fundamental theorem of demography, proof of

$\bullet$ First we will prove that there exist $m,M>0$ such that

$m\leq\frac{\|x_{k+1}\|}{\|x_{k}\|}\leq M$

(1)

for all $k$ , with $m$ and $M$ of the sequence. In to show this we use the primitivity of the matrices $A_{k}$ and $A_{\infty}$ . Primitivity of $A_{\infty}$ implies that there exists $l\in\mathbb{N}$ such that

$A^{l}_{\infty}\gg 0$

By continuity, this implies that there exists $k_{0}$ such that, for all $k\geq k_{0}$ , we have

$A_{k+l}A_{k+l-1}\cdots A_{k}\gg 0$

Let us then write $x_{k+l+1}$ as a function of $x_{k}$ :

$x_{k+l+1}=A_{k+l}\cdots A_{k}x_{k}$

We thus have

$\|x_{k+l+1}\|\leq C^{l+1}\|x_{k}\|$

(2)

But since the matrices $A_{k+l}$ ,…, $A_{k}$ are strictly positive for $k\geq k_{0}$ , there exists a $\varepsilon>0$ such that each of these matrices is superior or equal to $\varepsilon$ . From this we deduce that

$\|x_{k+l+1}\|\geq\varepsilon\|x_{k}\|$

for all $k\geq k_{0}$ . Applying (2), we then have that

$C^{l}\|x_{k+1}\|\geq\varepsilon\|x_{k}\|$

which yields

$\|x_{k+1}\|\geq\frac{\varepsilon}{C^{l}}\|x_{k}\|$

for all $k\geq 0$ , and so we indeed have (1).

$\bullet$ Let us denote by $e_{k}$ the (normalised) Perron eigenvector of $A_{k}$ . Thus

$A_{k}e_{k}=\lambda_{k}e_{k}\quad\|e_{k}\|=1$

Let us denote by $\pi_{k}$ the projection on the supplementary space of $\{e_{k}\}$ invariant by $A_{k}$ . Choosing a proper norm, we can find $\varepsilon>0$ such that

$\left|A_{k}\pi_{k}\right|\leq(\lambda_{k}-\varepsilon)$

for all $k$ . $\bullet$ We shall now prove that

$\frac{\left<e_{k+1}^{*},x_{k+1}\right>}{\left<e_{k}^{*},x_{k}\right>}\to% \lambda_{\infty}\textrm{ when }k\to\infty$

In order to do this, we compute the inner product of the sequence $x_{k+1}=A_{k}x_{k}$ with the $e_{k}$ ’s:

	$\displaystyle\left<e_{k+1}^{*},x_{k+1}\right>$	$\displaystyle=$	$\displaystyle\left<e_{k+1}^{}-e_{k}^{},A_{k}x_{k}\right>+\lambda_{k}\left<e_% {k}^{*},x_{k}\right>$
		$\displaystyle=$	$\displaystyle o\left(\left<e_{k}^{},x_{k}\right>\right)+\lambda_{k}\left<e_{k% }^{},x_{k}\right>$

Therefore we have

$\frac{\left<e_{k+1}^{*},x_{k+1}\right>}{\left<e_{k}^{*},x_{k}\right>}=o(1)+% \lambda_{k}$

$\bullet$ Now assume

$u_{k}=\frac{\pi_{k}x_{k}}{\left<e_{k}^{*},x_{k}\right>}$

We will verify that $u_{k}\to 0$ when $k\to\infty$ . We have

$\displaystyle u_{k+1}$

$\displaystyle=$

$\displaystyle(\pi_{k+1}-\pi_{k})A_{k}\frac{x_{k}}{\left<e_{k+1}^{*},x_{k+1}% \right>}+\frac{\left<e_{k}^{*},x_{k}\right>}{\left<e_{k}^{*},x_{k+1}\right>}A_% {k}\pi_{k}\frac{x_{k}}{\left<e_{k}^{*},x_{k}\right>}$

and so

$|u_{k+1}|\leq|\pi_{k+1}-\pi_{k}|C^{\prime}+\frac{\left<e_{k}^{*},x_{k}\right>}% {\left<e_{k+1}^{*},x_{k+1}\right>}(\lambda_{k}-\varepsilon)|u_{k}|$

We deduce that there exists $k_{1}\geq k_{0}$ such that, for all $k\geq k_{1}$

$|u_{k+1}|\leq\delta_{k}+(\lambda_{\infty}-\frac{\varepsilon}{2})|u_{k}|$

where we have noted

$\delta_{k}=(\pi_{k+1}-\pi_{k})C^{\prime}$

We have $\delta_{k}\to 0$ when $t\to\infty$ , we thus finally deduce that

$|u_{k}|\to 0\textrm{ when }k\to\infty$

Remark that this also implies that

$z_{k}=\frac{\pi_{k}x_{k}}{\|x_{k}\|}\to 0\textrm{ when }k\to\infty$

$\bullet$ We have $z_{k}\to 0$ when $k\to\infty$ , and $x_{k}/\|x_{k}\|$ can be written

$\frac{x_{k}}{\|x_{k}\|}=\alpha_{k}e_{k}+z_{k}$

Therefore, we have $\alpha_{k}e_{k}\to 1$ when $k\to\infty$ , which implies that $\alpha_{k}$ tends to 1, since we have chosen $e_{k}$ to be normalised (i.e., $\|e_{k}\|=1$ ).

We then can conclude that

$\frac{x_{k}}{\|x_{k}\|}\to e_{\infty}\textrm{ when }k\to\infty$

and the proof is done.

Title	fundamental theorem of demography, proof of
Canonical name	FundamentalTheoremOfDemographyProofOf
Date of creation	2013-03-22 13:24:42
Last modified on	2013-03-22 13:24:42
Owner	aplant (12431)
Last modified by	aplant (12431)
Numerical id	10
Author	aplant (12431)
Entry type	Proof
Classification	msc 92D25
Classification	msc 37A30