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Help in an exercise on Markov chain
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I'm doing an exercise on Markov chain and I'm in doubt if I'm doing it right.
Below is the image in the question and soon after my response.
MY QUESTION.
In the first item my answer was as follows
$$0.08333333, 0.2132867, 0.3868007, 0.2226836, 0.145979$$
Where I multiplied the matrix $P$ by $P$ and made $alpha P^2$,
My doubts are as follows. The sum of my lines is not $1$. And when you tried trying to do item $B$ by doing the matrix $P ^ 100$ for example this is going up the values ​​pro infinito.
I would like help if I am doing something wrong. Thanks in advance.
probability stochastic-processes markov-chains
edited Jul 17 at 21:01
Antinous
5,47041949
asked Jul 17 at 20:48
Lucas Freitas
84
 |Â
show 4 more comments
up vote
-2
down vote
favorite
I'm doing an exercise on Markov chain and I'm in doubt if I'm doing it right.
Below is the image in the question and soon after my response.
MY QUESTION.
In the first item my answer was as follows
$$0.08333333, 0.2132867, 0.3868007, 0.2226836, 0.145979$$
Where I multiplied the matrix $P$ by $P$ and made $alpha P^2$,
My doubts are as follows. The sum of my lines is not $1$. And when you tried trying to do item $B$ by doing the matrix $P ^ 100$ for example this is going up the values ​​pro infinito.
I would like help if I am doing something wrong. Thanks in advance.
probability stochastic-processes markov-chains
edited Jul 17 at 21:01
Antinous
5,47041949
asked Jul 17 at 20:48
Lucas Freitas
84
Just a thought, I'm not sure, but could it be you need to perform $P^2alpha^T$ maybe, where $alpha^T$ is the transpose of $alpha$ (i.e. a column vector rather than a row vector). PS - it would have been better if you set your question out rather than referring to a picture. That could be why the down votes...
– Antinous
Jul 17 at 21:03
Part b should be done by computing the left eigenvector with eigenvalue $1$, not by simply computing large powers of $P$. I can't tell why your $alpha P^2$ sums to something so far off from $1$.
– Ian
Jul 17 at 21:05
@Antinous No, the matrix is left-stochastic so the answer to part a is indeed obtained by computing $alpha P^2$.
– Ian
Jul 17 at 21:05
@Ian But 0.08333333 + 0.2132867 + 0.3868007 +0.2226836 + 0.145979 is not 1, is this right?
– Lucas Freitas
Jul 17 at 21:07
@LucasFreitas No, I didn't notice until I actually put it in a calculator that the error is quite large (the sum is something like 1.05), so you made some mistake (probably a data entry mistake).
– Ian
Jul 17 at 21:08
 |Â
show 4 more comments
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
I'm doing an exercise on Markov chain and I'm in doubt if I'm doing it right.
Below is the image in the question and soon after my response.
MY QUESTION.
In the first item my answer was as follows
$$0.08333333, 0.2132867, 0.3868007, 0.2226836, 0.145979$$
Where I multiplied the matrix $P$ by $P$ and made $alpha P^2$,
My doubts are as follows. The sum of my lines is not $1$. And when you tried trying to do item $B$ by doing the matrix $P ^ 100$ for example this is going up the values ​​pro infinito.
I would like help if I am doing something wrong. Thanks in advance.
probability stochastic-processes markov-chains
edited Jul 17 at 21:01
Antinous
5,47041949
asked Jul 17 at 20:48
Lucas Freitas
84
I'm doing an exercise on Markov chain and I'm in doubt if I'm doing it right.
Below is the image in the question and soon after my response.
MY QUESTION.
In the first item my answer was as follows
$$0.08333333, 0.2132867, 0.3868007, 0.2226836, 0.145979$$
Where I multiplied the matrix $P$ by $P$ and made $alpha P^2$,
My doubts are as follows. The sum of my lines is not $1$. And when you tried trying to do item $B$ by doing the matrix $P ^ 100$ for example this is going up the values ​​pro infinito.
