Dear readers,

The variables and constants are more or less based on the definitons in my previous thread: Rewrite variable in given constants and variables

There exists a stationary population C, so:
\frac{C_{j+1}(t+1)}{C_j(t+1)} = \frac{C_{j+1}(t)}{C_j(t)}, where j \in \{0, ..., l-1\} and t = \{0, 1, ...\}
( C_j(t) means the amount of people in age group j at time t. There are l+1 age groups, labeled from 0 to l)


Let \phi = (\phi_0, \phi_1, ..., \phi_l) with:

\phi_i = \phi_i(t) = \frac{C_i(t)}{\sum^l_{k=0} C_k(t)}, i \in \{0, 1, ..., l\}


Now I have to show for a arbitrary population A that is converges to the stationary distribution \phi. To prove that, let A and B be 2 general populations.

And let m(t) = \min_{n=0,...,r} \frac{A_n(t)}{B_n(t)}
( 0 < r < l. There are r + 1 fertile age groups, labeled 0 to r)


------


Show that m(t) \leq m(t+1) by using the properties below:

A_{j+1}(t+1) = A_j(t)p_j with j \in \{0, ..., l - 1\}
A_0(t+1) = \sum^r_{k=0} A_k(t)f_k
(these also yield for B)

and

\frac{A_0(t+1)}{B_0(t+1)} = \frac{\sum^r_{k=0} A_k(t)f_k}{\sum^r_{k=0} B_k(t)f_k} = \frac{\sum^r_{k=0} \frac{A_k(t)}{B_k(t)}B_k(t)f_k}{\sum^r_{k=0} B_k(t)f_k}

p_j is the probability of survival for people in age group j. So the amount of people A_{j+1}(t+1) is equal to the amount of people A_j(t) multiplied by the probablity of survival p_j. Hence, A_{j+1}(t+1) = A_j(t)p_j

This is different for the amount of people A_0, because there are no things like A_{-1} and p_{-1}. So this is a special case. f_k is the fertility factor of the people in age group k. k is between 0 and r, where r < l, because people stop 'producing' children when they get past a certain age.

-----


My reasoning:

m(t) = \min_{n=0,...,r} \frac{A_n(t)}{B_n(t)} = \min\left\{\frac{A_0(t)}{B_0(t)}, ..., \frac{A_r(t)}{B_r(t)}\right\}

m(t+1) = \min_{n=0,...,r} \frac{A_n(t+1)}{B_n(t+1)} = \min\left\{\frac{A_0(t+1)}{B_0(t+1)}, ..., \frac{A_r(t+1)}{B_r(t+1)}\right\} =
= \min\left\{\frac{\sum^r_{k=0} A_k(t)f_k}{\sum^r_{k=0} B_k(t)f_k}, \frac{A_0(t)p_0}{B_0(t)p_0}, ..., \frac{A_{r-1}(t)p_{r-1}}{B_{r-1}(t)p_{r-1}}\right\}

The survival probabilities of m(t+1) are in the numerator and denominator, so they cancel each other out. So the elements in the minimumfunction of m(t) and m(t+1) are mostly the same.

The only two differences are that m(t) has got the fraction \frac{A_r(t)}{B_r(t)} and m(t+1) hasn't.
On the other hand m(t+1) has got \frac{\sum^r_{k=0} A_k(t)f_k}{\sum^r_{k=0} B_k(t)f_k} and m(t) hasn't.


In short, when \frac{A_r(t)}{B_r(t)} \leq \frac{\sum^r_{k=0} A_k(t)f_k}{\sum^r_{k=0} B_k(t)f_k} for all t \in \{0, 1, ...\}, then m(t) \leq m(t+1).


I thought that:
\frac{\sum^r_{k=0} A_k(t)f_k}{\sum^r_{k=0} B_k(t)f_k} = \sum^r_{k=0} \frac{A_k(t)}{B_k(t)}, but that doesn't help me either.



-----


Anyone who can lend me a hand?
- Fruitbowl.