What exactly does it mean to multiply a vector by the transition matrix of a Markov process?
up vote
0
down vote
favorite
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
add a comment |
up vote
0
down vote
favorite
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
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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
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 at 4:25
user3491700
485
485
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
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.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
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.
add a comment |
up vote
1
down vote
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.
add a comment |
up vote
1
down vote
up vote
1
down vote
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.
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 at 4:36
angryavian
37.8k13180
37.8k13180
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023627%2fwhat-exactly-does-it-mean-to-multiply-a-vector-by-the-transition-matrix-of-a-mar%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown