>>58833>Also I want this formula to be applied to Maya immediatelyThat's exactly the reason why I've made this thread (I'm obsessed or something.)
Let's divide the genes of a child into two parts: one equivalent to the genes shared by the parent with an unrelated, random person of the parent's own race, and the other one (which I will call
additional genetic benefit for the lack of better term) representing the genes shared with the parent by the child in excess to what is shared with the parent by an unrelated person.
We are interested in the additional genetic benefit as it is exactly what makes parents invest in their children more than they invest in unrelated people, i.e. it is the part that makes children special from the genetic point of view.
To estimate the additional genetic benefit, denoted by the symbol \(\Delta_p\), let's subtract from the expected number of genes shared by the parent with his child (which, due to the fact that genes are Bernoulli distributed - meaning they are either present with some probability or not - is equal to the probability of the child having those genes) the expected number of genes shared by the parent with an unrelated person of the same ethnic origin:
\(\Delta_p = p_p - p_r = 1/4 + 3/4 F_{ST} +3/4 p - 3/4 F_{ST}p - F_{ST} - p + F_{ST}p = 1/4 (1 - F_{ST} - p + F_{ST}p)\).
Here, \(p_p\) means the probability of having the same genes with a child (or proportion of the shared genes, as it is the same), \(p_r\) is the same thing, but with regards to a random, unrelated person.
In the case of a mixed child we do the same thing, the only difference is in the formula for probability of having the same genes (refer to OP).
\(\Delta_m = p_m - p_r = 1/4 - 1/4 F_{ST} + 3/4 p + 1/4 F_{ST}p - F_{ST} - p + F_{ST}p = 1/4(1 - 5F_{ST} - p + 5 F_{ST}p)\) (\(m\) stands for mixed.)
Now we want to know how many mixed children given a fixation index \(F_{ST}\) do we need to have for it to be equivalent to having a single non-mixed child in terms of additional genetic benefit.
We get the formula \(n \cdot\ \Delta_m = \Delta_{p}\), from which we get \(n = \frac{\Delta_p}{\Delta_m}\). Let's substitute and simplify:
• \(n = \frac{1 - F_{ST} - p + F_{ST}p}{1 - 5F_{ST} - p + 5 F_{ST}p}\);
• \(n = \frac{(1 - F_{ST})(1 - p)}{(1 - 5F_{ST})(1 - p)}\);
• \(n = \frac{(1 - F_{ST})}{(1 - 5F_{ST})}\).
The formula has no \(p\) in it, which is amazing, because I had no idea it will be like that when I started and it allows to actually calculate the equivalent using only \(F_{ST}\) values, which can easily be found in tables.
Let's try with some example values. First let's assume \(F_{ST} = 0.2\). Then we have \(n = \frac{1 - 0.2}{1 - 5\cdot0.2} = \frac{0.8}{0}\), which means that no finite number of children with a person from a population distant by \(0.2\) would provide the additional genetic benefit, the number of required children goes to infinity.
For \(F_{ST} = 0.1\), roughly the distance between Whites and Asians, we get \(n = \frac{0.9}{0.5} = 1.8\).
Now back to the initial question. If you met an IRL Maya, how many children would you need to have with her for it to be equivalent to having a one non-mixed child? Maya is 1/4 Chinese, 1/4 Indonesian, 1/4 Indian and 1/4 Dutch, meaning that the her distance to a White person is \(1/4 \cdot 0.03 + 1/4 \cdot 0.14 + 1/4 \cdot 0.13 = 0.075\). Thus, the required number of children is \(n = \frac{1 - 0.075}{1 - 5 \cdot 0.075} = \frac{0.925}{0.625} = 1.48\), or, in other words, one child with Maya is worth about 0.68 White children.