What exactly does it mean to multiply a vector by the transition matrix of a Markov process?
$begingroup$
I know that given a stationary distribution and 2 state transition matrix that
$begin{pmatrix}
Pi _{1} & Pi _{2}
end{pmatrix}begin{pmatrix}
P_{00} & P_{01}\
P_{10}& P_{11}
end{pmatrix}
=begin{pmatrix}
Pi_{1} &Pi_{2}
end{pmatrix}$
is true. But what exactly is the operation of multiplication by the transition matrix doing? And why is it that multiplying a vector that has its nth component made up of the probability that the state is in that nth position (i.e. the stationary distribution) gives you the same vector? What is special about it such that the multiplication operation gives the same output? Or am I mistaken in that multiplication by the transition matrix means nothing special outside of the stationary distribution/invariant vector?
markov-chains markov-process stationary-processes
asked Dec 3 '18 at 4:25
user3491700user3491700
535
$endgroup$
add a comment |
$begingroup$
I know that given a stationary distribution and 2 state transition matrix that
$begin{pmatrix}
Pi _{1} & Pi _{2}
end{pmatrix}begin{pmatrix}
P_{00} & P_{01}\
P_{10}& P_{11}
end{pmatrix}
=begin{pmatrix}
Pi_{1} &Pi_{2}
end{pmatrix}$
is true. But what exactly is the operation of multiplication by the transition matrix doing? And why is it that multiplying a vector that has its nth component made up of the probability that the state is in that nth position (i.e. the stationary distribution) gives you the same vector? What is special about it such that the multiplication operation gives the same output? Or am I mistaken in that multiplication by the transition matrix means nothing special outside of the stationary distribution/invariant vector?
markov-chains markov-process stationary-processes
asked Dec 3 '18 at 4:25
user3491700user3491700
535
$endgroup$
add a comment |
$begingroup$
I know that given a stationary distribution and 2 state transition matrix that
$begin{pmatrix}
Pi _{1} & Pi _{2}
end{pmatrix}begin{pmatrix}
P_{00} & P_{01}\
P_{10}& P_{11}
end{pmatrix}
=begin{pmatrix}
Pi_{1} &Pi_{2}
end{pmatrix}$
is true. But what exactly is the operation of multiplication by the transition matrix doing? And why is it that multiplying a vector that has its nth component made up of the probability that the state is in that nth position (i.e. the stationary distribution) gives you the same vector? What is special about it such that the multiplication operation gives the same output? Or am I mistaken in that multiplication by the transition matrix means nothing special outside of the stationary distribution/invariant vector?
markov-chains markov-process stationary-processes
asked Dec 3 '18 at 4:25
user3491700user3491700
535
$endgroup$
I know that given a stationary distribution and 2 state transition matrix that
$begin{pmatrix}
Pi _{1} & Pi _{2}
end{pmatrix}begin{pmatrix}
P_{00} & P_{01}\
P_{10}& P_{11}
end{pmatrix}
=begin{pmatrix}
Pi_{1} &Pi_{2}
end{pmatrix}$
is true. But what exactly is the operation of multiplication by the transition matrix doing? And why is it that multiplying a vector that has its nth component made up of the probability that the state is in that nth position (i.e. the stationary distribution) gives you the same vector? What is special about it such that the multiplication operation gives the same output? Or am I mistaken in that multiplication by the transition matrix means nothing special outside of the stationary distribution/invariant vector?
markov-chains markov-process stationary-processes
markov-chains markov-process stationary-processes
asked Dec 3 '18 at 4:25
user3491700user3491700
535
asked Dec 3 '18 at 4:25
user3491700user3491700
535
asked Dec 3 '18 at 4:25
user3491700user3491700
535
asked Dec 3 '18 at 4:25
user3491700user3491700
535
asked Dec 3 '18 at 4:25
user3491700user3491700
535
535
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Given an initial distribution $pi$ (row vector) and the transition matrix $P$, the row vector $pi P$ contains the probabilities of ending up in the $2$ states after one step, if the initial state follows the initial distribution. More generally, $pi P^n$ contains the probabilities of ending up in the $2$ states after $n$ steps, if starting according to the initial distribution.
If $pi$ is a stationary distribution, then $pi P^n = pi$, which means that after any number of steps, the distribution of the state you land on is the same as the initial distribution.
answered Dec 3 '18 at 4:36
angryavianangryavian
40.4k23280
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1 Answer
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1 Answer
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$begingroup$
Given an initial distribution $pi$ (row vector) and the transition matrix $P$, the row vector $pi P$ contains the probabilities of ending up in the $2$ states after one step, if the initial state follows the initial distribution. More generally, $pi P^n$ contains the probabilities of ending up in the $2$ states after $n$ steps, if starting according to the initial distribution.
If $pi$ is a stationary distribution, then $pi P^n = pi$, which means that after any number of steps, the distribution of the state you land on is the same as the initial distribution.
answered Dec 3 '18 at 4:36
angryavianangryavian
40.4k23280
$endgroup$
add a comment |
$begingroup$
Given an initial distribution $pi$ (row vector) and the transition matrix $P$, the row vector $pi P$ contains the probabilities of ending up in the $2$ states after one step, if the initial state follows the initial distribution. More generally, $pi P^n$ contains the probabilities of ending up in the $2$ states after $n$ steps, if starting according to the initial distribution.
If $pi$ is a stationary distribution, then $pi P^n = pi$, which means that after any number of steps, the distribution of the state you land on is the same as the initial distribution.
answered Dec 3 '18 at 4:36
angryavianangryavian
40.4k23280
$endgroup$
add a comment |
$begingroup$
Given an initial distribution $pi$ (row vector) and the transition matrix $P$, the row vector $pi P$ contains the probabilities of ending up in the $2$ states after one step, if the initial state follows the initial distribution. More generally, $pi P^n$ contains the probabilities of ending up in the $2$ states after $n$ steps, if starting according to the initial distribution.
If $pi$ is a stationary distribution, then $pi P^n = pi$, which means that after any number of steps, the distribution of the state you land on is the same as the initial distribution.
answered Dec 3 '18 at 4:36
angryavianangryavian
40.4k23280
$endgroup$
Given an initial distribution $pi$ (row vector) and the transition matrix $P$, the row vector $pi P$ contains the probabilities of ending up in the $2$ states after one step, if the initial state follows the initial distribution. More generally, $pi P^n$ contains the probabilities of ending up in the $2$ states after $n$ steps, if starting according to the initial distribution.
If $pi$ is a stationary distribution, then $pi P^n = pi$, which means that after any number of steps, the distribution of the state you land on is the same as the initial distribution.
answered Dec 3 '18 at 4:36
angryavianangryavian
40.4k23280
answered Dec 3 '18 at 4:36
angryavianangryavian
40.4k23280
answered Dec 3 '18 at 4:36
angryavianangryavian
40.4k23280
answered Dec 3 '18 at 4:36
angryavianangryavian
40.4k23280
40.4k23280
add a comment |
add a comment |
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