Are first and last initials independent?

Rick Wicklin asks: which initials are most common?
This is followed with a simulation of the birthday problem. There are 676 different pairs of initials, so you might expect that to have probability 1/2 of two people in a group having the same initials, you’d need 31 people. This is the smallest k for which (676)!/((676-k)! 676^k), the probability that k people have different initials if all initials are chosen independently and uniformly at random, is less than one half. From the simulation, though, we see that it only takes 18.

A question that Wicklin doesn’t try to answer, though, is whether first and last initials are independent. In Wicklin’s data set of 4,502 employees at SAS (where he works), 403 people have first name starting with M, and 224 people have last name starting with L. 14 have the initials ML. From those marginal frequencies you’d expect (403)(224)/(4502) = 20; apparently in this sample parents with L last names are somewhat less likely to give their children M first names than in the population at large, although the effect is not statistically significant. (Spare me the usual data-mining caveats. ML is chosen not for any special properties it may have, but because they are my initials. If you looked at the data you know you started with your initials.)


inits = read.csv("C:/Users/Michael/Desktop/blog/initials.csv")
inits = inits[with(inits,order(I1,I2)),]
x=matrix(inits$COUNT,26,26,T)

A chi-squared test for independence (chisq.test(x)) gives \chi^2 = 915.4 with 625 degrees of freedom, and R reports p = 2.559 \times 10^{-13} but warns that “chi-squared approximation may be incorrect”. Some of the cell counts are quite small (in fact, zero is common!), which is the reason. So we resort to Monte Carlo methods. The call

chisq.test(x, simulate.p.value=T, B=10^6)

simulates a million contingency tables with the distributions of first and last initials given in Wicklin’s data — assuming those are independent — and reports the proportion of those tables which have χ2 larger than 915.4.

When I ran this simulation I got p = 0.005138, still significant but much less extreme than the former results. And as Wicklin points out, SAS is ethnically heterogeneous; it might be that in homogeneous populations first and last initials are independent, and dependence comes from aggregation. For an extreme example, say that there are two kinds of people, red and blue; red people’s first and last initials are independent and uniformly distributed over the first half of the alphabet, and blue people’s over the second half. But I don’t have a big ethnically homogeneous data set to test that on.

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2 thoughts on “Are first and last initials independent?

  1. The Monte Carlo method you mention above is interesting to me. If I am concerned about bias against my null hypothesis, would reporting the percentage of Monte Carlo simulations reporting a lower P value be appropriate in an academic paper? It’s a Master’s level public policy thesis, not intended to be published so I am just hoping to avoid doing something foolish.

    I have ran Monte Carlo simulations using randomly generated independent variables and I have strong evidence of bias against the null using a chi2 test. I’d love to keep the tests and just report the monte Carlo rankings in addition to the p-values. Any advice would be appreciated. Thanks.

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