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PostPosted: Sun, 6 May 2012 01:26:18 UTC 
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Joined: Sun, 6 May 2012 01:08:03 UTC
Posts: 5
Hi all hoping you can help me with a couple of questions:

Firstly I have a two-state transition probability matrix:
\bf{ T} = \bordermatrix{
& 0 & 1 \cr
0 & (1-a) & a \cr
1 & b & (1-b) \cr}
where 0<a,b<1. I need to prove that
P(X_{n} = 0 | X_{n-1} =0,X_{n+1} =0) = \frac{(1-a)^{2}}{ (1-a)^{2} + ab}

I know that to go from 0 \rightarrow 0  \rightarrow 0 would be (1-a)^{2} and the only other option is 0  \rightarrow 1  \rightarrow 0 which is ab so is it simply that this probability is just \frac{P(0 \rightarrow 0  \rightarrow 0)} {P(0 \rightarrow 0  \rightarrow 0) + P(0 \rightarrow 1  \rightarrow 0)} = \frac{(1-a)^{2}}{ (1-a)^{2} + ab}. This seems far too straightforward! :?

Secondly for another two-state transition probability matrix:
\bf{ T} = \bordermatrix{
& 0 & 1 \cr
0 & \frac{2}{3} & \frac{1}{3} \cr
1 & \frac{1}{4} & \frac{1}{4} \cr}
Given the chain is in state 0 at time n, what is the probability it was in state 0 at time n-1?

I wasn't sure at all about this either, the only thing I could think of was calculating \frac{\frac{2}{3}}{\frac{2}{3}+\frac{1}{4}}=\frac{8}{11} which sounds reasonable but could be a complete fallacy :lol: Could anyone please give me a thumbs up (or down) on this too?

Many thanks!


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PostPosted: Sun, 6 May 2012 02:13:02 UTC 
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jjgoth wrote:
Hi all hoping you can help me with a couple of questions:

Firstly I have a two-state transition probability matrix:
\bf{ T} = \bordermatrix{
& 0 & 1 \cr
0 & (1-a) & a \cr
1 & b & (1-b) \cr}
where 0<a,b<1. I need to prove that
P(X_{n} = 0 | X_{n-1} =0,X_{n+1} =0) = \frac{(1-a)^{2}}{ (1-a)^{2} + ab}

I know that to go from 0 \rightarrow 0  \rightarrow 0 would be (1-a)^{2} and the only other option is 0  \rightarrow 1  \rightarrow 0 which is ab so is it simply that this probability is just \frac{P(0 \rightarrow 0  \rightarrow 0)} {P(0 \rightarrow 0  \rightarrow 0) + P(0 \rightarrow 1  \rightarrow 0)} = \frac{(1-a)^{2}}{ (1-a)^{2} + ab}. This seems far too straightforward! :?

Secondly for another two-state transition probability matrix:
\bf{ T} = \bordermatrix{
& 0 & 1 \cr
0 & \frac{2}{3} & \frac{1}{3} \cr
1 & \frac{1}{4} & \frac{1}{4} \cr}
Given the chain is in state 0 at time n, what is the probability it was in state 0 at time n-1?

I wasn't sure at all about this either, the only thing I could think of was calculating \frac{\frac{2}{3}}{\frac{2}{3}+\frac{1}{4}}=\frac{8}{11} which sounds reasonable but could be a complete fallacy :lol: Could anyone please give me a thumbs up (or down) on this too?

Many thanks!


That seems pretty reasonable to me, why do you think it isn't what you should do?

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PostPosted: Sun, 6 May 2012 02:27:01 UTC 
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Joined: Sun, 6 May 2012 01:08:03 UTC
Posts: 5
It's just I haven't answered questions of this form so just used intuition so could easily be wrong! That's cool though if it's right, thanks :D


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PostPosted: Sun, 6 May 2012 02:34:55 UTC 
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:)

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PostPosted: Mon, 7 May 2012 10:23:38 UTC 
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Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
Shadow wrote:
jjgoth wrote:
Hi all hoping you can help me with a couple of questions:

Firstly I have a two-state transition probability matrix:
\bf{ T} = \bordermatrix{
& 0 & 1 \cr
0 & (1-a) & a \cr
1 & b & (1-b) \cr}
where 0<a,b<1. I need to prove that
P(X_{n} = 0 | X_{n-1} =0,X_{n+1} =0) = \frac{(1-a)^{2}}{ (1-a)^{2} + ab}

I know that to go from 0 \rightarrow 0  \rightarrow 0 would be (1-a)^{2} and the only other option is 0  \rightarrow 1  \rightarrow 0 which is ab so is it simply that this probability is just \frac{P(0 \rightarrow 0  \rightarrow 0)} {P(0 \rightarrow 0  \rightarrow 0) + P(0 \rightarrow 1  \rightarrow 0)} = \frac{(1-a)^{2}}{ (1-a)^{2} + ab}. This seems far too straightforward! :?

