hence for any valid starting distribution vector ##\mathbf v##

##\lim_{k \to \infty } \mathbf v^T P^k \to\mathbf v^T \big(\mathbf{1 \pi}^T\big) = \big(\mathbf v^T \mathbf 1 \big) \mathbf \pi^T = \mathbf \pi^T##
since ##\mathbf v^T \mathbf 1 = 1##

hence the stable distribution is the limiting distribution.

Limits need not exist for periodic chains, and discussions of limits get muddled with reducible chains. These concepts are better illustrated with larger matrices.
- - - - --
process-wise for this example, ##P^2 = P## so ##P## is idempotent and already the limitting form. You can check that ##P## is rank one and maybe written as ##P=\mathbf{1\pi}^T##

For the second part you'd be smarter to run gaussian elimination on ##\big(P-I\big)^T## and find the sole non-zero vector in the nullspace (then rescale as needed).
You didn't comment on the math solution.
 

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