Variance of positively recurrent Markov chain hitting time
$begingroup$
Consider the following Markov chain:
How to compute the variance of $T_{1,0}$ the time to get from state $1$ to state $0$?
Now I can get the stationary distribution $pi_i=frac{1}{3}(frac{2}{3})^i$. Thus we know the the time to get from state k to state k is $m_{kk}=frac{1}{pi_k}$.
Also from $m_{0,0}=0.6*m_{0,0}+0.4*(1+m_{1,0})$ we can get $E[T_{1,0}]$.
I think it can be solve using $Var[X]=E[X^2]-(E[X])^2$ but how to get $E[T_{1,0}^2]$?
probability-theory stochastic-processes markov-chains
edited Oct 15 '17 at 19:42
Did
247k23223459
asked Oct 15 '17 at 4:19
helenhelen
111
$endgroup$
add a comment |
$begingroup$
Consider the following Markov chain:
How to compute the variance of $T_{1,0}$ the time to get from state $1$ to state $0$?
Now I can get the stationary distribution $pi_i=frac{1}{3}(frac{2}{3})^i$. Thus we know the the time to get from state k to state k is $m_{kk}=frac{1}{pi_k}$.
Also from $m_{0,0}=0.6*m_{0,0}+0.4*(1+m_{1,0})$ we can get $E[T_{1,0}]$.
I think it can be solve using $Var[X]=E[X^2]-(E[X])^2$ but how to get $E[T_{1,0}^2]$?
probability-theory stochastic-processes markov-chains
edited Oct 15 '17 at 19:42
Did
247k23223459
asked Oct 15 '17 at 4:19
helenhelen
111
$endgroup$
add a comment |
$begingroup$
Consider the following Markov chain:
How to compute the variance of $T_{1,0}$ the time to get from state $1$ to state $0$?
Now I can get the stationary distribution $pi_i=frac{1}{3}(frac{2}{3})^i$. Thus we know the the time to get from state k to state k is $m_{kk}=frac{1}{pi_k}$.
Also from $m_{0,0}=0.6*m_{0,0}+0.4*(1+m_{1,0})$ we can get $E[T_{1,0}]$.
I think it can be solve using $Var[X]=E[X^2]-(E[X])^2$ but how to get $E[T_{1,0}^2]$?
probability-theory stochastic-processes markov-chains
edited Oct 15 '17 at 19:42
Did
247k23223459
asked Oct 15 '17 at 4:19
helenhelen
111
$endgroup$
Consider the following Markov chain:
How to compute the variance of $T_{1,0}$ the time to get from state $1$ to state $0$?
Now I can get the stationary distribution $pi_i=frac{1}{3}(frac{2}{3})^i$. Thus we know the the time to get from state k to state k is $m_{kk}=frac{1}{pi_k}$.
Also from $m_{0,0}=0.6*m_{0,0}+0.4*(1+m_{1,0})$ we can get $E[T_{1,0}]$.
I think it can be solve using $Var[X]=E[X^2]-(E[X])^2$ but how to get $E[T_{1,0}^2]$?
probability-theory stochastic-processes markov-chains
probability-theory stochastic-processes markov-chains
edited Oct 15 '17 at 19:42
Did
247k23223459
asked Oct 15 '17 at 4:19
helenhelen
111
edited Oct 15 '17 at 19:42
Did
247k23223459
asked Oct 15 '17 at 4:19
helenhelen
111
edited Oct 15 '17 at 19:42
Did
247k23223459
edited Oct 15 '17 at 19:42
Did
247k23223459
edited Oct 15 '17 at 19:42
Did
247k23223459
247k23223459
asked Oct 15 '17 at 4:19
helenhelen
111
asked Oct 15 '17 at 4:19
helenhelen
111
asked Oct 15 '17 at 4:19
helenhelen
111
111
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
(2017-10-24) Amusing downvote, purely for mathematical reasons, I am sure...
To reach the higher moments of $T_{1,0}$, one can use the Markov property of the chain after one step, which translates into the identity in distribution $$T_{1,0}=1+Bcdot(T_{2,1}+T'_{1,0})$$ where $B$ is Bernoulli with $P(B=1)=b$ and $P(B=0)=1-b$ for $b=.4$, and $T'_{1,0}$ is distributed like $T_{1,0}$ and independent of $T_{2,1}$.
