Random walk on non-negative integers - Transcience Recurrence etc
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Good evening. I have a problem concerning the random walk on non-negative integers. Suppose that $ p_{0,1}=1 , p_{i,i+1}=1-p , p_{i,i-1}=p $ . I would like to know for which value of $ p $ this random walk is transient/recurrent. I tried to apply the classical theorem about calculating the probabilities $ p_{0,0}^{(2n)} $ and then try to tell if the series over $ n $ converges, but it seems really hard to find an explicit formula about these n-th transition probabilities.
Does anybody have any idea about this?
markov-chains
asked Jan 24 at 16:25
Petros KarajanPetros Karajan
113
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add a comment |
$begingroup$
Good evening. I have a problem concerning the random walk on non-negative integers. Suppose that $ p_{0,1}=1 , p_{i,i+1}=1-p , p_{i,i-1}=p $ . I would like to know for which value of $ p $ this random walk is transient/recurrent. I tried to apply the classical theorem about calculating the probabilities $ p_{0,0}^{(2n)} $ and then try to tell if the series over $ n $ converges, but it seems really hard to find an explicit formula about these n-th transition probabilities.
Does anybody have any idea about this?
markov-chains
asked Jan 24 at 16:25
Petros KarajanPetros Karajan
113
$endgroup$
$begingroup$
This is asked before on some link. But a simple start is to assume you are in a state $i>0$ and define $q$ as the probability that you ever (eventually) reach a point to your left (i.e., $i-1$). You can write a quadratic equation for $q$ in terms of $p$ by simply considering the two possibilities associated with the first step.
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– Michael
Jan 24 at 16:38
$begingroup$
You mean to define $ T_{i-1}= inf { j geq 1 : X_j = i-1 } $ and then calculate $ P [ T_{i-1} < infty | X_0 = i ] $ by conditioning on the first step ??
$endgroup$
– Petros Karajan
Jan 24 at 16:58
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Yes. You will notice that $T_i$ and $T_{i+1}$ have the same distribution for $i geq 1$.
$endgroup$
– Michael
Jan 24 at 18:06
add a comment |
$begingroup$
Good evening. I have a problem concerning the random walk on non-negative integers. Suppose that $ p_{0,1}=1 , p_{i,i+1}=1-p , p_{i,i-1}=p $ . I would like to know for which value of $ p $ this random walk is transient/recurrent. I tried to apply the classical theorem about calculating the probabilities $ p_{0,0}^{(2n)} $ and then try to tell if the series over $ n $ converges, but it seems really hard to find an explicit formula about these n-th transition probabilities.
Does anybody have any idea about this?
markov-chains
asked Jan 24 at 16:25
Petros KarajanPetros Karajan
113
$endgroup$
Good evening. I have a problem concerning the random walk on non-negative integers. Suppose that $ p_{0,1}=1 , p_{i,i+1}=1-p , p_{i,i-1}=p $ . I would like to know for which value of $ p $ this random walk is transient/recurrent. I tried to apply the classical theorem about calculating the probabilities $ p_{0,0}^{(2n)} $ and then try to tell if the series over $ n $ converges, but it seems really hard to find an explicit formula about these n-th transition probabilities.
Does anybody have any idea about this?
markov-chains
markov-chains
asked Jan 24 at 16:25
Petros KarajanPetros Karajan
113
asked Jan 24 at 16:25
Petros KarajanPetros Karajan
113
asked Jan 24 at 16:25
Petros KarajanPetros Karajan
113
asked Jan 24 at 16:25
Petros KarajanPetros Karajan
113
asked Jan 24 at 16:25
Petros KarajanPetros Karajan
113
113
$begingroup$
This is asked before on some link. But a simple start is to assume you are in a state $i>0$ and define $q$ as the probability that you ever (eventually) reach a point to your left (i.e., $i-1$). You can write a quadratic equation for $q$ in terms of $p$ by simply considering the two possibilities associated with the first step.
$endgroup$
– Michael
Jan 24 at 16:38
$begingroup$
You mean to define $ T_{i-1}= inf { j geq 1 : X_j = i-1 } $ and then calculate $ P [ T_{i-1} < infty | X_0 = i ] $ by conditioning on the first step ??
$endgroup$
– Petros Karajan
Jan 24 at 16:58
$begingroup$
Yes. You will notice that $T_i$ and $T_{i+1}$ have the same distribution for $i geq 1$.
$endgroup$
– Michael
Jan 24 at 18:06
add a comment |
$begingroup$
This is asked before on some link. But a simple start is to assume you are in a state $i>0$ and define $q$ as the probability that you ever (eventually) reach a point to your left (i.e., $i-1$). You can write a quadratic equation for $q$ in terms of $p$ by simply considering the two possibilities associated with the first step.
$endgroup$
– Michael
Jan 24 at 16:38
$begingroup$
You mean to define $ T_{i-1}= inf { j geq 1 : X_j = i-1 } $ and then calculate $ P [ T_{i-1} < infty | X_0 = i ] $ by conditioning on the first step ??
$endgroup$
– Petros Karajan
Jan 24 at 16:58
$begingroup$
Yes. You will notice that $T_i$ and $T_{i+1}$ have the same distribution for $i geq 1$.
$endgroup$
– Michael
Jan 24 at 18:06
$begingroup$
This is asked before on some link. But a simple start is to assume you are in a state $i>0$ and define $q$ as the probability that you ever (eventually) reach a point to your left (i.e., $i-1$). You can write a quadratic equation for $q$ in terms of $p$ by simply considering the two possibilities associated with the first step.
$endgroup$
– Michael
Jan 24 at 16:38
$begingroup$
This is asked before on some link. But a simple start is to assume you are in a state $i>0$ and define $q$ as the probability that you ever (eventually) reach a point to your left (i.e., $i-1$). You can write a quadratic equation for $q$ in terms of $p$ by simply considering the two possibilities associated with the first step.
$endgroup$
– Michael
Jan 24 at 16:38
$begingroup$
You mean to define $ T_{i-1}= inf { j geq 1 : X_j = i-1 } $ and then calculate $ P [ T_{i-1} < infty | X_0 = i ] $ by conditioning on the first step ??
$endgroup$
– Petros Karajan
Jan 24 at 16:58
$begingroup$
You mean to define $ T_{i-1}= inf { j geq 1 : X_j = i-1 } $ and then calculate $ P [ T_{i-1} < infty | X_0 = i ] $ by conditioning on the first step ??
$endgroup$
– Petros Karajan
Jan 24 at 16:58
$begingroup$
Yes. You will notice that $T_i$ and $T_{i+1}$ have the same distribution for $i geq 1$.
$endgroup$
– Michael
Jan 24 at 18:06
$begingroup$
Yes. You will notice that $T_i$ and $T_{i+1}$ have the same distribution for $i geq 1$.
$endgroup$
– Michael
Jan 24 at 18:06
add a comment |
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$begingroup$
This is asked before on some link. But a simple start is to assume you are in a state $i>0$ and define $q$ as the probability that you ever (eventually) reach a point to your left (i.e., $i-1$). You can write a quadratic equation for $q$ in terms of $p$ by simply considering the two possibilities associated with the first step.
$endgroup$
– Michael
Jan 24 at 16:38
$begingroup$
You mean to define $ T_{i-1}= inf { j geq 1 : X_j = i-1 } $ and then calculate $ P [ T_{i-1} < infty | X_0 = i ] $ by conditioning on the first step ??
$endgroup$
– Petros Karajan
Jan 24 at 16:58
$begingroup$
Yes. You will notice that $T_i$ and $T_{i+1}$ have the same distribution for $i geq 1$.
$endgroup$
– Michael
Jan 24 at 18:06