Results 1 to 2 of 2

Math Help - continous Markov Chain

  1. #1
    Senior Member
    Joined
    Jan 2008
    From
    Montreal
    Posts
    311
    Awards
    1

    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 =<br />
\left\{ \begin{array}{rcl}<br />
\lambda+\theta, & \mbox{if} & \mbox{population} <K \\<br />
\lambda, & \mbox{if} & \mbox{population} \geq K<br />
\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]}<br />
-\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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Nov 2008
    Posts
    58
    Thanks
    1
    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 =<br />
\left\{ \begin{array}{rcl}<br />
\lambda+\theta, & \mbox{if} & \mbox{i} <K \\<br />
\lambda, & \mbox{if} & \mbox{i} \geq K<br />
\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]}<br />
-\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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Markov Chain of random variables from a primitive markov chain
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: October 19th 2011, 09:12 AM
  2. Markov chain
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: November 20th 2010, 02:40 AM
  3. Markov chain
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: March 24th 2010, 10:58 AM
  4. Replies: 2
    Last Post: October 28th 2008, 07:32 PM
  5. Markov chain
    Posted in the Advanced Statistics Forum
    Replies: 7
    Last Post: March 15th 2008, 12:37 PM

Search Tags


/mathhelpforum @mathhelpforum