Total time spent in state $i$ before reaching state $i+1$
Let $T_k$ be the total cumulative time spent in state $i$ before reaching state $i+1$ in a continuous time birth-death process, with birth and death rates $lambda_k$ and $mu_k$. What is the distribution of $T_k$?
To solve this problem, I tried to set up the equation:
$P(T_k < z) = sum_{n=0}^infty P(T_k< z | N_k=n)*P(N_k=n)$
But, I ran into a problem calculating both $P(N_k=n)$, which should be geometrically distributed with $p=1-rho_{k,k}$, where $rho_{k,k}$ is the probability of returning to state k from state k in finite time. $P(T_k< z | N_k=n)$ having a gamma distribution also seems to complicate the problem.
probability probability-distributions markov-chains birth-death-process
asked Nov 27 at 3:35
jrstevens
84
add a comment |
Let $T_k$ be the total cumulative time spent in state $i$ before reaching state $i+1$ in a continuous time birth-death process, with birth and death rates $lambda_k$ and $mu_k$. What is the distribution of $T_k$?
To solve this problem, I tried to set up the equation:
$P(T_k < z) = sum_{n=0}^infty P(T_k< z | N_k=n)*P(N_k=n)$
But, I ran into a problem calculating both $P(N_k=n)$, which should be geometrically distributed with $p=1-rho_{k,k}$, where $rho_{k,k}$ is the probability of returning to state k from state k in finite time. $P(T_k< z | N_k=n)$ having a gamma distribution also seems to complicate the problem.
probability probability-distributions markov-chains birth-death-process
asked Nov 27 at 3:35
jrstevens
84
$rho_{k,k}$ is really the probability to return to $k$ starting from $k$ without hitting $k+1$ first. Finite time doesn't really have anything to do with it because the event of oscillating strictly below $k+1$ forever will have zero probability.
– Ian
Nov 27 at 3:39
I see. So then $rho_{k,k}$ is really the probability that we move from $k$ to $k-1$, or $frac{mu_k}{mu_k+lambda_k}$
– jrstevens
Nov 27 at 13:41
That is correct.
– Ian
Nov 27 at 14:10
add a comment |
Let $T_k$ be the total cumulative time spent in state $i$ before reaching state $i+1$ in a continuous time birth-death process, with birth and death rates $lambda_k$ and $mu_k$. What is the distribution of $T_k$?
To solve this problem, I tried to set up the equation:
$P(T_k < z) = sum_{n=0}^infty P(T_k< z | N_k=n)*P(N_k=n)$
But, I ran into a problem calculating both $P(N_k=n)$, which should be geometrically distributed with $p=1-rho_{k,k}$, where $rho_{k,k}$ is the probability of returning to state k from state k in finite time. $P(T_k< z | N_k=n)$ having a gamma distribution also seems to complicate the problem.
probability probability-distributions markov-chains birth-death-process
asked Nov 27 at 3:35
jrstevens
84
Let $T_k$ be the total cumulative time spent in state $i$ before reaching state $i+1$ in a continuous time birth-death process, with birth and death rates $lambda_k$ and $mu_k$. What is the distribution of $T_k$?
To solve this problem, I tried to set up the equation:
$P(T_k < z) = sum_{n=0}^infty P(T_k< z | N_k=n)*P(N_k=n)$
But, I ran into a problem calculating both $P(N_k=n)$, which should be geometrically distributed with $p=1-rho_{k,k}$, where $rho_{k,k}$ is the probability of returning to state k from state k in finite time. $P(T_k< z | N_k=n)$ having a gamma distribution also seems to complicate the problem.
probability probability-distributions markov-chains birth-death-process
probability probability-distributions markov-chains birth-death-process
asked Nov 27 at 3:35
jrstevens
84
asked Nov 27 at 3:35
jrstevens
84
asked Nov 27 at 3:35
jrstevens
84
asked Nov 27 at 3:35
jrstevens
84
asked Nov 27 at 3:35
jrstevens
84
84
$rho_{k,k}$ is really the probability to return to $k$ starting from $k$ without hitting $k+1$ first. Finite time doesn't really have anything to do with it because the event of oscillating strictly below $k+1$ forever will have zero probability.
– Ian
Nov 27 at 3:39
I see. So then $rho_{k,k}$ is really the probability that we move from $k$ to $k-1$, or $frac{mu_k}{mu_k+lambda_k}$
– jrstevens
Nov 27 at 13:41
That is correct.
– Ian
Nov 27 at 14:10
add a comment |
$rho_{k,k}$ is really the probability to return to $k$ starting from $k$ without hitting $k+1$ first. Finite time doesn't really have anything to do with it because the event of oscillating strictly below $k+1$ forever will have zero probability.
– Ian
Nov 27 at 3:39
I see. So then $rho_{k,k}$ is really the probability that we move from $k$ to $k-1$, or $frac{mu_k}{mu_k+lambda_k}$
– jrstevens
Nov 27 at 13:41
That is correct.
– Ian
Nov 27 at 14:10
$rho_{k,k}$ is really the probability to return to $k$ starting from $k$ without hitting $k+1$ first. Finite time doesn't really have anything to do with it because the event of oscillating strictly below $k+1$ forever will have zero probability.
– Ian
Nov 27 at 3:39
$rho_{k,k}$ is really the probability to return to $k$ starting from $k$ without hitting $k+1$ first. Finite time doesn't really have anything to do with it because the event of oscillating strictly below $k+1$ forever will have zero probability.
– Ian
Nov 27 at 3:39
I see. So then $rho_{k,k}$ is really the probability that we move from $k$ to $k-1$, or $frac{mu_k}{mu_k+lambda_k}$
– jrstevens
Nov 27 at 13:41
I see. So then $rho_{k,k}$ is really the probability that we move from $k$ to $k-1$, or $frac{mu_k}{mu_k+lambda_k}$
– jrstevens
Nov 27 at 13:41
That is correct.
– Ian
Nov 27 at 14:10
That is correct.
– Ian
Nov 27 at 14:10
add a comment |
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$rho_{k,k}$ is really the probability to return to $k$ starting from $k$ without hitting $k+1$ first. Finite time doesn't really have anything to do with it because the event of oscillating strictly below $k+1$ forever will have zero probability.
– Ian
Nov 27 at 3:39
I see. So then $rho_{k,k}$ is really the probability that we move from $k$ to $k-1$, or $frac{mu_k}{mu_k+lambda_k}$
– jrstevens
Nov 27 at 13:41
That is correct.
– Ian
Nov 27 at 14:10