"My simulated sister is smarter than me"

Apologies that this is a bit techie - and it won't make sense without reading the previous post.

Yesterday I did some simple stats to show that my sister is most likely 8 IQ points smarter than me (to be fair and by symmetry, the converse could also be true).

Health warnings:

1. expected value only when averaged over large numbers of copies of my sister and myself;
2. equally true for my brother and myself - the stats are gender-blind.

How far is one sibling likely to be from the parental midpoint average?

Intuitively, you wouldn't expect every sibling to be exactly the average (they're not clones) but over a large family the pluses and minus would sort of average out to the mid-parental mean. But what about if we're just considering the deviation from average, without caring about the sign?

We seem to have a choice: halve the expected difference between two siblings, or find the average (absolute) deviation from the mean. As we saw yesterday, these give different answers.

I therefore decided to run an Excel simulation using the built-in RAND() function. Here's the four coin-set (taking values from 0 to 4):

IF(RAND()>0.5,1,0) + IF(RAND()>0.5,1,0) + IF(RAND()>0.5,1,0) + IF(RAND()>0.5,1,0)

and here is the last part of the spreadsheet model showing 100 tosses of two 4-coin sets (random variables X and Y) showing the number of heads.

If you like, you can consider this a simple four gene model for intelligence, with each gene presenting as two alleles, each of which code up or down for IQ by 7.5 points.

I ran each 100 toss simulation ten times and noted the results in the table on the right.

• The heading "Mean-IQ" refers to ten runs of the IQ (7.5) value in the "abs(X-Y)" column on the left, showing the mean difference in IQ between the two siblings; 


• the heading "Dec-IQ" refers to ten runs of the IQ (7.5) value in the "abs(X-2)" column on the left, showing the average deviation (+ and -) of a single sibling's IQ from the parental-midpoint mean.

From yesterday's post the computed values are respectively 8.2 and 5.625.

If we go back to selecting embryos for implantation, which is the right statistic to use to measure our likely IQ gain over the biological default of just taking what comes?

The leftmost statistic, 5.625 IQ points above the mean, would be sort of accurate if we were conceptually considering two embryos, one randomly varying and the other always exactly on the parental midpoint mean. But it wouldn't work, not least because the random embryo might well be below the mean but we're counting all variation as positive. So it's not realistic.

The statistic we get by halving the expected inter-sibling gap of 8.2 IQ points is better as we always select the smarter of the two embryos. However, since both X and Y are varying freely on the range {0,1,2,3,4} it's a bit difficult to correlate the abs(X-Y)/2 gap with the range-midpoint (mean) of 2. At this point we handwave and mutter about symmetry.

And what do you do when the presented embryos are all below the expected average?*

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* Which with two embryos will occur 25% of the time. I feel like spending some more money and genotyping a few more ...