joint distribution of number of visits of a Markov chain












3














Let ${X_{n}}_{n geq 0}$ be a Markov chain on the state space $S = {1,dots,r}$, $r geq 2$. Let $alpha = (alpha_{1},dots,alpha_{r})$ be the initial distribution and $P$ the transition matrix. For $i = 1,dots,r$ define the following random variables



$$
L_{i,K} = sum_{j=0}^{K-1}mathbb{1}_{{X_{j} = i}}
$$
i.e. $L_{i,K}$ is the number of visits to state $i$ up to time $K-1$. I want to determine the joint distribution



$$
mathbb{P}(L_{1,K} = n_{1},dots,L_{r,K}=n_{r})
$$
where $n_{1} + dots + n_{r} = K$. My attempt is the following. First I have determined the generating function of $(L_{1,K} = n_{1},dots,L_{r,K}=n_{r})$. The result is given by



$$
F_{n}(x_{1},dots,x_{r}) := sum_{n_{1} + cdots + n_{r} = n}mathbb{P}left( L_{1,n} = n_{1},dots,L_{r,n} = n_{r} right)x_{1}^{n_{1}} cdots x_{r}^{n_{r}} = [alpha_{1}x_{1},dots,alpha_{r}x_{r}] left( PDright)^{n-1} e,
$$
where
$$
D =
begin{bmatrix}
x_{1} && \
& ddots &\
& & x_{r}
end{bmatrix}.
$$
and $e = (1,dots,1) in mathbb{R}^{r}$. Is there a way to extract the probabilities?






probability markov-chains







share|cite|improve this question




















  • 1




    There is no known (simple) way to do this: If you take a general graph with $r$ nodes and define the Markov chain by a random walk over the graph (starting in state 1) then $P[L_{1,r}=1, L_{2,r}=1, ..., L_{r,r}=1]>0$ iff there is a Hamiltonian path starting at state 1.
    – Michael
    Dec 10 '18 at 8:43










  • I have edit my question. I hope the answer is now more accessible
    – wayne
    Dec 10 '18 at 12:16






  • 1




    In general you can extract the $(n_1, ..., n_r)=(1,1,...,1)$ term from a transform by taking $(partial/partial x_1)(partial/partial x_2)...(partial/partial x_n)$ and setting $x_i=0$ for all $i$. But I don't think your edits are going to solve an NP-hard problem.
    – Michael
    Dec 10 '18 at 15:21










  • so you mean that this is the best expression to compute the probabilities? There is no easy way to extract the probabilities?
    – wayne
    Dec 10 '18 at 18:15
















3














Let ${X_{n}}_{n geq 0}$ be a Markov chain on the state space $S = {1,dots,r}$, $r geq 2$. Let $alpha = (alpha_{1},dots,alpha_{r})$ be the initial distribution and $P$ the transition matrix. For $i = 1,dots,r$ define the following random variables



$$
L_{i,K} = sum_{j=0}^{K-1}mathbb{1}_{{X_{j} = i}}
$$
i.e. $L_{i,K}$ is the number of visits to state $i$ up to time $K-1$. I want to determine the joint distribution



$$
mathbb{P}(L_{1,K} = n_{1},dots,L_{r,K}=n_{r})
$$
where $n_{1} + dots + n_{r} = K$. My attempt is the following. First I have determined the generating function of $(L_{1,K} = n_{1},dots,L_{r,K}=n_{r})$. The result is given by



$$
F_{n}(x_{1},dots,x_{r}) := sum_{n_{1} + cdots + n_{r} = n}mathbb{P}left( L_{1,n} = n_{1},dots,L_{r,n} = n_{r} right)x_{1}^{n_{1}} cdots x_{r}^{n_{r}} = [alpha_{1}x_{1},dots,alpha_{r}x_{r}] left( PDright)^{n-1} e,
$$
where
$$
D =
begin{bmatrix}
x_{1} && \
& ddots &\
& & x_{r}
end{bmatrix}.
$$
and $e = (1,dots,1) in mathbb{R}^{r}$. Is there a way to extract the probabilities?






