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cl85 · #2 December 2nd, 2008, 10:12 AM

Quote:

Originally Posted by lllll

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 $1-P_0-P_1$

The following users thank cl85 for this useful post:
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