"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 ...

"My sister is smarter than me"

Well, maybe.

Your brother or sister could be exactly the same intelligence as yourself, but that's not the way to bet. Yesterday I wrote about the Carl Shulman and Nick Bostrom paper, "Embryo Selection for Cognitive Enhancement: Curiosity or Game-changer?". They said that the average distance in intelligence between two siblings was 8.4 IQ points.

Hmm.

When I saw that figure I was mightily impressed by their mathematical skills. After much thinking I decided that what they had done was this. They had taken the Gaussian distribution for the IQ of offspring with

mean = parental midpoint IQ; standard deviation = 7.5,

and considered two random variables from this distribution. Let's call them, with a bow to genetics, X for daughter and Y for son.

They had then computed the expected value of abs(X-Y). The absolute value is important here: the expected value of the mere difference (X-Y) is zero, as one sibling taken at random will on average be no brighter or dimmer than the other.

I hate these messy integrals incorporating the bivariate Gaussian distribution with weird boundaries! And apparently so do Carl and Nick: they got their answers by simulation.

But there is another way where we can still be analytic. The answer is in the power of four.

Take four coins and toss them. Count the number of heads - there are five possible outcomes: 0. 1. 2, 3, 4. The respective probabilities are 1/16, 4/16, 6/16, 4/16. 1/16.

The mean number of heads is 2 and the standard deviation of the number of heads is 1. How convenient.

[SD = √(npq) = √(4 * 0.5 * 0.5) = 1].

When you draw the histogram you get something like the above, which is not a million miles away from looking rather ... normal.

If we wanted to make this about sibling intelligence, map the number of heads into IQ scores:

2 heads = IQ 100
3 heads = IQ 107.5
4 heads = IQ 115.

Warning, IQ can go down as well as up.

Now imagine if we had one set of four coins to our left "the daugher IQ" and another set to our right "the son IQ". Throw simultaneously and count the number of left heads and right heads.

The difference (abs, remember, absolute value) in the number of heads is the IQ difference - each head counts as 7.5 IQ points. What do we get?

There's a spreadsheet for that.

Each cell in the bottom table is the product of the corresponding cells in the top two

Now these three tables are either self-evident after a little consideration, or too long to explain. I simply draw your attention to the figure in the bottom right in bold. The sum of all the entries in the third table = 1.09375.

This is the expected value of the difference in number of heads between left and right, "daughter" and "son" coin sets.

To change it into an IQ figure we multiply by 7.5 giving 8.2 ("my sister is 8 IQ points smarter than me!" ... or maybe conversely ...).

If we want the gap between each sibling's expected IQ from the mean expected IQ (over all possible siblings) we halve it to get 4.1. Think about it - don't you just love stats *.

This compares to Carl's and Nick's number of 4.2 based on ten million simulations (table below). Not a bad approximation to the normal distribution and you can see what's going on.

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Admittedly, extending to the case of 'best of ten siblings' seems a few coin sets too far.

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* and then I worry about this ...

which equates to 0.75 * 7.5 = 5.625 IQ points.

Time for a simulation ...