I would like help if I am doing something wrong. Thanks in advance.
probability stochastic-processes markov-chains
edited Jul 17 at 21:01
Antinous
5,47041949
asked Jul 17 at 20:48
Lucas Freitas
84
edited Jul 17 at 21:01
Antinous
5,47041949
edited Jul 17 at 21:01
Antinous
5,47041949
edited Jul 17 at 21:01
Antinous
5,47041949
5,47041949
asked Jul 17 at 20:48
Lucas Freitas
84
asked Jul 17 at 20:48
Lucas Freitas
84
asked Jul 17 at 20:48
Lucas Freitas
84
84
Just a thought, I'm not sure, but could it be you need to perform $P^2alpha^T$ maybe, where $alpha^T$ is the transpose of $alpha$ (i.e. a column vector rather than a row vector). PS - it would have been better if you set your question out rather than referring to a picture. That could be why the down votes...
– Antinous
Jul 17 at 21:03
Part b should be done by computing the left eigenvector with eigenvalue $1$, not by simply computing large powers of $P$. I can't tell why your $alpha P^2$ sums to something so far off from $1$.
– Ian
Jul 17 at 21:05
@Antinous No, the matrix is left-stochastic so the answer to part a is indeed obtained by computing $alpha P^2$.
– Ian
Jul 17 at 21:05
@Ian But 0.08333333 + 0.2132867 + 0.3868007 +0.2226836 + 0.145979 is not 1, is this right?
– Lucas Freitas
Jul 17 at 21:07
@LucasFreitas No, I didn't notice until I actually put it in a calculator that the error is quite large (the sum is something like 1.05), so you made some mistake (probably a data entry mistake).
– Ian
Jul 17 at 21:08
 |Â
show 4 more comments
Just a thought, I'm not sure, but could it be you need to perform $P^2alpha^T$ maybe, where $alpha^T$ is the transpose of $alpha$ (i.e. a column vector rather than a row vector). PS - it would have been better if you set your question out rather than referring to a picture. That could be why the down votes...
– Antinous
Jul 17 at 21:03
Part b should be done by computing the left eigenvector with eigenvalue $1$, not by simply computing large powers of $P$. I can't tell why your $alpha P^2$ sums to something so far off from $1$.
– Ian
Jul 17 at 21:05
@Antinous No, the matrix is left-stochastic so the answer to part a is indeed obtained by computing $alpha P^2$.
– Ian
Jul 17 at 21:05
@Ian But 0.08333333 + 0.2132867 + 0.3868007 +0.2226836 + 0.145979 is not 1, is this right?
– Lucas Freitas
Jul 17 at 21:07
@LucasFreitas No, I didn't notice until I actually put it in a calculator that the error is quite large (the sum is something like 1.05), so you made some mistake (probably a data entry mistake).
– Ian
Jul 17 at 21:08
Just a thought, I'm not sure, but could it be you need to perform $P^2alpha^T$ maybe, where $alpha^T$ is the transpose of $alpha$ (i.e. a column vector rather than a row vector). PS - it would have been better if you set your question out rather than referring to a picture. That could be why the down votes...
– Antinous
Jul 17 at 21:03
Just a thought, I'm not sure, but could it be you need to perform $P^2alpha^T$ maybe, where $alpha^T$ is the transpose of $alpha$ (i.e. a column vector rather than a row vector). PS - it would have been better if you set your question out rather than referring to a picture. That could be why the down votes...
– Antinous
Jul 17 at 21:03
Part b should be done by computing the left eigenvector with eigenvalue $1$, not by simply computing large powers of $P$. I can't tell why your $alpha P^2$ sums to something so far off from $1$.
– Ian
Jul 17 at 21:05
Part b should be done by computing the left eigenvector with eigenvalue $1$, not by simply computing large powers of $P$. I can't tell why your $alpha P^2$ sums to something so far off from $1$.
– Ian
Jul 17 at 21:05
@Antinous No, the matrix is left-stochastic so the answer to part a is indeed obtained by computing $alpha P^2$.
– Ian
Jul 17 at 21:05
@Antinous No, the matrix is left-stochastic so the answer to part a is indeed obtained by computing $alpha P^2$.
– Ian
Jul 17 at 21:05
@Ian But 0.08333333 + 0.2132867 + 0.3868007 +0.2226836 + 0.145979 is not 1, is this right?
– Lucas Freitas
Jul 17 at 21:07
@Ian But 0.08333333 + 0.2132867 + 0.3868007 +0.2226836 + 0.145979 is not 1, is this right?
– Lucas Freitas
Jul 17 at 21:07
@LucasFreitas No, I didn't notice until I actually put it in a calculator that the error is quite large (the sum is something like 1.05), so you made some mistake (probably a data entry mistake).
– Ian
Jul 17 at 21:08
@LucasFreitas No, I didn't notice until I actually put it in a calculator that the error is quite large (the sum is something like 1.05), so you made some mistake (probably a data entry mistake).
– Ian
Jul 17 at 21:08
 |Â
show 4 more comments
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
You probably missed some number at the time of computing
answered Jul 18 at 3:16
Lucas Oliveira Freitas
194
add a comment |Â
up vote
0
down vote
Your approach is correct, but you’re right to have doubts about your result. If the row sums (and the sum of the elements of the resulting distribution) are not all equal to $1$, then you’ve made a computational error somewhere. With this particular $alpha$, $alpha P^2$ is just the last row of $P^2$. The only element that’s incorrect is the first one, so recheck your work for the lower-left element of $P^2$. Most of the first column of $P$ is zero, so there’s not a lot of room for making errors there.
For the second part, there’s no need to diagonalize $P$. From the wording of the problem, you can safely assume that there is an unique steady-state distribution and so as Ian mentioned, you just need a left eigenvector of $1$. That is, solve the equation $mathbfpi(P-I)=0$ with the additional condition that the elements of $mathbfpi$ all lie in $[0,1]$ and sum to $1$. You can, of course, find any solution and then normalize it to meet this condition.
answered Jul 18 at 1:44
amd
25.9k2943
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
You probably missed some number at the time of computing
answered Jul 18 at 3:16
Lucas Oliveira Freitas
194
add a comment |Â
up vote
0
down vote
accepted
You probably missed some number at the time of computing
answered Jul 18 at 3:16
Lucas Oliveira Freitas
194
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
You probably missed some number at the time of computing
answered Jul 18 at 3:16
Lucas Oliveira Freitas
194
You probably missed some number at the time of computing
answered Jul 18 at 3:16
Lucas Oliveira Freitas
194
answered Jul 18 at 3:16
Lucas Oliveira Freitas
194
answered Jul 18 at 3:16
Lucas Oliveira Freitas
194
answered Jul 18 at 3:16
Lucas Oliveira Freitas
194
194
add a comment |Â
add a comment |Â
up vote
0
down vote
Your approach is correct, but you’re right to have doubts about your result. If the row sums (and the sum of the elements of the resulting distribution) are not all equal to $1$, then you’ve made a computational error somewhere. With this particular $alpha$, $alpha P^2$ is just the last row of $P^2$. The only element that’s incorrect is the first one, so recheck your work for the lower-left element of $P^2$. Most of the first column of $P$ is zero, so there’s not a lot of room for making errors there.
For the second part, there’s no need to diagonalize $P$. From the wording of the problem, you can safely assume that there is an unique steady-state distribution and so as Ian mentioned, you just need a left eigenvector of $1$. That is, solve the equation $mathbfpi(P-I)=0$ with the additional condition that the elements of $mathbfpi$ all lie in $[0,1]$ and sum to $1$. You can, of course, find any solution and then normalize it to meet this condition.
answered Jul 18 at 1:44
amd
25.9k2943
add a comment |Â
up vote
0
down vote
Your approach is correct, but you’re right to have doubts about your result. If the row sums (and the sum of the elements of the resulting distribution) are not all equal to $1$, then you’ve made a computational error somewhere. With this particular $alpha$, $alpha P^2$ is just the last row of $P^2$. The only element that’s incorrect is the first one, so recheck your work for the lower-left element of $P^2$. Most of the first column of $P$ is zero, so there’s not a lot of room for making errors there.
For the second part, there’s no need to diagonalize $P$. From the wording of the problem, you can safely assume that there is an unique steady-state distribution and so as Ian mentioned, you just need a left eigenvector of $1$. That is, solve the equation $mathbfpi(P-I)=0$ with the additional condition that the elements of $mathbfpi$ all lie in $[0,1]$ and sum to $1$. You can, of course, find any solution and then normalize it to meet this condition.
answered Jul 18 at 1:44
amd
25.9k2943
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Your approach is correct, but you’re right to have doubts about your result. If the row sums (and the sum of the elements of the resulting distribution) are not all equal to $1$, then you’ve made a computational error somewhere. With this particular $alpha$, $alpha P^2$ is just the last row of $P^2$. The only element that’s incorrect is the first one, so recheck your work for the lower-left element of $P^2$. Most of the first column of $P$ is zero, so there’s not a lot of room for making errors there.
For the second part, there’s no need to diagonalize $P$. From the wording of the problem, you can safely assume that there is an unique steady-state distribution and so as Ian mentioned, you just need a left eigenvector of $1$. That is, solve the equation $mathbfpi(P-I)=0$ with the additional condition that the elements of $mathbfpi$ all lie in $[0,1]$ and sum to $1$. You can, of course, find any solution and then normalize it to meet this condition.
answered Jul 18 at 1:44
amd
25.9k2943
Your approach is correct, but you’re right to have doubts about your result. If the row sums (and the sum of the elements of the resulting distribution) are not all equal to $1$, then you’ve made a computational error somewhere. With this particular $alpha$, $alpha P^2$ is just the last row of $P^2$. The only element that’s incorrect is the first one, so recheck your work for the lower-left element of $P^2$. Most of the first column of $P$ is zero, so there’s not a lot of room for making errors there.
For the second part, there’s no need to diagonalize $P$. From the wording of the problem, you can safely assume that there is an unique steady-state distribution and so as Ian mentioned, you just need a left eigenvector of $1$. That is, solve the equation $mathbfpi(P-I)=0$ with the additional condition that the elements of $mathbfpi$ all lie in $[0,1]$ and sum to $1$. You can, of course, find any solution and then normalize it to meet this condition.
answered Jul 18 at 1:44
amd
25.9k2943
answered Jul 18 at 1:44
amd
25.9k2943
answered Jul 18 at 1:44
amd
25.9k2943
answered Jul 18 at 1:44
amd
25.9k2943
25.9k2943
add a comment |Â
add a comment |Â
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Just a thought, I'm not sure, but could it be you need to perform $P^2alpha^T$ maybe, where $alpha^T$ is the transpose of $alpha$ (i.e. a column vector rather than a row vector). PS - it would have been better if you set your question out rather than referring to a picture. That could be why the down votes...
– Antinous
Jul 17 at 21:03
Part b should be done by computing the left eigenvector with eigenvalue $1$, not by simply computing large powers of $P$. I can't tell why your $alpha P^2$ sums to something so far off from $1$.
– Ian
Jul 17 at 21:05
@Antinous No, the matrix is left-stochastic so the answer to part a is indeed obtained by computing $alpha P^2$.
– Ian
Jul 17 at 21:05
@Ian But 0.08333333 + 0.2132867 + 0.3868007 +0.2226836 + 0.145979 is not 1, is this right?
– Lucas Freitas
Jul 17 at 21:07
@LucasFreitas No, I didn't notice until I actually put it in a calculator that the error is quite large (the sum is something like 1.05), so you made some mistake (probably a data entry mistake).
– Ian
Jul 17 at 21:08