# Inequality with minimum function

• Jan 31st 2013, 10:22 AM
Fruitbowl
Inequality with minimum function

The variables and constants are more or less based on the definitons in my previous thread: http://mathhelpforum.com/advanced-ap...variables.html

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 $r$)

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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

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

Quote:

$p_j$ 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.

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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.

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Anyone who can lend me a hand?
- Fruitbowl.