Each individual in a biological population is assumed to give birth at an exponential rate λ and to die at an exponential rate µ. In addition, there is an exponential rate of increase θ due to immigration. However, immigration is not allowed when the population size is N or larger.a) Set this up as a birth and death model.b) If N=3, λ=θ=1, and μ=2, determine the proportion of time that immigration is restricted.

Respuesta :

Answer:

π0 = 0.208

Step-by-step explanation:

check the attachment for explicit explanations.

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Thus, the proportion of  time is restricted to:

[tex]\pi = 0.208[/tex]

Given that an exponential rate [tex]\lambda[/tex],  at an expoential rate [tex]\mu[/tex]. An exponential rate of increase [tex]\theta[/tex] : the popuilation size is N.

Hence the total birth rate where there are n persons in the system is: [tex]n\lambda+\theta[/tex].  are assume to accury at an expoential rate [tex]\mu[/tex] for each member of the population. So: [tex]n\mu = \mu n[/tex]

Let x(t) denote the population size at time t. Suppose: [tex]x(0)= ix(0)= i[/tex] and let [tex]M(t)= E[x(t)][/tex]. So they will determine M(t) by deriving and then solving a differential equation that is :

We start by deriving on equation for M(t+h) by conditioning on X(t) this yelds:

[tex]M(t+h)= E[x(t+h)]= E(E[x(t+h)] x(t))[/tex].

Now, given the size of the population at the time then, ignoring events whose probability is [tex]\theta (h)[/tex]. The population at time t+h. Will be either increase in size by 1. If a birth or immigration occurs in (t, t+h) or decrease by 1 if a death occurs in this internal, or remain the same if neither of these two possibilities occurs that is given [tex]x(t)*x(t)[/tex]:

[tex]x(t+h)x(t+h)=x(t)+1[/tex]  with pobability

[tex][\theta +x(t)x]h +\theta h \\[/tex]

X(t)-1 with probability [tex]x(t)\mu h +\theta (h)[/tex]

x(t) with probability [tex]1-[\theta + x(t) \lambda + x(t) \mu ] h + \theta(h)[/tex]

[tex]E[x(t+h)x(t)]= x(t)+[\theta+ x(t)\lambda-x(t)\mu]h+\theta(h)[/tex]

Since the probabilities have to add up to 1, that determines. The value of the probabiliy measure for he best case.

For [tex]0\leq i\leq N[/tex]

we have [tex]\lambda i = i \lambda + \theta[/tex]

For [tex]i \geq N[/tex] we have [tex]\mu i = i \mu[/tex]

[tex]\pi[/tex] has [tex]\pi \theta = \frac{1}{1+ \sum \phi n }[/tex]

where [tex]\phi n = \frac{\lambda}{\mu}[/tex]

When [tex]\theta=\lambda[/tex]

For [tex]n\geq 3[/tex] we see [tex]\phi n + \frac{1}{\phi n } = \frac{n \lambda}{(n+\mu) \mu}[/tex]

This leads to [tex]\pi = 0.208[/tex]

Learn more: brainly.com/question/17001402