If you have two independent things with kurtoses k1,k2 and corresponding variances v1,v2 then their sum (i.e., the convolution of the probability distributions) has kurtosis (v1v1+v2)2k1+(v2v1+v2)2k2+6v1v2(v1+v2)2 (in general there are two more cross-terms involving “cokurtosis” values that equal 0 in this case, and the last term involves another cokurtosis that equals 1 in this case).
We can rewrite this as (v1v1+v2)2(k1−3)+(v2v1+v2)2(k2−3)+3((v1v1+v2)2+2v1v2(v1+v2)2+(v2v1+v2)2) which equals (v1v1+v2)2(k1−3)+(v2v1+v2)2(k2−3)+3. So if both kurtoses differ from 3 by at most δ then the new kurtosis differs from 3 by at most v21+v22(v1+v2)2δ which is at most δ, and strictly less provided both variances are nonzero. If v1=v2 then indeed the factor is exactly 1⁄2.
So Maxwell’s suspicions and Bucky’s calibrated eyeball are both correct.
If you have two independent things with kurtoses k1,k2 and corresponding variances v1,v2 then their sum (i.e., the convolution of the probability distributions) has kurtosis (v1v1+v2)2k1+(v2v1+v2)2k2+6v1v2(v1+v2)2 (in general there are two more cross-terms involving “cokurtosis” values that equal 0 in this case, and the last term involves another cokurtosis that equals 1 in this case).
We can rewrite this as (v1v1+v2)2(k1−3)+(v2v1+v2)2(k2−3)+3((v1v1+v2)2+2v1v2(v1+v2)2+(v2v1+v2)2) which equals (v1v1+v2)2(k1−3)+(v2v1+v2)2(k2−3)+3. So if both kurtoses differ from 3 by at most δ then the new kurtosis differs from 3 by at most v21+v22(v1+v2)2δ which is at most δ, and strictly less provided both variances are nonzero. If v1=v2 then indeed the factor is exactly 1⁄2.
So Maxwell’s suspicions and Bucky’s calibrated eyeball are both correct.
Wow! Cool—thanks!