Since $T_{2,1}$ is also distributed like $T_{1,0}$, one is after the solution of the identity
$$Tstackrel d=1+Bcdot(T+T')tag{$ast$}$$
This allows to recover the first moment $E(T_{1,0})$ since $(ast)$ implies $$E(T)=1+2bcdot E(T)$$ hence, using $b<frac12$, $$E(T)=frac1{1-2b}$$ but also the second moment since $(ast)$ also implies $$E(T^2)=1+4bcdot E(T)+bcdot(2E(T^2)+2E(T))$$ hence, using again $b<frac12$, $$E(T^2)=frac{1+6bcdot E(T)}{1-2b}=frac{1+4b}{(1-2b)^2}$$ and finally,
$$mathrm{var}(T)=frac{4b}{(1-2b)^2}$$
Note that, more generally, turning to generating functions rather than moments, $(ast)$ would allow to reach distributions since, for every $s$ in $[0,1]$, $$E(s^T)=scdot(1-b+bcdot E(s^T)^2)$$ hence
$$E(s^T)=frac{1-sqrt{1-4b(1-b)s^2}}{2bs}$$
from which the full distribution of $T_{1,0}$ follows.
answered Oct 15 '17 at 19:32
DidDid
247k23223459
$endgroup$
$begingroup$
Could you explain a little bit more on how you imply the second moment equation: $E(T^2)=1+4b⋅E(T)+b⋅(2E(T^2)+2E(T))$ ?
$endgroup$
– helen
Oct 15 '17 at 20:30
$begingroup$
Guys. I think there's a typo. I think the last term is $b^2(2E(T^2) + 2E(T))$. Maybe that's what's confusing?
$endgroup$
– Wilmer E. Henao
Oct 16 '17 at 0:01
$begingroup$
@helen. If you're interested in the derivation of the moment generation function here. You can find it in the book Markov Chains by Norris. Page 21 has a very detailed derivation of this. :)
$endgroup$
– Wilmer E. Henao
Oct 16 '17 at 0:07
$begingroup$
@WilmerE.Henao Think harder: $B^2=B$ hence $E(B^2)=b$, not $b^2$.
$endgroup$
– Did
Oct 16 '17 at 7:37
$begingroup$
@helen Sorry but what is not direct in $$T^2=^d1+2Bcdot(T+T')+Bcdot(T^2+2TT'+T'^2)$$ which implies $$E(T^2)=1+2E(B)cdot(2E(T))+E(B)cdot(2E(T^2)+2E(T)^2) ?$$
$endgroup$
– Did
Oct 16 '17 at 7:39
add a comment |
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1 Answer
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$begingroup$
(2017-10-24) Amusing downvote, purely for mathematical reasons, I am sure...
To reach the higher moments of $T_{1,0}$, one can use the Markov property of the chain after one step, which translates into the identity in distribution $$T_{1,0}=1+Bcdot(T_{2,1}+T'_{1,0})$$ where $B$ is Bernoulli with $P(B=1)=b$ and $P(B=0)=1-b$ for $b=.4$, and $T'_{1,0}$ is distributed like $T_{1,0}$ and independent of $T_{2,1}$.
Since $T_{2,1}$ is also distributed like $T_{1,0}$, one is after the solution of the identity
$$Tstackrel d=1+Bcdot(T+T')tag{$ast$}$$
This allows to recover the first moment $E(T_{1,0})$ since $(ast)$ implies $$E(T)=1+2bcdot E(T)$$ hence, using $b<frac12$, $$E(T)=frac1{1-2b}$$ but also the second moment since $(ast)$ also implies $$E(T^2)=1+4bcdot E(T)+bcdot(2E(T^2)+2E(T))$$ hence, using again $b<frac12$, $$E(T^2)=frac{1+6bcdot E(T)}{1-2b}=frac{1+4b}{(1-2b)^2}$$ and finally,
$$mathrm{var}(T)=frac{4b}{(1-2b)^2}$$
Note that, more generally, turning to generating functions rather than moments, $(ast)$ would allow to reach distributions since, for every $s$ in $[0,1]$, $$E(s^T)=scdot(1-b+bcdot E(s^T)^2)$$ hence
$$E(s^T)=frac{1-sqrt{1-4b(1-b)s^2}}{2bs}$$
from which the full distribution of $T_{1,0}$ follows.
answered Oct 15 '17 at 19:32
DidDid
247k23223459
$endgroup$
$begingroup$
Could you explain a little bit more on how you imply the second moment equation: $E(T^2)=1+4b⋅E(T)+b⋅(2E(T^2)+2E(T))$ ?
$endgroup$
– helen
Oct 15 '17 at 20:30
$begingroup$
Guys. I think there's a typo. I think the last term is $b^2(2E(T^2) + 2E(T))$. Maybe that's what's confusing?
$endgroup$
– Wilmer E. Henao
Oct 16 '17 at 0:01
$begingroup$
@helen. If you're interested in the derivation of the moment generation function here. You can find it in the book Markov Chains by Norris. Page 21 has a very detailed derivation of this. :)
$endgroup$
– Wilmer E. Henao
Oct 16 '17 at 0:07
$begingroup$
@WilmerE.Henao Think harder: $B^2=B$ hence $E(B^2)=b$, not $b^2$.
$endgroup$
– Did
Oct 16 '17 at 7:37
$begingroup$
@helen Sorry but what is not direct in $$T^2=^d1+2Bcdot(T+T')+Bcdot(T^2+2TT'+T'^2)$$ which implies $$E(T^2)=1+2E(B)cdot(2E(T))+E(B)cdot(2E(T^2)+2E(T)^2) ?$$
$endgroup$
– Did
Oct 16 '17 at 7:39
add a comment |
$begingroup$
(2017-10-24) Amusing downvote, purely for mathematical reasons, I am sure...
To reach the higher moments of $T_{1,0}$, one can use the Markov property of the chain after one step, which translates into the identity in distribution $$T_{1,0}=1+Bcdot(T_{2,1}+T'_{1,0})$$ where $B$ is Bernoulli with $P(B=1)=b$ and $P(B=0)=1-b$ for $b=.4$, and $T'_{1,0}$ is distributed like $T_{1,0}$ and independent of $T_{2,1}$.
Since $T_{2,1}$ is also distributed like $T_{1,0}$, one is after the solution of the identity
$$Tstackrel d=1+Bcdot(T+T')tag{$ast$}$$
This allows to recover the first moment $E(T_{1,0})$ since $(ast)$ implies $$E(T)=1+2bcdot E(T)$$ hence, using $b<frac12$, $$E(T)=frac1{1-2b}$$ but also the second moment since $(ast)$ also implies $$E(T^2)=1+4bcdot E(T)+bcdot(2E(T^2)+2E(T))$$ hence, using again $b<frac12$, $$E(T^2)=frac{1+6bcdot E(T)}{1-2b}=frac{1+4b}{(1-2b)^2}$$ and finally,
$$mathrm{var}(T)=frac{4b}{(1-2b)^2}$$
Note that, more generally, turning to generating functions rather than moments, $(ast)$ would allow to reach distributions since, for every $s$ in $[0,1]$, $$E(s^T)=scdot(1-b+bcdot E(s^T)^2)$$ hence
$$E(s^T)=frac{1-sqrt{1-4b(1-b)s^2}}{2bs}$$
from which the full distribution of $T_{1,0}$ follows.
answered Oct 15 '17 at 19:32
DidDid
247k23223459
$endgroup$
$begingroup$
Could you explain a little bit more on how you imply the second moment equation: $E(T^2)=1+4b⋅E(T)+b⋅(2E(T^2)+2E(T))$ ?
$endgroup$
– helen
Oct 15 '17 at 20:30
$begingroup$
Guys. I think there's a typo. I think the last term is $b^2(2E(T^2) + 2E(T))$. Maybe that's what's confusing?
$endgroup$
– Wilmer E. Henao
Oct 16 '17 at 0:01
$begingroup$
@helen. If you're interested in the derivation of the moment generation function here. You can find it in the book Markov Chains by Norris. Page 21 has a very detailed derivation of this. :)
$endgroup$
– Wilmer E. Henao
Oct 16 '17 at 0:07
$begingroup$
@WilmerE.Henao Think harder: $B^2=B$ hence $E(B^2)=b$, not $b^2$.
$endgroup$
– Did
Oct 16 '17 at 7:37
$begingroup$
@helen Sorry but what is not direct in $$T^2=^d1+2Bcdot(T+T')+Bcdot(T^2+2TT'+T'^2)$$ which implies $$E(T^2)=1+2E(B)cdot(2E(T))+E(B)cdot(2E(T^2)+2E(T)^2) ?$$
$endgroup$
– Did
Oct 16 '17 at 7:39
add a comment |
$begingroup$
(2017-10-24) Amusing downvote, purely for mathematical reasons, I am sure...
