# 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

 $\frac{\left}{\left}\to% \lambda_{\infty}\textrm{ when }k\to\infty$

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:

 $\displaystyle\left$ $\displaystyle=$ $\displaystyle\left+\lambda_{k}\left$ $\displaystyle=$ $\displaystyle o\left(\left\right)+\lambda_{k}\left$

Therefore we have

 $\frac{\left}{\left}=o(1)+% \lambda_{k}$

$\bullet$ Now assume

 $u_{k}=\frac{\pi_{k}x_{k}}{\left}$

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

 $\displaystyle u_{k+1}$ $\displaystyle=$ $\displaystyle(\pi_{k+1}-\pi_{k})A_{k}\frac{x_{k}}{\left}+\frac{\left}{\left}A_% {k}\pi_{k}\frac{x_{k}}{\left}$

and so

 $|u_{k+1}|\leq|\pi_{k+1}-\pi_{k}|C^{\prime}+\frac{\left}% {\left}(\lambda_{k}-\varepsilon)|u_{k}|$

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 FundamentalTheoremOfDemographyProofOf 2013-03-22 13:24:42 2013-03-22 13:24:42 aplant (12431) aplant (12431) 10 aplant (12431) Proof msc 92D25 msc 37A30