Thread: Markov chains/limiting matrix

Hello. Can someone please help in finding the limiting matrix of the following transitional matrix of an absorbing Markov chain:

P=

1	0	0
.1	.6	.3
.2	.2	.6

I have tried finding the limiting matrix myself but my answer does not seem to be right. Can you please check if it's alright?

The matrix P is in standard form, which is the form

P=

Identity matrix	0
R	Q

And I know that the limiting matrix must be in this form:

Limiting matrix of P =

Identity matrix	0
FR	0

F is the fundamental matrix for P. It is equal to (I - Q)^-1.

1	0
0	1

minus

.6	.3
.2	.6

is

.4	-.3
-.2	.4

.4	-.3
-.2	.4

to the power of -1 (or in inverse) is

4	3
2	4

Now, FR is equal to

4	3
2	4

multiplied by

.1

.2

The product is

1

1

This makes the limiting matrix seem like

1	0	0
1	0	0
1	0	0

This matrix just seems strange O_o.

Hello, Yoodle15!

I arrived at the same result.

Find the limiting matrix of the following transitional matrix of an absorbing Markov chain:

. .

I was taught this method . . .



From [1], we have: .

And we have: .

These simplify to: .
. . And we have [2]: . . .

Solve the system of equations: .

Therefore: .


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


This makes sense if we consider the transition probabilities.

We have three states:

Matrix gives us these probabilities:

. .

Thank you for your help, Soroban! I am relieved that the answer is alright! I cross-checked the answer many times and couldn't find another answer but still felt, for some reason, suspicious of it.