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Thread: continous Markov Chain

  1. #1
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    continous Markov Chain

    Assume that in a city the population regenerates itself at an exponential rate $\displaystyle \lambda$, and dies at an exponential rate $\displaystyle \mu$. Furthermore assume that immigrants arrive according to an exponential rate $\displaystyle \theta$; when the population in the given city reaches size $\displaystyle K$ no further immigrants are permitted. If $\displaystyle K=2, \ \lambda =1, \ \theta=1$ and $\displaystyle \mu=2$ for how long will the city stop accepting new immigrants.

    I figure that this is a birth and death process with different birth rates at different population sizes, thus we would have a birth rate of:

    $\displaystyle \lambda_i =
    \left\{ \begin{array}{rcl}
    \lambda+\theta, & \mbox{if} & \mbox{population} <K \\
    \lambda, & \mbox{if} & \mbox{population} \geq K
    \end{array}\right.$

    and the death rate would simply be $\displaystyle \mu$

    if I wanted to find the expected time the city would reject immigrants I would $\displaystyle 1-E[K(<2)]$, which would give me:

    so for $\displaystyle 1-\bigg{(}E[K=1] +E[K=0]\bigg{)} = 1-\overbrace{\bigg{(} \frac{1}{\lambda_i}+\frac{\mu}{\lambda}(E[T_{i-1}]) \bigg{)}}^{E[T_1]}
    -\overbrace{\bigg{(}\frac{1}{\lambda_i}\bigg{)}}^{E[T_0]}$

    calculating the brackets yields:

    $\displaystyle \bigg{(} \frac{1}{1+1}+\frac{2}{1}(E[T_{1-1}])\bigg{)}-\bigg{(}\frac{1}{1+1}\bigg{)}= \bigg{(} \frac{1}{2}+2\left(\frac{1}{2}\right)\bigg{)}-\bigg{(}\frac{1}{2}\bigg{)}=1$

    so my final solution would be: 1-1=0 which doesn't seem right.
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  2. #2
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    Quote Originally Posted by lllll View Post
    Assume that in a city the population regenerates itself at an exponential rate $\displaystyle \lambda$, and dies at an exponential rate $\displaystyle \mu$. Furthermore assume that immigrants arrive according to an exponential rate $\displaystyle \theta$; when the population in the given city reaches size $\displaystyle K$ no further immigrants are permitted. If $\displaystyle K=2, \ \lambda =1, \ \theta=1$ and $\displaystyle \mu=2$ for how long will the city stop accepting new immigrants. Do you mean long run proportion of time?

    I figure that this is a birth and death process with different birth rates at different population sizes, thus we would have a birth rate of:

    $\displaystyle \lambda_i =
    \left\{ \begin{array}{rcl}
    \lambda+\theta, & \mbox{if} & \mbox{i} <K \\
    \lambda, & \mbox{if} & \mbox{i} \geq K
    \end{array}\right.$

    and the death rate would simply be $\displaystyle \mu$

    if I wanted to find the expected time the city would reject immigrants I would $\displaystyle 1-E[K(<2)]$, which would give me:

    so for $\displaystyle 1-\bigg{(}E[K=1] +E[K=0]\bigg{)} = 1-\overbrace{\bigg{(} \frac{1}{\lambda_i}+\frac{\mu}{\lambda}(E[T_{i-1}]) \bigg{)}}^{E[T_1]}
    -\overbrace{\bigg{(}\frac{1}{\lambda_i}\bigg{)}}^{E[T_0]}$

    calculating the brackets yields:

    $\displaystyle \bigg{(} \frac{1}{1+1}+\frac{2}{1}(E[T_{1-1}])\bigg{)}-\bigg{(}\frac{1}{1+1}\bigg{)}= \bigg{(} \frac{1}{2}+2\left(\frac{1}{2}\right)\bigg{)}-\bigg{(}\frac{1}{2}\bigg{)}=1$

    so my final solution would be: 1-1=0 which doesn't seem right.
    You have to draw out the Markov chain and solve for the stationary distribution for each state $\displaystyle P_n$, then the long run proportion of times that the city will stop accepting immigrant is $\displaystyle 1-P_0-P_1$
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