Question

Let \({{\bf{z}}_{\scriptstyle{\bf{1}}\atop\scriptstyle\,}}{\bf{,}}{{\bf{z}}_{\scriptstyle{\bf{2}}\atop\scriptstyle\,}}....\) from a Markov chain, and assume that distribution of \({z_{\scriptstyle1\atop\scriptstyle\,}}\)is the stationary distribution. Show that the joint distribution \(\left( {{{\bf{z}}_{\scriptstyle{\bf{1}}\atop\scriptstyle\,}}{\bf{,}}{{\bf{z}}_{\scriptstyle{\bf{2}}\atop\scriptstyle\,}}} \right)\)of is the same as the joint distribution of \(\left( {{{\bf{z}}_{\scriptstyle{\bf{i}}\atop\scriptstyle\,}}{\bf{,}}{{\bf{z}}_{\scriptstyle{\bf{i + 1}}\atop\scriptstyle\,}}} \right)\) for all\(i > 1\) convenience, you may assume that the Markov chain has finite state space, but the result holds in general.

Short answer

From a Markov chain, assume that distribution of is the stationary distribution.

The joint probability mass function \(\,\left( {{z_1}\,\,,{z_2}} \right)\)

\({g_{1,2}}\left( {{z_1}\,\,,{z_2}} \right) = g\left( {{z_1}} \right)h\left( {{z_2}\,\,\mid {z_1}} \right)\,\,\)

Use that Z1 has the stationary distribution.

Step-by-step solution

Definition of the stationary distribution

A stationary distribution is a specific entity that remains unaffected by the effect of a matrix or operator.

The distribution of \({z_1}\,\) is the stationary distribution. By proving that \({z_i}\,\,\) it has the stationary distribution for every i, it follows that \(\,\left( {{z_1}\,\,,{z_2}} \right)\) have the same distribution as \(\left( {{z_1}\,\,,{z_{i + 1}}} \right)\)

The joint probability mass function \(\,\left( {{z_1}\,\,,{z_2}} \right)\)

\({g_{1,2}}\left( {{z_1}\,\,,{z_2}} \right) = g\left( {{z_1}} \right)h\left( {{z_2}\,\,\mid {z_1}} \right)\,\,\)

Proven by simple induction

where g is p.f. or p.d.f. of\({z_1}\,\)

And his conditional p.f. or p.d.f. given that\(\,{Z_1}\, = {z_1}\)

which is true of the fact that \({z_1}\,\)it has stationary distribution.

Also, it follows that\({z_i}\,\) a stationary distribution for all i can be proven by simple induction when for\(n = 1\,\) \({Z_1}\,\)is stationary is the base. Next, because we now\({z_i}\,\) have stationary distribution, it follows that

\({g_{i,i + 1}}\left( {{z_i}\,\,,{z_{i + 1}}} \right) = g\left( {{z_i}} \right)h\left( {{z_{i + 1}}\,\,{z_i}\,} \right) = {g_{1,2}}\left( {{z_i}\,\,,{z_{i + 1}}} \right)\)

For arbitrary i, which was to be seen.

Hence,

Use that Z1 has the stationary distribution.