probability markov-chains







share|cite|improve this question




















  • 1




    There is no known (simple) way to do this: If you take a general graph with $r$ nodes and define the Markov chain by a random walk over the graph (starting in state 1) then $P[L_{1,r}=1, L_{2,r}=1, ..., L_{r,r}=1]>0$ iff there is a Hamiltonian path starting at state 1.
    – Michael
    Dec 10 '18 at 8:43










  • I have edit my question. I hope the answer is now more accessible
    – wayne
    Dec 10 '18 at 12:16






  • 1




    In general you can extract the $(n_1, ..., n_r)=(1,1,...,1)$ term from a transform by taking $(partial/partial x_1)(partial/partial x_2)...(partial/partial x_n)$ and setting $x_i=0$ for all $i$. But I don't think your edits are going to solve an NP-hard problem.
    – Michael
    Dec 10 '18 at 15:21










  • so you mean that this is the best expression to compute the probabilities? There is no easy way to extract the probabilities?
    – wayne
    Dec 10 '18 at 18:15














3












3








3







Let ${X_{n}}_{n geq 0}$ be a Markov chain on the state space $S = {1,dots,r}$, $r geq 2$. Let $alpha = (alpha_{1},dots,alpha_{r})$ be the initial distribution and $P$ the transition matrix. For $i = 1,dots,r$ define the following random variables



$$
L_{i,K} = sum_{j=0}^{K-1}mathbb{1}_{{X_{j} = i}}
$$
i.e. $L_{i,K}$ is the number of visits to state $i$ up to time $K-1$. I want to determine the joint distribution



$$
mathbb{P}(L_{1,K} = n_{1},dots,L_{r,K}=n_{r})
$$
where $n_{1} + dots + n_{r} = K$. My attempt is the following. First I have determined the generating function of $(L_{1,K} = n_{1},dots,L_{r,K}=n_{r})$. The result is given by



$$
F_{n}(x_{1},dots,x_{r}) := sum_{n_{1} + cdots + n_{r} = n}mathbb{P}left( L_{1,n} = n_{1},dots,L_{r,n} = n_{r} right)x_{1}^{n_{1}} cdots x_{r}^{n_{r}} = [alpha_{1}x_{1},dots,alpha_{r}x_{r}] left( PDright)^{n-1} e,
$$
where
$$
D =
begin{bmatrix}
x_{1} && \
& ddots &\
& & x_{r}
end{bmatrix}.
$$
and $e = (1,dots,1) in mathbb{R}^{r}$. Is there a way to extract the probabilities?






probability markov-chains







share|cite|improve this question















Let ${X_{n}}_{n geq 0}$ be a Markov chain on the state space $S = {1,dots,r}$, $r geq 2$. Let $alpha = (alpha_{1},dots,alpha_{r})$ be the initial distribution and $P$ the transition matrix. For $i = 1,dots,r$ define the following random variables



$$
L_{i,K} = sum_{j=0}^{K-1}mathbb{1}_{{X_{j} = i}}
$$
i.e. $L_{i,K}$ is the number of visits to state $i$ up to time $K-1$. I want to determine the joint distribution



$$
mathbb{P}(L_{1,K} = n_{1},dots,L_{r,K}=n_{r})
$$
where $n_{1} + dots + n_{r} = K$. My attempt is the following. First I have determined the generating function of $(L_{1,K} = n_{1},dots,L_{r,K}=n_{r})$. The result is given by



$$
F_{n}(x_{1},dots,x_{r}) := sum_{n_{1} + cdots + n_{r} = n}mathbb{P}left( L_{1,n} = n_{1},dots,L_{r,n} = n_{r} right)x_{1}^{n_{1}} cdots x_{r}^{n_{r}} = [alpha_{1}x_{1},dots,alpha_{r}x_{r}] left( PDright)^{n-1} e,
$$
where
$$
D =
begin{bmatrix}
x_{1} && \
& ddots &\
& & x_{r}
end{bmatrix}.
$$
and $e = (1,dots,1) in mathbb{R}^{r}$. Is there a way to extract the probabilities?






