# Nahuatl names and natality in aztec empire

Budelberger, Richard budelberger.richard at TISCALI.FR
Wed Jan 21 02:41:00 UTC 2004

```30 nivôse an CCXII (le 21 janvier 2004 d. c.-d. c. g.), 03h34.

----- Message d'origine -----
De : Davius Sanctex <davius_sanctex at hotmail.com>
À : <nahuat-l at server2.umt.edu>
Envoyé : samedi 30 décembre 2000 01:09
Objet : Nahuatl names and natality in aztec empire

> A nice mathematical argument relating name "Teyacapan"
> to average number for children in Aztec Empire.
> ____________________________
> Bob McCaa points that in a census of 1540 in a group
> of 1205 women, 313 were named "Teyacapan" (= first
> born), i.e. 25,04%:
>
>
>
> I will show that these data implies that the average number per family
> was at most 5,135
> and that 6,88%

0,588 %

> of couples had not any child!
>
> _________________________________________
> This number can be related to the average number of
> children in a family. We assume:
>
> 1) Nearly all first born female babies was named
>    "Teyacapan"
> 2) The probability of borns by unit of time remain
>    uniform for a community and population is stationary.
>
> First step:
> Second hypothesis implies borns can be well modelized by
> a Poisson distribution, thus the probability of a couple
> to have k kids is P(k):
>
> P(k)= exp(-m)*(m^k)/k!
>
> [Where m is the average number of children]
>
> Second step:
> Thus if the probability of a child to belong to a familiy
> with exactly k kids is p(k):
>
> p(k) = P(k)/(1-P(0))
>
> [p(0) is the % of families that have no kid].
>
> Thirst step:
> If we take a woman at random the probability of being the
> first kid in a family of k kids is just 1/k [= q(k)].
> And thus the probability that a woman to be the first baby
> of a family is r:
>
> r = p(1)*q(1) + p(2)*q(2) + ...+ p(k)*q(k) + ... =
>   = P(1)/1 + p(2)/2 + ...+ p(k)/k + ... = 25,04%
>
> This last equation enable us to evaluate m. For m = 5,135
>
> k    P(k)     p(k)   p(k)/k
>
> 0  0,00588     __
> 1  0,03022  0,03039  0,03039
> 2  0,07759  0,07805  0,03902
> 3  0,13282  0,13361  0,04453
> 4  0,17053  0,17154  0,04288
> 5  0,17515  0,17618  0,03523
> 6  0,14991  0,15079  0,02513
> 7  0,10997  0,11063  0,01580
> 8  0,07059  0,07101  0,00887
> 9  0,04028  0,04052  0,00450
>
> sum                  0,2498 = 24,98%
>
> This shows that the average number must be of order 5,135.
> Moreover, of this table we deduce that 5,88 % = P(0)

P(0) = 0,588 %

> of couples have not babies and the majority (17,51%) have 5 babies.
> The number of families with 9 is 4,02% ...
> _________________________________________________________________________
> Get Your Private, Free E-mail from MSN Hotmail at http://www.hotmail.com.

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