## Sunday, June 14, 2009

### Long tail solution

Some time ago I posted a puzzle and promised a solution later. The solution appears below.

The problem concerns distributions with tails obeying a power law. This means values above some bound b the density of values x is proportional to 1/x**a for some value a. Let us compute the mean and median values of the tail.

The value of the mean is (integral(b to infinity)(x/x**a)dx)/(integral(b to infinity)(1/x**a)dx) = ((1-a)/(2-a))*b.

The value of the median is that value of c for which (integral(c to infinity)(1/x**a)dx) = .5*(integral(b to infinity)(1/x**a)dx) => c**(1-a) = .5*b**(1-a) => c=b*2**(1/(a-1)).

So the ratio of the mean to the median is (a-1)/((a-2)*2**(1/(a-1))).

For the given puzzle the mean is \$75000 and the median is \$22000. So the ratio is 72/22 = 3.40909... and we wish to find the value of a for which (a-1)/((a-2)*2**(1/(a-1))) equals this ratio.

A simple program finds a=2.196695263+. We are looking for b which will be ((a-2)/(a-1))*\$75000 = \$12327.40.