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Examples of homogeneous Markov chains

Equilibrium distributions of Markov ChainsManipulating ergodic Markov chains in order to make them non-ergodicWhen the sum of independent Markov chains is a Markov chain?Markov Chains that preserve an ordering of the state spaceBayesian Estimator and Markov ChainsMarkov Process, Markov ChainNonhomogenous Chainsexistence of two intercommunicating states that have the same period in a homogeneous markov chainDetermining whether the following are Markov ChainsSumming two independent Markov Chains on a discrete-time and space environment

0

$begingroup$

I have tried to solve this exercises...
Let ${S_{n}}$ be a randon walk (in $Z$), where $S_{0} = 0$ with $p$ to move to rigth, and $q = 1-p$ to move to left.

I want to know if the following processes are homogeneous Markov chains.

$textbf{1. $X_{n} = |S_{n}$|}$
I'm confused because $P(X_{2} = 2 | X_{1}=1)= p+q = 1$, because:

$P(S_{2} = -2|S_{1}=-1) =q$ and $P(S_{2} = 2|S_{1}=1) = p$

but also $P(X_{2} = 0) = 1 $ for the same reason, so I don't know what to do here.

$textbf{2. $Z_{n} = S_{n}-S_{n-1}$} $

I think there is a Markov Chain, but it's not homogeneous because:
Here the space of states is $ {-1, 1 } $ so

$P(Z_{2} = 1| Z_{1}=-1) = P(S_{3}=-1|S_{2}=-2) + P(S_{3}=1|S_{2}=0) = p+p=2p$

and

$P(Z_{1} = 1| Z_{0}=-1) = P(S_{2}=0|S_{1}=-1) = p$

Then it is not homogeneous.

$textbf{3. $Y_{n} = M_{n} - S_{n} $|}$ where $M_{n} = Max { S_k| k epsilon {0,...,n} }$

$P(Y_{2} = 1| Y_{1}=0) = q$ and

$P(Y_{3} = 1| Y_{2}=0) = 2q$

So it is not homogeneous.

Can anybody tell me...I am fine?
And can anybody help me with exercise 1 please

probability stochastic-processes markov-chains markov-process

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asked yesterday

JazzJazz

919

$endgroup$

add a comment | 

0

$begingroup$

I have tried to solve this exercises...
Let ${S_{n}}$ be a randon walk (in $Z$), where $S_{0} = 0$ with $p$ to move to rigth, and $q = 1-p$ to move to left.

I want to know if the following processes are homogeneous Markov chains.

$textbf{1. $X_{n} = |S_{n}$|}$
I'm confused because $P(X_{2} = 2 | X_{1}=1)= p+q = 1$, because:

$P(S_{2} = -2|S_{1}=-1) =q$ and $P(S_{2} = 2|S_{1}=1) = p$

but also $P(X_{2} = 0) = 1 $ for the same reason, so I don't know what to do here.

$textbf{2. $Z_{n} = S_{n}-S_{n-1}$} $

I think there is a Markov Chain, but it's not homogeneous because:
Here the space of states is $ {-1, 1 } $ so

$P(Z_{2} = 1| Z_{1}=-1) = P(S_{3}=-1|S_{2}=-2) + P(S_{3}=1|S_{2}=0) = p+p=2p$

and

$P(Z_{1} = 1| Z_{0}=-1) = P(S_{2}=0|S_{1}=-1) = p$

Then it is not homogeneous.

$textbf{3. $Y_{n} = M_{n} - S_{n} $|}$ where $M_{n} = Max { S_k| k epsilon {0,...,n} }$

$P(Y_{2} = 1| Y_{1}=0) = q$ and

$P(Y_{3} = 1| Y_{2}=0) = 2q$

So it is not homogeneous.

Can anybody tell me...I am fine?
And can anybody help me with exercise 1 please

probability stochastic-processes markov-chains markov-process

share|cite|improve this question

asked yesterday

JazzJazz

919

$endgroup$

add a comment | 

$begingroup$

I have tried to solve this exercises...
Let ${S_{n}}$ be a randon walk (in $Z$), where $S_{0} = 0$ with $p$ to move to rigth, and $q = 1-p$ to move to left.

I want to know if the following processes are homogeneous Markov chains.

$textbf{1. $X_{n} = |S_{n}$|}$
I'm confused because $P(X_{2} = 2 | X_{1}=1)= p+q = 1$, because:

$P(S_{2} = -2|S_{1}=-1) =q$ and $P(S_{2} = 2|S_{1}=1) = p$

but also $P(X_{2} = 0) = 1 $ for the same reason, so I don't know what to do here.

$textbf{2. $Z_{n} = S_{n}-S_{n-1}$} $

I think there is a Markov Chain, but it's not homogeneous because:
Here the space of states is $ {-1, 1 } $ so

$P(Z_{2} = 1| Z_{1}=-1) = P(S_{3}=-1|S_{2}=-2) + P(S_{3}=1|S_{2}=0) = p+p=2p$

and

$P(Z_{1} = 1| Z_{0}=-1) = P(S_{2}=0|S_{1}=-1) = p$

Then it is not homogeneous.

$textbf{3. $Y_{n} = M_{n} - S_{n} $|}$ where $M_{n} = Max { S_k| k epsilon {0,...,n} }$

$P(Y_{2} = 1| Y_{1}=0) = q$ and

$P(Y_{3} = 1| Y_{2}=0) = 2q$

So it is not homogeneous.

Can anybody tell me...I am fine?
And can anybody help me with exercise 1 please

probability stochastic-processes markov-chains markov-process

share|cite|improve this question

asked yesterday

JazzJazz

919

$endgroup$

share|cite|improve this question

asked yesterday

JazzJazz

919

share|cite|improve this question

asked yesterday

JazzJazz

919

asked yesterday

JazzJazz

919

asked yesterday

JazzJazz

919

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I think we have to careful about the exact statement we are trying to show here. I assume that we are taking the definition of homogenous Markov chain to be that ${X_n}_{n geq 1}$ satisfies $P(X_n mid X_1, dots, X_{n-1}) = P(X_n mid X_{n-1})$. Using that definition, we have $S_n = sum_{i=1}^n X_i$, where $X_i sim U({-1,1})$ and the $X_i$ are all independent. Then, we naturally have
$$P(S_n mid S_1, dots, S_{n-1}) = P(sum_{i=1}^n X_i mid X_1, X_1 + X_2, dots, sum_{i=1}^{n-1} X_i) = P(sum_{i=1}^n X_i mid sum_{i=1}^{n-1} X_i)$$

With that in mind, part (1) follows similarly
$$P(|S_n| mid |S_1|, |S_2|, dots, |S_{n-1}|) = P(|sum_{i=1}^n X_i| mid |X_1|, |X_1 + X_2|, dots, |sum_{i=1}^{n-1} X_i|) $$
$$= P(|sum_{i=1}^n X_i| mid |sum_{i=1}^{n-1} X_i|) = P(|S_n| mid |S_{n-1}|)$$

Part (2) follows from the fact that $S_n - S_{n-1} = X_n$ and the fact that the $X_i$ are independent from each other (i.e. they form a trivial Markov chain with no memory).

Hope that helps!

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answered yesterday

silesile

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$begingroup$

I think we have to careful about the exact statement we are trying to show here. I assume that we are taking the definition of homogenous Markov chain to be that ${X_n}_{n geq 1}$ satisfies $P(X_n mid X_1, dots, X_{n-1}) = P(X_n mid X_{n-1})$. Using that definition, we have $S_n = sum_{i=1}^n X_i$, where $X_i sim U({-1,1})$ and the $X_i$ are all independent. Then, we naturally have
$$P(S_n mid S_1, dots, S_{n-1}) = P(sum_{i=1}^n X_i mid X_1, X_1 + X_2, dots, sum_{i=1}^{n-1} X_i) = P(sum_{i=1}^n X_i mid sum_{i=1}^{n-1} X_i)$$

