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Old November 30th, 2008, 03:47 PM
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Default continous Markov Chain

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

\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 \mu

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

so for 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:

\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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Old December 2nd, 2008, 10:12 AM
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Quote:
Originally Posted by lllll View Post
Assume that in a city the population regenerates itself at an exponential rate \lambda, and dies at an exponential rate \mu. Furthermore assume that immigrants arrive according to an exponential rate \theta; when the population in the given city reaches size K no further immigrants are permitted. If K=2, \ \lambda =1, \ \theta=1 and \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:

\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 \mu

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

so for 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:

\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 P_n, then the long run proportion of times that the city will stop accepting immigrant is 1-P_0-P_1
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