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In a population that grows, would a disparity in having a boy/girl probability cause the ratio of males to females tend to the same? e.g. if the probability of having boys was $.49$ and having a girl was $.51$ would that mean a growing population would tend to be $49$% male and $51$% female?
What about a non growing population?, would that mean finally one gender will be the only gender left after sufficient amount of time?
It's not possible for one gender to "go extinct", as it were. That would happen if there were two different species whose rates of reproduction were different; that is, if girls produced only girls and boys produced only boys.
Here, in each generation you expect 49% boys and 51% girls to be born. No matter what the initial population, once the first generation has died you will be left with an expected proportion of 49% males to 51% females. This holds whether or not the population is growing or fixed -- but only assuming the death rates for males and females are the same.
The accepted answer here is correct only under some very narrow unstated assumptions about the mortality rates in the population. The answer assumes that each sex lives for the same deterministic length of time, which is what gives the result. This assumption is false in practice, as can be seen by the different life-expectancies of the sexes. In reality, the long-run population proportion of sexes in a closed population is affected by both the initial birth proportions and the mortality rates for the two sexes. Ceteris paribus, if one sex lives longer than the other, its proportion in the population will be higher, since its members are present for longer.
In view of this, it is not necessarily the case that a birth proportion $p$ of males (e.g., $p = 0.49$) leads to a long-run population proportion of $p$ in a closed population. This conjecture is false in the case where mortality rates for the two sexes are allowed to differ (as they do in real life).
Example of a reproduction process with different mortality rates: As a simple counter-example to demonstrate the above conclusion, we can consider a simple discrete-time process where each male lives for one time period and each female lives for at least one time period, and lives for an additional time period with probability $0 \leqslant \alpha \leqslant 1$ (i.e., we can set this to allow females to live longer than males). We will assume that each female reproduces in the first time period, but she is too old to reproduce in the second period. We assume that each female has an expected number of children equal to $\lambda$, and we let $p$ be the probability that a child is male; the sexes of children are independent.
Let $M_{k,t}$ and $F_{k,t}$ denote the number of males and females of ages $k = 1,2$ at time $t = 0,1,2,...$ and let $N_{t+1}$ denote the number of children that are born between times $t$ and $t+1$ (i.e., they are the number of children born to the generation of age $k=1$ at time $t$). Then the stochastic process evolves according to the recursive equations:
$$\begin{equation} \begin{aligned} N_{t+1} &\sim \text{Poisson}(\lambda F_{1,t} ), \\[6pt] M_{1, t+1} &\sim \text{Binomial}(N_{t+1}, p), \\[6pt] M_{2, t+1} &= 0, \\[6pt] F_{1, t+1} &= N_{t+1} - M_{1,t+1}, \\[6pt] F_{2, t+1} &\sim \text{Binomial}(F_{1,t}, \alpha). \\[6pt] \end{aligned} \end{equation}$$
The expected values for this process are:
$$\begin{equation} \begin{aligned} \mathbb{E}(N_{t+1}) &= \lambda F_{t,1}, \\[6pt] \mathbb{E}(M_{1, t+1}) &= \lambda p F_{t,1}, \\[6pt] M_{2, t+1} &= 0, \\[6pt] \mathbb{E}(F_{1, t+1}) &= \lambda (1-p) F_{t,1}, \\[6pt] F_{2, t+1} &= \alpha F_{1,t}. \\[6pt] \end{aligned} \end{equation}$$
The steady state proportion of males in the population is:
$$\text{Steady-state proportion of Males} = \frac{\lambda}{(\alpha + \lambda)} \cdot p.$$
In the special case where $\alpha = 0$, both sexes live for the same amount of time and so the steady-state proportion of males in the population is equal to the birth-proportion of males. However, if $\alpha > 0$ then females live longer than males and so the steady-state proportion of males in the population is less than the birth-proportion of males. This is in accordance with intuition, since the greater longevity of females means there are "old ladies" in the population that do not contribute to the new births, but still contribute to the population numbers.
It is worth noting that this model can easily be extended to allow females to live for longer than two periods, with $\alpha$ being the expected number of additional periods they live. In this case the same equation holds for the steady-state proportion of males in the population. Taking $\alpha \rightarrow \infty$ means that the proportion of males in the population shrinks to zero, even though their birth-proportion is still $p$. (Note that we might have to impose some additional requirement in the model to ensure that reproduction requires at least one male in the population. This is a separate issue that does not affect the present phenomenon.)
Observe that this phenomenon occurs regardless of whether or not there is population growth over time. The expected population is given recursively by $\mathbb{E}(\text{Pop}_{t+1}) = \lambda (1-p) \mathbb{E}(\text{Pop}_{t+1})$, so the rate of population growth is $\lambda(1-p)-1$. To get a stable expected-population we set the birth-rate to $\lambda = \tfrac{1}{1-p}$. Since $\alpha$ is still a free parameter we can still vary this to vary the steady-state proportion of males in the population.
