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

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?

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By the way, I'm perfectly aware that the explanation is present on the Wikipedia article; I just thought it would be nice for the question to be here as well, for people looking for it. – Joe Z. Feb 13 '14 at 19:57
If the mean existed, it would be $0$ by a symmetry argument. – André Nicolas Feb 13 '14 at 21:05

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.

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indeed for the last paragraph, one of my friend actually sampling them and looked at the mean to prove to someone else who insisted it was 0... it might for a while looks like it is going to 0, then you get massive jumps all over the place. – Lost1 Feb 13 '14 at 23:31

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.

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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.

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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.

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but standard deviation does not matter. there exist distributions such that it has finite mean but infinite standard deviation – Lost1 Feb 14 '14 at 16:19