# fundamental theorem of demography, proof of

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

$$m\le \frac{\parallel {x}_{k+1}\parallel}{\parallel {x}_{k}\parallel}\le M$$ | (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 ${A}_{\mathrm{\infty}}$. Primitivity of ${A}_{\mathrm{\infty}}$ implies that there exists $l\in \mathbb{N}$ such that

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

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

$${A}_{k+l}{A}_{k+l-1}\mathrm{\cdots}{A}_{k}\gg 0$$ |

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

$${x}_{k+l+1}={A}_{k+l}\mathrm{\cdots}{A}_{k}{x}_{k}$$ |

We thus have

$$\parallel {x}_{k+l+1}\parallel \le {C}^{l+1}\parallel {x}_{k}\parallel $$ | (2) |

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

$$\parallel {x}_{k+l+1}\parallel \ge \epsilon \parallel {x}_{k}\parallel $$ |

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

$${C}^{l}\parallel {x}_{k+1}\parallel \ge \epsilon \parallel {x}_{k}\parallel $$ |

which yields

$$\parallel {x}_{k+1}\parallel \ge \frac{\epsilon}{{C}^{l}}\parallel {x}_{k}\parallel $$ |

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

$\bullet $ Let us denote by ${e}_{k}$ the (normalised) Perron eigenvector^{} of
${A}_{k}$. Thus

$${A}_{k}{e}_{k}={\lambda}_{k}{e}_{k}\mathit{\hspace{1em}}\parallel {e}_{k}\parallel =1$$ |

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 $\epsilon >0$ such that

$$\left|{A}_{k}{\pi}_{k}\right|\le ({\lambda}_{k}-\epsilon )$$ |

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

$$ |

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:

$$ | $=$ | $$ | ||

$=$ | $$ |

Therefore we have

$$ |

$\bullet $ Now assume

$$ |

We will verify that ${u}_{k}\to 0$ when $k\to \mathrm{\infty}$. We have

${u}_{k+1}$ | $=$ | $$ |

and so

$$ |

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

$$|{u}_{k+1}|\le {\delta}_{k}+({\lambda}_{\mathrm{\infty}}-\frac{\epsilon}{2})|{u}_{k}|$$ |

where we have noted

$${\delta}_{k}=({\pi}_{k+1}-{\pi}_{k}){C}^{\prime}$$ |

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

$$|{u}_{k}|\to 0\text{when}k\to \mathrm{\infty}$$ |

Remark that this also implies that

$${z}_{k}=\frac{{\pi}_{k}{x}_{k}}{\parallel {x}_{k}\parallel}\to 0\text{when}k\to \mathrm{\infty}$$ |

$\bullet $ We have ${z}_{k}\to 0$ when $k\to \mathrm{\infty}$, and ${x}_{k}/\parallel {x}_{k}\parallel $ can be written

$$\frac{{x}_{k}}{\parallel {x}_{k}\parallel}={\alpha}_{k}{e}_{k}+{z}_{k}$$ |

Therefore, we have ${\alpha}_{k}{e}_{k}\to 1$ when $k\to \mathrm{\infty}$, which implies
that ${\alpha}_{k}$ tends to 1, since we have chosen ${e}_{k}$ to be
normalised (*i.e.*,$\parallel {e}_{k}\parallel =1$).

We then can conclude that

$$\frac{{x}_{k}}{\parallel {x}_{k}\parallel}\to {e}_{\mathrm{\infty}}\text{when}k\to \mathrm{\infty}$$ |

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 |