The mean of the mean is the mean

Posted on 9 April 2014 by John

There’s a theorem in statistics that says

$E( \bar{X} ) = \mu$

You could read this aloud as “the mean of the mean is the mean.” More explicitly, it says that the expected value of the average of some number of samples from some distribution is equal to the expected value of the distribution itself. The shorter reading is confusing since “mean” refers to three different things in the same sentence. In reverse order, these are:

The mean of the distribution, defined by an integral.
The sample mean, calculated by averaging samples from the distribution.
The mean of the sample mean as a random variable.

The hypothesis of this theorem is that the underlying distribution has a mean. Lets see where things break down if the distribution does not have a mean.

It’s tempting to say that the Cauchy distribution has mean 0. Or some might want to say that the mean is infinite. But if we take any value to be the mean of a Cauchy distribution — 0, ∞, 42, etc. — then the theorem above would be false. The mean of n samples from a Cauchy has the same distribution as the original Cauchy! The variability does not decrease with n, as it would with samples from a normal, for example. The sample mean doesn’t converge to any value as n increases. It just keeps wandering around with the same distribution, no matter how large the sample. That’s because the mean of the Cauchy distribution simply doesn’t exist.

Patches and specs

Posted on 5 April 2014 by John

From Leslie Lamport:

Every time code is patched, it becomes a little uglier, harder to understand, harder to maintain, bugs get introduced.

If you don’t start with a spec, every piece of code you write is a patch.

Which means the program starts out from Day One being ugly, hard to understand, and hard to maintain.

Quintic root

Posted on 3 April 2014 by John

Here’s a curious result I ran across the other day. Suppose you have a quintic equation of the form z x⁵ – x – 1 = 0. (It’s possible to reduce a general quintic equation to this form, known as Bring-Jerrard normal form.) There is no elementary formula for the roots of this equation, but the following infinite series does give a root as a function of the leading coefficient z:

$\sum_{n=0}^\infty {5n \choose n} \frac{z^n}{4n+1}$

One reason this is interesting is that the series above has a special form that makes it a hypergeometric function of z. You can read more about it here.

I could imagine situations where having such an expression for a root is useful, though I doubt the series would be much use if you just wanted to find the roots of a fifth degree polynomial numerically. Direct application of something like Newton’s method would be much simpler.