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Suppose that children always inherit their last names from their father (which implies that no new last names are ever created).

Pick a last name of interest (e.g. Smith), and let $X_n \in \left[0, 1\right]$ denote the fraction of men with last name Smith in generation $n$. Assuming the probability of getting married, the expected number of children, etc, is independent of one's last name (and independent across couples), is it true that $X_n$ is a martingale, i.e. that $$\mathbf{E}\left[X_{n+1} \,|\, X_n = x\right] = x$$

and, if the above is true, does it also follow that, since $X_n$ is bounded both above and below (by 1 and 0, respectively), $X_n$ must converge to one of its bounds as $n \rightarrow \infty$?

Edit: convergence to the bounds does not follow in general (see https://math.stackexchange.com/questions/1125320/polyas-urn-model-limit-distribution), so I'll add some additional assumptions. First, observe that 0 and 1 are both absorbing states for $X_n$, i.e. $\mathbf{Pr}\left[X_{n+1} = 1 \,|\, X_n = 1\right] = 1$ and $\mathbf{Pr}\left[X_{n+1} = 0 \,|\, X_n = 0\right] = 1$. For "interior" states, i.e. states $x \in \left(0, 1\right)$, I assume $$\mathbf{Var}\left[X_{n+1} \,|\, X_n = x\right] > 0$$

to rule out deterministic situations (where every man has exactly one son and one daughter, for example). Furthermore, I assume that

$$\mathbf{Pr}\left[X_{n+1} \in \left\{0, 1\right\} \,|\, X_n = x\right] > 0 \; \forall x$$

i.e. there is always a positive probability of last names going extinct (because all Smiths in generation $n$ have only daughters, or all non-Smiths in generation $n$ have only daughters, for example).

Under those assumptions, do we have

$$lim_{n \rightarrow \infty} \, \mathbf{Pr}\left[X_n \in \left\{0, 1\right\}\right] = 1$$

meaning that $X_n$ reaches one of the absorbing states almost surely? (If not, is there anything we can say about the limiting distribution for $X_n$?)

In other words, is it correct to say that (under the assumption that no new last names are ever created), the entire human population will almost surely have the same last name at some distant future date (because all other last names become extinct)? Does the answer depend on the population growth rate?

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    $\begingroup$ Are you talking about Lévy's zero–one law? This is not what it means. Consider the case where every man has one son. Or where every man has at least one son (a kind of Polya urn model) $\endgroup$ – Henry Mar 17 at 22:49
  • $\begingroup$ @Henry yes, I have a vague memory from a probability class I took many years ago that a martingale that both (a) is bounded both above and below, and (b) has positive variance for any interior state (i.e. away from the bounds) will almost surely converge to either one of the bounds. But I may be misremembering. $\endgroup$ – Adrian Mar 18 at 1:15
  • $\begingroup$ stats.stackexchange.com/questions/79972/… is related (it's about a supermartingale bounded above by 1 and below by 0) $\endgroup$ – Adrian Mar 18 at 19:16
  • $\begingroup$ Another related question: math.stackexchange.com/questions/1125320/… $\endgroup$ – Adrian Mar 21 at 0:44
  • $\begingroup$ Related: en.wikipedia.org/wiki/Galton%E2%80%93Watson_process $\endgroup$ – Adrian Mar 23 at 0:47
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All that matters for the analysis are the numbers of Smiths and non-Smiths in the population at any time, say $k.$ Let these be $m$ and $n,$ respectively, so that $X_k=m/(m+n).$

The model supposes the number of descendants of each person $i,$ $1\le i\le m+n,$ is given by a random variable $Z_i$ and these variables are independent. Let the first $m$ of these indexes indicate the Smiths.

We must assume these random variables have no probability of being zero -- for otherwise, there is a chance the entire population is wiped out in the next generation and the fraction of Smiths becomes undefined. This would violate the Martingale assumption that $E[|X_k|]\lt \infty.$

The number of Smiths in the next generation is $\sum_{i=1}^m Z_i$ while the total size of the next generation is $\sum_{i=1}^{m+n}Z_i.$ Compute the expected fraction of Smiths using the iid assumption as

$$E[X_{k+1}\mid X_k] = E\left[\frac{\sum_{i=1}^m Z_i}{\sum_{i=1}^{m+n} Z_i}\right] = m E\left[\frac{ Z_1}{\sum_{i=1}^{m+n} Z_i}\right].$$

(If the appearance of $m$ and $n$ on the right hand side make you uncomfortable, let $N$ be the unknown population size at time $k$ and replace "$m$" by "$NX_k$" and "$n$" by "$N-NX_k.$")

The fraction of non-Smiths is computed in the same way and seen to equal $n/m$ times the fraction of Smiths, whence the expectation equals $m/(m+n)=X_k:$ under the foregoing assumptions, the sequence is a Martingale.

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  • $\begingroup$ Thank you for this excellent answer (+1). Do you know whether $X_k$ will almost surely reach one of its absorbing states (i.e. 0 or 1)? $\endgroup$ – Adrian Mar 18 at 3:55

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