probability markov-chains




probability markov-chains






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 12:13

























asked Dec 9 '18 at 20:58









wayne

492112




492112








  • 1




    There is no known (simple) way to do this: If you take a general graph with $r$ nodes and define the Markov chain by a random walk over the graph (starting in state 1) then $P[L_{1,r}=1, L_{2,r}=1, ..., L_{r,r}=1]>0$ iff there is a Hamiltonian path starting at state 1.
    – Michael
    Dec 10 '18 at 8:43










  • I have edit my question. I hope the answer is now more accessible
    – wayne
    Dec 10 '18 at 12:16






  • 1




    In general you can extract the $(n_1, ..., n_r)=(1,1,...,1)$ term from a transform by taking $(partial/partial x_1)(partial/partial x_2)...(partial/partial x_n)$ and setting $x_i=0$ for all $i$. But I don't think your edits are going to solve an NP-hard problem.
    – Michael
    Dec 10 '18 at 15:21










  • so you mean that this is the best expression to compute the probabilities? There is no easy way to extract the probabilities?
    – wayne
    Dec 10 '18 at 18:15














  • 1




    There is no known (simple) way to do this: If you take a general graph with $r$ nodes and define the Markov chain by a random walk over the graph (starting in state 1) then $P[L_{1,r}=1, L_{2,r}=1, ..., L_{r,r}=1]>0$ iff there is a Hamiltonian path starting at state 1.
    – Michael
    Dec 10 '18 at 8:43










  • I have edit my question. I hope the answer is now more accessible
    – wayne
    Dec 10 '18 at 12:16






  • 1




    In general you can extract the $(n_1, ..., n_r)=(1,1,...,1)$ term from a transform by taking $(partial/partial x_1)(partial/partial x_2)...(partial/partial x_n)$ and setting $x_i=0$ for all $i$. But I don't think your edits are going to solve an NP-hard problem.
    – Michael
    Dec 10 '18 at 15:21










  • so you mean that this is the best expression to compute the probabilities? There is no easy way to extract the probabilities?
    – wayne
    Dec 10 '18 at 18:15








1




1




There is no known (simple) way to do this: If you take a general graph with $r$ nodes and define the Markov chain by a random walk over the graph (starting in state 1) then $P[L_{1,r}=1, L_{2,r}=1, ..., L_{r,r}=1]>0$ iff there is a Hamiltonian path starting at state 1.
– Michael
Dec 10 '18 at 8:43




There is no known (simple) way to do this: If you take a general graph with $r$ nodes and define the Markov chain by a random walk over the graph (starting in state 1) then $P[L_{1,r}=1, L_{2,r}=1, ..., L_{r,r}=1]>0$ iff there is a Hamiltonian path starting at state 1.
– Michael
Dec 10 '18 at 8:43












I have edit my question. I hope the answer is now more accessible
– wayne
Dec 10 '18 at 12:16




I have edit my question. I hope the answer is now more accessible
– wayne
Dec 10 '18 at 12:16




1




1




In general you can extract the $(n_1, ..., n_r)=(1,1,...,1)$ term from a transform by taking $(partial/partial x_1)(partial/partial x_2)...(partial/partial x_n)$ and setting $x_i=0$ for all $i$. But I don't think your edits are going to solve an NP-hard problem.
– Michael
Dec 10 '18 at 15:21




In general you can extract the $(n_1, ..., n_r)=(1,1,...,1)$ term from a transform by taking $(partial/partial x_1)(partial/partial x_2)...(partial/partial x_n)$ and setting $x_i=0$ for all $i$. But I don't think your edits are going to solve an NP-hard problem.
– Michael
Dec 10 '18 at 15:21












so you mean that this is the best expression to compute the probabilities? There is no easy way to extract the probabilities?
– wayne
Dec 10 '18 at 18:15




so you mean that this is the best expression to compute the probabilities? There is no easy way to extract the probabilities?
– wayne
Dec 10 '18 at 18:15















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