Simulation of Markov chain
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I wrote this algorithm on matlab to simulate a markov chain :
function X=markov_chain(n,cl1,cl2,A,p10,p20)
X=;
first_term=[p10 p20];
for x=1:n
X(x)=simulate(first_term(1),first_term(2),cl1,cl2);
first_term=first_term*A;
end
end
n is the desired length of the Markov chain cl1 and cl2 two values, A the transition matrix and p10 (resp. p20) the probability of being in the state 1 at t=0 (resp. state 2)
the simulate function returns either cl1 with probability first_term(1) or cl2 with probability first_term(2).
It seems like there is something wrong with what I've done. Indeed, my final goal is to estimate the transition matrix given a certain Markov chain. I found contradictory results and it seems like my code here doesn't really simulate a Markov chain, but I can't see why.
statistics markov-chains
edited Dec 5 '18 at 23:38
Bernard
120k740114
asked Dec 5 '18 at 23:30
yjntyjnt
63
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add a comment |
$begingroup$
I wrote this algorithm on matlab to simulate a markov chain :
function X=markov_chain(n,cl1,cl2,A,p10,p20)
X=;
first_term=[p10 p20];
for x=1:n
X(x)=simulate(first_term(1),first_term(2),cl1,cl2);
first_term=first_term*A;
end
end
n is the desired length of the Markov chain cl1 and cl2 two values, A the transition matrix and p10 (resp. p20) the probability of being in the state 1 at t=0 (resp. state 2)
the simulate function returns either cl1 with probability first_term(1) or cl2 with probability first_term(2).
It seems like there is something wrong with what I've done. Indeed, my final goal is to estimate the transition matrix given a certain Markov chain. I found contradictory results and it seems like my code here doesn't really simulate a Markov chain, but I can't see why.
statistics markov-chains
edited Dec 5 '18 at 23:38
Bernard
120k740114
asked Dec 5 '18 at 23:30
yjntyjnt
63
$endgroup$
$begingroup$
Welcome to MSE. Please read this text about how to ask a good question.
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– José Carlos Santos
Dec 6 '18 at 0:03
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You’re using the $n$-step transition probabilities $A^n$ instead of the single-step probabilities $A$.
$endgroup$
– amd
Dec 6 '18 at 0:05
$begingroup$
why so? isn't the relation between the states at different times $t$: $x_t=x_{t-1}A$?
$endgroup$
– yjnt
Dec 6 '18 at 0:11
add a comment |
$begingroup$
I wrote this algorithm on matlab to simulate a markov chain :
function X=markov_chain(n,cl1,cl2,A,p10,p20)
X=;
first_term=[p10 p20];
for x=1:n
X(x)=simulate(first_term(1),first_term(2),cl1,cl2);
first_term=first_term*A;
end
end
n is the desired length of the Markov chain cl1 and cl2 two values, A the transition matrix and p10 (resp. p20) the probability of being in the state 1 at t=0 (resp. state 2)
the simulate function returns either cl1 with probability first_term(1) or cl2 with probability first_term(2).
It seems like there is something wrong with what I've done. Indeed, my final goal is to estimate the transition matrix given a certain Markov chain. I found contradictory results and it seems like my code here doesn't really simulate a Markov chain, but I can't see why.
statistics markov-chains
edited Dec 5 '18 at 23:38
Bernard
120k740114
asked Dec 5 '18 at 23:30
yjntyjnt
63
$endgroup$
I wrote this algorithm on matlab to simulate a markov chain :
function X=markov_chain(n,cl1,cl2,A,p10,p20)
X=;
first_term=[p10 p20];
for x=1:n
X(x)=simulate(first_term(1),first_term(2),cl1,cl2);
first_term=first_term*A;
end
end
n is the desired length of the Markov chain cl1 and cl2 two values, A the transition matrix and p10 (resp. p20) the probability of being in the state 1 at t=0 (resp. state 2)
the simulate function returns either cl1 with probability first_term(1) or cl2 with probability first_term(2).
It seems like there is something wrong with what I've done. Indeed, my final goal is to estimate the transition matrix given a certain Markov chain. I found contradictory results and it seems like my code here doesn't really simulate a Markov chain, but I can't see why.
statistics markov-chains
statistics markov-chains
edited Dec 5 '18 at 23:38
Bernard
120k740114
asked Dec 5 '18 at 23:30
yjntyjnt
63
edited Dec 5 '18 at 23:38
Bernard
120k740114
asked Dec 5 '18 at 23:30
yjntyjnt
63
edited Dec 5 '18 at 23:38
Bernard
120k740114
edited Dec 5 '18 at 23:38
Bernard
120k740114
edited Dec 5 '18 at 23:38
Bernard
120k740114
120k740114
asked Dec 5 '18 at 23:30
yjntyjnt
63
asked Dec 5 '18 at 23:30
yjntyjnt
63
asked Dec 5 '18 at 23:30
yjntyjnt
63
63
$begingroup$
Welcome to MSE. Please read this text about how to ask a good question.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 0:03
$begingroup$
You’re using the $n$-step transition probabilities $A^n$ instead of the single-step probabilities $A$.
$endgroup$
– amd
Dec 6 '18 at 0:05
$begingroup$
why so? isn't the relation between the states at different times $t$: $x_t=x_{t-1}A$?
$endgroup$
– yjnt
Dec 6 '18 at 0:11
add a comment |
$begingroup$
Welcome to MSE. Please read this text about how to ask a good question.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 0:03
$begingroup$
You’re using the $n$-step transition probabilities $A^n$ instead of the single-step probabilities $A$.
$endgroup$
– amd
Dec 6 '18 at 0:05
$begingroup$
why so? isn't the relation between the states at different times $t$: $x_t=x_{t-1}A$?
$endgroup$
– yjnt
Dec 6 '18 at 0:11
$begingroup$
Welcome to MSE. Please read this text about how to ask a good question.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 0:03
$begingroup$
Welcome to MSE. Please read this text about how to ask a good question.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 0:03
$begingroup$
You’re using the $n$-step transition probabilities $A^n$ instead of the single-step probabilities $A$.
$endgroup$
– amd
Dec 6 '18 at 0:05
$begingroup$
You’re using the $n$-step transition probabilities $A^n$ instead of the single-step probabilities $A$.
$endgroup$
– amd
Dec 6 '18 at 0:05
$begingroup$
why so? isn't the relation between the states at different times $t$: $x_t=x_{t-1}A$?
$endgroup$
– yjnt
Dec 6 '18 at 0:11
$begingroup$
why so? isn't the relation between the states at different times $t$: $x_t=x_{t-1}A$?
$endgroup$
– yjnt
Dec 6 '18 at 0:11
add a comment |
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$begingroup$
Welcome to MSE. Please read this text about how to ask a good question.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 0:03
$begingroup$
You’re using the $n$-step transition probabilities $A^n$ instead of the single-step probabilities $A$.
$endgroup$
– amd
Dec 6 '18 at 0:05
$begingroup$
why so? isn't the relation between the states at different times $t$: $x_t=x_{t-1}A$?
$endgroup$
– yjnt
Dec 6 '18 at 0:11