Secondly for another two-state transition probability matrix:
\bf{ T} = \bordermatrix{
& 0 & 1 \cr
0 & \frac{2}{3} & \frac{1}{3} \cr
1 & \frac{1}{4} & \frac{1}{4} \cr}
Given the chain is in state 0 at time n, what is the probability it was in state 0 at time n-1?

I wasn't sure at all about this either, the only thing I could think of was calculating \frac{\frac{2}{3}}{\frac{2}{3}+\frac{1}{4}}=\frac{8}{11} which sounds reasonable but could be a complete fallacy :lol: Could anyone please give me a thumbs up (or down) on this too?

Many thanks!


That seems pretty reasonable to me, why do you think it isn't what you should do?


The second one is wrong --- you don't even have a stochastic matrix, and you should use Bayes' theorem.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Mon, 7 May 2012 13:27:32 UTC 
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outermeasure wrote:
Shadow wrote:
jjgoth wrote:
Hi all hoping you can help me with a couple of questions:

Firstly I have a two-state transition probability matrix:
\bf{ T} = \bordermatrix{
& 0 & 1 \cr
0 & (1-a) & a \cr
1 & b & (1-b) \cr}
where 0<a,b<1. I need to prove that
P(X_{n} = 0 | X_{n-1} =0,X_{n+1} =0) = \frac{(1-a)^{2}}{ (1-a)^{2} + ab}

I know that to go from 0 \rightarrow 0  \rightarrow 0 would be (1-a)^{2} and the only other option is 0  \rightarrow 1  \rightarrow 0 which is ab so is it simply that this probability is just \frac{P(0 \rightarrow 0  \rightarrow 0)} {P(0 \rightarrow 0  \rightarrow 0) + P(0 \rightarrow 1  \rightarrow 0)} = \frac{(1-a)^{2}}{ (1-a)^{2} + ab}. This seems far too straightforward! :?

Secondly for another two-state transition probability matrix:
\bf{ T} = \bordermatrix{
& 0 & 1 \cr
0 & \frac{2}{3} & \frac{1}{3} \cr
1 & \frac{1}{4} & \frac{1}{4} \cr}
Given the chain is in state 0 at time n, what is the probability it was in state 0 at time n-1?

I wasn't sure at all about this either, the only thing I could think of was calculating \frac{\frac{2}{3}}{\frac{2}{3}+\frac{1}{4}}=\frac{8}{11} which sounds reasonable but could be a complete fallacy :lol: Could anyone please give me a thumbs up (or down) on this too?

Many thanks!


That seems pretty reasonable to me, why do you think it isn't what you should do?


The second one is wrong --- you don't even have a stochastic matrix, and you should use Bayes' theorem.


Right, didn't check on the second one, should have said that for the op's benefit.

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PostPosted: Tue, 8 May 2012 06:27:43 UTC 
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Joined: Sun, 6 May 2012 01:08:03 UTC
Posts: 5
My bad, lack of sleep is getting to me...

\bf{ T} = \bordermatrix{
& 0 & 1 \cr
0 & \frac{2}{3} & \frac{1}{3} \cr
1 & \frac{1}{4} & \frac{3}{4} \cr}
Given the chain is in state 0 at time n, what is the probability it was in state 0 at time n-1?

So using Bayes' theorem I get:
P(n-1=0|n=0) = \dfrac{P(n=0|n-1=0) \times P (n-1=0)}{P(n=0)}= \dfrac{\frac{2}{3} \times \frac{\frac{2}{3}}{\frac{2}{3} + \frac{3}{4}}}{1} = \frac{16}{51} \approx 0.3137

Is that better? :)


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PostPosted: Tue, 8 May 2012 06:49:32 UTC 
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6814
Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
jjgoth wrote:
My bad, lack of sleep is getting to me...

\bf{ T} = \bordermatrix{
& 0 & 1 \cr
0 & \frac{2}{3} & \frac{1}{3} \cr
1 & \frac{1}{4} & \frac{3}{4} \cr}
Given the chain is in state 0 at time n, what is the probability it was in state 0 at time n-1?

So using Bayes' theorem I get:
P(n-1=0|n=0) = \dfrac{P(n=0|n-1=0) \times P (n-1=0)}{P(n=0)}= \dfrac{\frac{2}{3} \times \frac{\frac{2}{3}}{\frac{2}{3} + \frac{3}{4}}}{1} = \frac{16}{51} \approx 0.3137

Is that better? :)


No.

Unless you assume your starting distribution \mathbb{P}(X_{n-1}=0)=..., you won't get \mathbb{P}(X_{n-1}=0\mid X_n=0)=\frac{16}{51}. However, there is no reason why you would want, a priori, X_{n-1} to have that distribution (and another problem is whether that is reachable from a distribution of X_0, for all n). On the other hand, if you start with the invariant distribution, and solve for your reverse time Markov chain ...

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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