To reach the higher moments of $T_{1,0}$, one can use the Markov property of the chain after one step, which translates into the identity in distribution $$T_{1,0}=1+Bcdot(T_{2,1}+T'_{1,0})$$ where $B$ is Bernoulli with $P(B=1)=b$ and $P(B=0)=1-b$ for $b=.4$, and $T'_{1,0}$ is distributed like $T_{1,0}$ and independent of $T_{2,1}$.
Since $T_{2,1}$ is also distributed like $T_{1,0}$, one is after the solution of the identity
$$Tstackrel d=1+Bcdot(T+T')tag{$ast$}$$
This allows to recover the first moment $E(T_{1,0})$ since $(ast)$ implies $$E(T)=1+2bcdot E(T)$$ hence, using $b<frac12$, $$E(T)=frac1{1-2b}$$ but also the second moment since $(ast)$ also implies $$E(T^2)=1+4bcdot E(T)+bcdot(2E(T^2)+2E(T))$$ hence, using again $b<frac12$, $$E(T^2)=frac{1+6bcdot E(T)}{1-2b}=frac{1+4b}{(1-2b)^2}$$ and finally,
$$mathrm{var}(T)=frac{4b}{(1-2b)^2}$$
Note that, more generally, turning to generating functions rather than moments, $(ast)$ would allow to reach distributions since, for every $s$ in $[0,1]$, $$E(s^T)=scdot(1-b+bcdot E(s^T)^2)$$ hence
$$E(s^T)=frac{1-sqrt{1-4b(1-b)s^2}}{2bs}$$
from which the full distribution of $T_{1,0}$ follows.
answered Oct 15 '17 at 19:32
DidDid
247k23223459
$endgroup$
(2017-10-24) Amusing downvote, purely for mathematical reasons, I am sure...
To reach the higher moments of $T_{1,0}$, one can use the Markov property of the chain after one step, which translates into the identity in distribution $$T_{1,0}=1+Bcdot(T_{2,1}+T'_{1,0})$$ where $B$ is Bernoulli with $P(B=1)=b$ and $P(B=0)=1-b$ for $b=.4$, and $T'_{1,0}$ is distributed like $T_{1,0}$ and independent of $T_{2,1}$.
Since $T_{2,1}$ is also distributed like $T_{1,0}$, one is after the solution of the identity
$$Tstackrel d=1+Bcdot(T+T')tag{$ast$}$$
This allows to recover the first moment $E(T_{1,0})$ since $(ast)$ implies $$E(T)=1+2bcdot E(T)$$ hence, using $b<frac12$, $$E(T)=frac1{1-2b}$$ but also the second moment since $(ast)$ also implies $$E(T^2)=1+4bcdot E(T)+bcdot(2E(T^2)+2E(T))$$ hence, using again $b<frac12$, $$E(T^2)=frac{1+6bcdot E(T)}{1-2b}=frac{1+4b}{(1-2b)^2}$$ and finally,
$$mathrm{var}(T)=frac{4b}{(1-2b)^2}$$
Note that, more generally, turning to generating functions rather than moments, $(ast)$ would allow to reach distributions since, for every $s$ in $[0,1]$, $$E(s^T)=scdot(1-b+bcdot E(s^T)^2)$$ hence
$$E(s^T)=frac{1-sqrt{1-4b(1-b)s^2}}{2bs}$$
from which the full distribution of $T_{1,0}$ follows.
answered Oct 15 '17 at 19:32
DidDid
247k23223459
edited Jan 12 at 11:04
answered Oct 15 '17 at 19:32
DidDid
247k23223459
answered Oct 15 '17 at 19:32
DidDid
247k23223459
answered Oct 15 '17 at 19:32
DidDid
247k23223459
247k23223459
$begingroup$
Could you explain a little bit more on how you imply the second moment equation: $E(T^2)=1+4b⋅E(T)+b⋅(2E(T^2)+2E(T))$ ?
$endgroup$
– helen
Oct 15 '17 at 20:30
$begingroup$
Guys. I think there's a typo. I think the last term is $b^2(2E(T^2) + 2E(T))$. Maybe that's what's confusing?
$endgroup$
– Wilmer E. Henao
Oct 16 '17 at 0:01
$begingroup$
@helen. If you're interested in the derivation of the moment generation function here. You can find it in the book Markov Chains by Norris. Page 21 has a very detailed derivation of this. :)
$endgroup$
– Wilmer E. Henao
Oct 16 '17 at 0:07
$begingroup$
@WilmerE.Henao Think harder: $B^2=B$ hence $E(B^2)=b$, not $b^2$.
