When is it necessary to solve Kolmogorov forward equations (KFE) for a Markov Chain?
$begingroup$
Say I have a continuous time markov chain, time homogeneous $X$ with a few states (say, 2). I want to know the distribution of where $X$ is at time $t$, call it $mu_t$, which will be a vector of 2 elements since there are two states. From the book Grimmet and Stirzaker, we have the Chapman Kolmogorov equations:
$$mu_{t+s}=mu_t*P_s, forall s,t ge 0$$
where $P_s$ is the infinitesimal probability transition matrix at time $s$.
I know we also have the Kolmogorov forward and Kolmogorov backward equations which tell us how $P_t$ evolves with time.
I'd like to when it is necessary to use the KFE equations. In every Markov Chain problem, we are given either:
- The probability transition matrix, $P_t$ explicitly, OR
- The generator matrix $G$
IF I have the generator matrix $G$ then like it shows on page 258, I can easily get the $P_t$ matrix by $p_ii=1+g_{ii}*h+o(h)$ and $p_ij=g_{ij}*h+o(h)$ (as shown on page 258)
and if I have $P_t$ matrix, then I can always use the Chapman-Kolmogorov equation in order to get $mu_t$ at an instant in time $t$: $mu_{t}=mu_0*P_t$ (as I showed above). Is that correct?
Then, given $mu_0$, the initial state distribution, I don’t see any reason to solve the KFE equations. When would I need to solve them?
stochastic-processes markov-chains markov-process stochastic-pde hidden-markov-models
asked Dec 9 '18 at 23:52
nundonundo
13018
$endgroup$
add a comment |
$begingroup$
Say I have a continuous time markov chain, time homogeneous $X$ with a few states (say, 2). I want to know the distribution of where $X$ is at time $t$, call it $mu_t$, which will be a vector of 2 elements since there are two states. From the book Grimmet and Stirzaker, we have the Chapman Kolmogorov equations:
$$mu_{t+s}=mu_t*P_s, forall s,t ge 0$$
where $P_s$ is the infinitesimal probability transition matrix at time $s$.
I know we also have the Kolmogorov forward and Kolmogorov backward equations which tell us how $P_t$ evolves with time.
I'd like to when it is necessary to use the KFE equations. In every Markov Chain problem, we are given either:
- The probability transition matrix, $P_t$ explicitly, OR
- The generator matrix $G$
IF I have the generator matrix $G$ then like it shows on page 258, I can easily get the $P_t$ matrix by $p_ii=1+g_{ii}*h+o(h)$ and $p_ij=g_{ij}*h+o(h)$ (as shown on page 258)
and if I have $P_t$ matrix, then I can always use the Chapman-Kolmogorov equation in order to get $mu_t$ at an instant in time $t$: $mu_{t}=mu_0*P_t$ (as I showed above). Is that correct?
Then, given $mu_0$, the initial state distribution, I don’t see any reason to solve the KFE equations. When would I need to solve them?
stochastic-processes markov-chains markov-process stochastic-pde hidden-markov-models
asked Dec 9 '18 at 23:52
nundonundo
13018
$endgroup$
1
$begingroup$
You have tagged this as stochastic-pde. Is there a stochastic-pde somewhere in your question?
$endgroup$
– Zachary Selk
Dec 30 '18 at 1:49
add a comment |
$begingroup$
Say I have a continuous time markov chain, time homogeneous $X$ with a few states (say, 2). I want to know the distribution of where $X$ is at time $t$, call it $mu_t$, which will be a vector of 2 elements since there are two states. From the book Grimmet and Stirzaker, we have the Chapman Kolmogorov equations:
$$mu_{t+s}=mu_t*P_s, forall s,t ge 0$$
where $P_s$ is the infinitesimal probability transition matrix at time $s$.
I know we also have the Kolmogorov forward and Kolmogorov backward equations which tell us how $P_t$ evolves with time.
I'd like to when it is necessary to use the KFE equations. In every Markov Chain problem, we are given either:
- The probability transition matrix, $P_t$ explicitly, OR
- The generator matrix $G$
IF I have the generator matrix $G$ then like it shows on page 258, I can easily get the $P_t$ matrix by $p_ii=1+g_{ii}*h+o(h)$ and $p_ij=g_{ij}*h+o(h)$ (as shown on page 258)
and if I have $P_t$ matrix, then I can always use the Chapman-Kolmogorov equation in order to get $mu_t$ at an instant in time $t$: $mu_{t}=mu_0*P_t$ (as I showed above). Is that correct?
Then, given $mu_0$, the initial state distribution, I don’t see any reason to solve the KFE equations. When would I need to solve them?
stochastic-processes markov-chains markov-process stochastic-pde hidden-markov-models
asked Dec 9 '18 at 23:52
nundonundo
13018
$endgroup$
Say I have a continuous time markov chain, time homogeneous $X$ with a few states (say, 2). I want to know the distribution of where $X$ is at time $t$, call it $mu_t$, which will be a vector of 2 elements since there are two states. From the book Grimmet and Stirzaker, we have the Chapman Kolmogorov equations:
$$mu_{t+s}=mu_t*P_s, forall s,t ge 0$$
where $P_s$ is the infinitesimal probability transition matrix at time $s$.
I know we also have the Kolmogorov forward and Kolmogorov backward equations which tell us how $P_t$ evolves with time.
I'd like to when it is necessary to use the KFE equations. In every Markov Chain problem, we are given either:
- The probability transition matrix, $P_t$ explicitly, OR
- The generator matrix $G$
IF I have the generator matrix $G$ then like it shows on page 258, I can easily get the $P_t$ matrix by $p_ii=1+g_{ii}*h+o(h)$ and $p_ij=g_{ij}*h+o(h)$ (as shown on page 258)
and if I have $P_t$ matrix, then I can always use the Chapman-Kolmogorov equation in order to get $mu_t$ at an instant in time $t$: $mu_{t}=mu_0*P_t$ (as I showed above). Is that correct?
Then, given $mu_0$, the initial state distribution, I don’t see any reason to solve the KFE equations. When would I need to solve them?
stochastic-processes markov-chains markov-process stochastic-pde hidden-markov-models
stochastic-processes markov-chains markov-process stochastic-pde hidden-markov-models
asked Dec 9 '18 at 23:52
nundonundo
13018
asked Dec 9 '18 at 23:52
nundonundo
13018
edited Dec 10 '18 at 0:00
nundo
asked Dec 9 '18 at 23:52
nundonundo
13018
asked Dec 9 '18 at 23:52
nundonundo
13018
asked Dec 9 '18 at 23:52
nundonundo
13018
13018
1
$begingroup$
You have tagged this as stochastic-pde. Is there a stochastic-pde somewhere in your question?
$endgroup$
– Zachary Selk
Dec 30 '18 at 1:49
add a comment |
1
$begingroup$
You have tagged this as stochastic-pde. Is there a stochastic-pde somewhere in your question?
$endgroup$
– Zachary Selk
Dec 30 '18 at 1:49
1
1
$begingroup$
You have tagged this as stochastic-pde. Is there a stochastic-pde somewhere in your question?
$endgroup$
– Zachary Selk
Dec 30 '18 at 1:49
$begingroup$
You have tagged this as stochastic-pde. Is there a stochastic-pde somewhere in your question?
$endgroup$
– Zachary Selk
Dec 30 '18 at 1:49
add a comment |
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$begingroup$
You have tagged this as stochastic-pde. Is there a stochastic-pde somewhere in your question?
$endgroup$
– Zachary Selk
Dec 30 '18 at 1:49