Why does the Cauchy distribution have no mean if it's symmetric around 0?

Question

Something that didn't make intuitive sense to me when learning about the Cauchy distribution was that there was no defined mean for the function, even though the function was clearly centered at zero and equally valued in both directions.

Is there any reason for this?

Answer 1

The mathematical answer is what Lost1 said, so I won't repeat it.

Morally, the mathematical answer is the correct one because the interesting objects in probability (to me anyway) tend to be idealized versions of things one encounters in experiments. Before I tell you what I mean by that, take a second and ask yourself what you would do if a person who had never encountered any higher mathematics before asked you "what is a mean?"

I would tell them that a mean is an average. If you repeat an experiment a lot of times, then average the results you get, the mean is that number. By the law of large numbers, we know that up to a little fuzziness and some regularity assumptions that answer is usually essentially correct.

Surely any definition of "mean" has to agree with the one I just gave. The problem with the Cauchy distribution is that if you had a bunch of genuine independent standard Cauchy distributed random variables and you averaged them, your limit wouldn't be all that close to zero. It would be some random number. In fact, its distribution would again be standard Cauchy.

In essence, I think the reason the mean of a Cauchy distribution isn't zero is that if you were to encounter a bunch of approximately independent, approximately Cauchy random variables, their empirical average probably would not be all that close to zero.

Answer 2

For the mean to exist, you need $\int^\infty_{-\infty}\frac{|x|}{1+x^2}\text{d}x$ to be finite. This is the same as the requirement of a function being integrable. A measurable function $f$ is integrable if $\int |f|\text{d}\mu<\infty$. As it has been pointed out by many people, if the mean exists, it is 0 by symmetry, but it does not.

Answer 3

The argument mentioned in the post, reformulated in a somewhat more standard terminology, is that the distribution of a Cauchy random variable is symmetric around zero. This suggests, and actually implies, that the median should be zero and says nothing about the mean.

Answer 4

Just to add a clear demonstration of what actually happens when you try to deal with the mean and higher moments of a Cauchy distribution, I ran a quick script to repeatedly take n samples of a standard Cauchy distribution (5 runs for each n):

n = 1000
    mean = -1.02224, sd = 22.0379
    mean = 0.443686, sd = 18.5603
    mean = -0.616193, sd = 20.8578
    mean = 0.544703, sd = 16.2545
    mean = 1.99947, sd = 56.7486
n = 10000
    mean = 0.20199, sd = 41.3423
    mean = 3.47629, sd = 364.8
    mean = -1.4106, sd = 80.6524
    mean = -0.441166, sd = 224.783
    mean = -0.674296, sd = 66.4877
n = 100000
    mean = 1.13362, sd = 413.799
    mean = -1.06265, sd = 228.098
    mean = 1.09204, sd = 317.432
    mean = 3.80845, sd = 1493.95
    mean = -0.377224, sd = 295.982
n = 1000000
    mean = -1.41118, sd = 3189.89
    mean = -1.66183, sd = 1797.63
    mean = -0.176471, sd = 422.138
    mean = 1.30805, sd = 2023.47
    mean = 0.723504, sd = 1575.73

You can plainly see how the mean and SD jump crazily all over the place, and show no sign of converging on any kind of meaningful value.