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$\frac{1}{\pi(1+x^2)}$ is a valid probability distribution - it integrates to $1.$

If I take its expectation, $\int_{-\infty}^\infty \frac x{\pi(1+x^2)} dx$, I get an unbounded value.

However, the distribution is symmetrical around zero, so intuitively the expectation should be zero.

How I do I reconcile those two facts? What is the expectation of the distribution?

David K
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dkv
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    It does not have a well-defined expectation. See for instance this, or you can have a look at the Wikipedia article. In particular, this means many things (such as the strong law of large numbers, etc) will not apply to Cauchy distributions: which can be checked empirically (e.g., by simulations). – Clement C. Feb 19 '15 at 17:43

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The mean is not defined. In reality you could compute the mean by computing the integral from -a to a and then taking $a\to \infty$, that is $\lim_{a\to \infty}\int_{-a}^a xf(x)dx = 0$ but you can also do: $\lim_{a\to 0} \int_{-2a}^a xf(x)dx \neq 0$ so in reality you can have any value. Empirically this means that if you generate random numbers from this distribution the empirical mean will not stabilize.

See http://en.wikipedia.org/wiki/Cauchy_distribution#Mean

Martingalo
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