Now, I can put limit outside this sum,

and we'll get that here we have the limit as n is turned into infinity.

Sum j from one to M. Pkj of n multiplied with Pji.

Here, we have a product of two matrices,

the first matrix is capital P(n) and the second matrix is

matrix P. You know that P and then bracket is the same as P n,

therefore it's the same as P n plus one,

and P n plus one is the same as P n plus one brackets.

This is according to the theorem which we have already proven.

And due to this representation,

we get that here we have an element of the matrix

P(n+1) element number K i.

And here we can use this formula, boxed formula,

once more until we will get this is equal to Pi star.

Therefore, because it started with element,

was the vector Pi star P and finally got the element of the vector Pi.

Therefore, these two vectors coincide as the first item is proven.

Now, let me continue with the second part of this corollary.

Now, let me prove the second part of this corollary.

According to our definitions,

we shall show that the limit of (Pi)j^n,

as n stands to infinity is equal to (Pi)j^*.

And here the initial distribution,

(Pi)j^0 can be arbitrary.

So we can start with any distribution and will get as limiting

distribution the distribution (Pi)J^*.

It's equal to the limit as n standing to infinity of the sum

(Pi)k^0 multiplied by (Pi)kj of n. Here the sum was

taken by k from 1 to capital M. Here we use the fact which was shown before that

a vector Pi^n is equal to the vector

(Pi)^0 multiplied by metrics P and (n) brace.

Once more as this fact was shown before,

n... What to do now,

we will change the places of the limit and the sum.

And since (Pi)k^0 also do not depend on

n became process limit before the probability of (Pi)kj^n.

This means the tool gives the following sum,

k from 1 to M by (Pi)k^0 multiplied by

the limit as n stands to infinity, (P)kj of (n).

Now let's return to this formula which is boxed.

And get that this limit

is equal to (Pi)j^*.

You see that this value of (Pi)j^* do not depend on k at all.

Therefore, we can put this (Pi)j^* outside the sum.

And finally, we will get that the sum is equal to (Pi)j^* multiplied by

with the sum k from 1 to capital M, (Pi)k^0.

And since, (Pi)1^0 and so on, (Pi)M^0,

in some probability distribution;

the sum of all those elements is equal to one.

And to get that this expression is equal to (Pi)j^*.

So finally, we conclude that the limit of (Pi)j^n under

any choice of initial distribution is equal to (Pi)j^*.

And this observation completes the proof.

So once more, what is important in this corollary,

that the values (Pi)j^* obtain as a limit of (P)ij^n.

All other things are not important at all.

Xt can be or not Ergodic.

And also it isn't important that this values are positive.

It can be any probability distribution.

And, what is really important here is that

this distribution is obtained as a limit of such elements.

Now I would like to provide an example how one can

use this corollaries in particular cases.

We consider as a full length example,

assumes that our Markov chain has two states,

and the probability to access 2 from 1 is equal to 0.8;

to go back is 0.6;

from 1 to 1 is 0.2,

and from 2 to 2 is 0.4.

So this Markov chain has a falling transition metrics

namely it's 0.2, 0.8, 0.4.

Oh sorry, 6, and 0.4.

Okay. Now we conclude that

this Markov chain is Ergodic because it consist of only one class,

all elements are current, and aperiodic.

So what can implies the Ergodic theorem,

so the limit of the corresponding (n)

state transition elements exist and all of them are positive.

And also both corollaries are fulfilled.

Okay, let me find these limits from the first corollary namely,

if I'll denote the elements of the vector (P)^* by (a b),

I can find this (a b) from the system which appears if I will write this

as the condition of stationary P by *P is equal to (Pi)^*.

The following form, so,

(a b) multiplied by 0.2,

0.8, 0.6, 0.4 is equal to (a b).

This is basically a system of two equations with two unknown variables a and b.

So have 0.2 multiplied by a,

plus 0.6 multiplied by b is equal to a.

And also, 0.8 multiplied by a,

plus 0.4 multiplied by b is equal to b.

If so this system we'll get that a is equal to 3

divided by 7 and b is equal to 4 divided by 7.