$endgroup$
– Did
Oct 16 '17 at 7:37
$begingroup$
@helen Sorry but what is not direct in $$T^2=^d1+2Bcdot(T+T')+Bcdot(T^2+2TT'+T'^2)$$ which implies $$E(T^2)=1+2E(B)cdot(2E(T))+E(B)cdot(2E(T^2)+2E(T)^2) ?$$
$endgroup$
– Did
Oct 16 '17 at 7:39
add a comment |
$begingroup$
Could you explain a little bit more on how you imply the second moment equation: $E(T^2)=1+4b⋅E(T)+b⋅(2E(T^2)+2E(T))$ ?
$endgroup$
– helen
Oct 15 '17 at 20:30
$begingroup$
Guys. I think there's a typo. I think the last term is $b^2(2E(T^2) + 2E(T))$. Maybe that's what's confusing?
$endgroup$
– Wilmer E. Henao
Oct 16 '17 at 0:01
$begingroup$
@helen. If you're interested in the derivation of the moment generation function here. You can find it in the book Markov Chains by Norris. Page 21 has a very detailed derivation of this. :)
$endgroup$
– Wilmer E. Henao
Oct 16 '17 at 0:07
$begingroup$
@WilmerE.Henao Think harder: $B^2=B$ hence $E(B^2)=b$, not $b^2$.
$endgroup$
– Did
Oct 16 '17 at 7:37
$begingroup$
@helen Sorry but what is not direct in $$T^2=^d1+2Bcdot(T+T')+Bcdot(T^2+2TT'+T'^2)$$ which implies $$E(T^2)=1+2E(B)cdot(2E(T))+E(B)cdot(2E(T^2)+2E(T)^2) ?$$
$endgroup$
– Did
Oct 16 '17 at 7:39
$begingroup$
Could you explain a little bit more on how you imply the second moment equation: $E(T^2)=1+4b⋅E(T)+b⋅(2E(T^2)+2E(T))$ ?
$endgroup$
– helen
Oct 15 '17 at 20:30
$begingroup$
Could you explain a little bit more on how you imply the second moment equation: $E(T^2)=1+4b⋅E(T)+b⋅(2E(T^2)+2E(T))$ ?
$endgroup$
– helen
Oct 15 '17 at 20:30
$begingroup$
Guys. I think there's a typo. I think the last term is $b^2(2E(T^2) + 2E(T))$. Maybe that's what's confusing?
$endgroup$
– Wilmer E. Henao
Oct 16 '17 at 0:01
$begingroup$
Guys. I think there's a typo. I think the last term is $b^2(2E(T^2) + 2E(T))$. Maybe that's what's confusing?
$endgroup$
– Wilmer E. Henao
Oct 16 '17 at 0:01
$begingroup$
@helen. If you're interested in the derivation of the moment generation function here. You can find it in the book Markov Chains by Norris. Page 21 has a very detailed derivation of this. :)
$endgroup$
– Wilmer E. Henao
Oct 16 '17 at 0:07
$begingroup$
@helen. If you're interested in the derivation of the moment generation function here. You can find it in the book Markov Chains by Norris. Page 21 has a very detailed derivation of this. :)
$endgroup$
– Wilmer E. Henao
Oct 16 '17 at 0:07
$begingroup$
@WilmerE.Henao Think harder: $B^2=B$ hence $E(B^2)=b$, not $b^2$.
$endgroup$
– Did
Oct 16 '17 at 7:37
$begingroup$
@WilmerE.Henao Think harder: $B^2=B$ hence $E(B^2)=b$, not $b^2$.
$endgroup$
– Did
Oct 16 '17 at 7:37
$begingroup$
@helen Sorry but what is not direct in $$T^2=^d1+2Bcdot(T+T')+Bcdot(T^2+2TT'+T'^2)$$ which implies $$E(T^2)=1+2E(B)cdot(2E(T))+E(B)cdot(2E(T^2)+2E(T)^2) ?$$
$endgroup$
– Did
Oct 16 '17 at 7:39
$begingroup$
@helen Sorry but what is not direct in $$T^2=^d1+2Bcdot(T+T')+Bcdot(T^2+2TT'+T'^2)$$ which implies $$E(T^2)=1+2E(B)cdot(2E(T))+E(B)cdot(2E(T^2)+2E(T)^2) ?$$
$endgroup$
– Did
Oct 16 '17 at 7:39
add a comment |
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