With that in mind, part (1) follows similarly
$$P(|S_n| mid |S_1|, |S_2|, dots, |S_{n-1}|) = P(|sum_{i=1}^n X_i| mid |X_1|, |X_1 + X_2|, dots, |sum_{i=1}^{n-1} X_i|) $$
$$= P(|sum_{i=1}^n X_i| mid |sum_{i=1}^{n-1} X_i|) = P(|S_n| mid |S_{n-1}|)$$

Part (2) follows from the fact that $S_n - S_{n-1} = X_n$ and the fact that the $X_i$ are independent from each other (i.e. they form a trivial Markov chain with no memory).

Hope that helps!

share|cite|improve this answer

answered yesterday

silesile

264

$endgroup$

add a comment | 

0

$begingroup$

I think we have to careful about the exact statement we are trying to show here. I assume that we are taking the definition of homogenous Markov chain to be that ${X_n}_{n geq 1}$ satisfies $P(X_n mid X_1, dots, X_{n-1}) = P(X_n mid X_{n-1})$. Using that definition, we have $S_n = sum_{i=1}^n X_i$, where $X_i sim U({-1,1})$ and the $X_i$ are all independent. Then, we naturally have
$$P(S_n mid S_1, dots, S_{n-1}) = P(sum_{i=1}^n X_i mid X_1, X_1 + X_2, dots, sum_{i=1}^{n-1} X_i) = P(sum_{i=1}^n X_i mid sum_{i=1}^{n-1} X_i)$$

With that in mind, part (1) follows similarly
$$P(|S_n| mid |S_1|, |S_2|, dots, |S_{n-1}|) = P(|sum_{i=1}^n X_i| mid |X_1|, |X_1 + X_2|, dots, |sum_{i=1}^{n-1} X_i|) $$
$$= P(|sum_{i=1}^n X_i| mid |sum_{i=1}^{n-1} X_i|) = P(|S_n| mid |S_{n-1}|)$$

Part (2) follows from the fact that $S_n - S_{n-1} = X_n$ and the fact that the $X_i$ are independent from each other (i.e. they form a trivial Markov chain with no memory).

Hope that helps!

share|cite|improve this answer

answered yesterday

silesile

264

$endgroup$

add a comment | 

$begingroup$

I think we have to careful about the exact statement we are trying to show here. I assume that we are taking the definition of homogenous Markov chain to be that ${X_n}_{n geq 1}$ satisfies $P(X_n mid X_1, dots, X_{n-1}) = P(X_n mid X_{n-1})$. Using that definition, we have $S_n = sum_{i=1}^n X_i$, where $X_i sim U({-1,1})$ and the $X_i$ are all independent. Then, we naturally have
$$P(S_n mid S_1, dots, S_{n-1}) = P(sum_{i=1}^n X_i mid X_1, X_1 + X_2, dots, sum_{i=1}^{n-1} X_i) = P(sum_{i=1}^n X_i mid sum_{i=1}^{n-1} X_i)$$

With that in mind, part (1) follows similarly
$$P(|S_n| mid |S_1|, |S_2|, dots, |S_{n-1}|) = P(|sum_{i=1}^n X_i| mid |X_1|, |X_1 + X_2|, dots, |sum_{i=1}^{n-1} X_i|) $$
$$= P(|sum_{i=1}^n X_i| mid |sum_{i=1}^{n-1} X_i|) = P(|S_n| mid |S_{n-1}|)$$

Part (2) follows from the fact that $S_n - S_{n-1} = X_n$ and the fact that the $X_i$ are independent from each other (i.e. they form a trivial Markov chain with no memory).

Hope that helps!

share|cite|improve this answer

answered yesterday

silesile

264

$endgroup$

share|cite|improve this answer

answered yesterday

silesile

264

answered yesterday

silesile

264

answered yesterday

silesile

264