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Let $X$ be random variable with pdf $f_X(x) = dfrac{1}{pi(1+x^2)}$. I understand that mathematically, the improper integral, $displaystyleintlimits_{-infty}^{infty}dfrac{x}{pi(1+x^2)}dx$ does not exist ($underset{T_1to-infty}{lim}underset{T_2toinfty}{lim}displaystyleintlimits_{T_1}^{T_2}dfrac{x}{pi(1+x^2)}dx = frac{ln(1+T_2^2) - ln(1+T_1^2)}{2pi} = infty - infty) implies$ undefined. However I am unable to understand the intuition, why $E[X] ne 0$ since the pdf is an even function which takes positive and negative values with equal probability. Hence large enough samples should produce mean close to 0. Please help?

There are several ways to look at it:

• Let $f(x):=frac{pi^{-1}}{1+x^2}$ so $int_{-infty}^c xf(x) dx=-infty,,int_d^infty xf(x) dx$ for any $c,,dinmathbb{R}$. So if we choose $c<d$, you could argue the mean is $-infty+int_c^d xf(x) dx+infty$. In theory, you can get any value you like if you think the infinities cancel.


• "But I'm integrating an odd function! That has to give me $0$!" Yes, if the two pieces you're cancelling are both finite. But $infty-infty$ is an indefinite form, so you can't use that theorem.


• The characteristic function is $varphi(t):=exp -left|tright|$. If we average $n$ samples, the result has characteristic function $varphi^n(t/n)=varphi(t)$. It's immune to the CLT. Nor should you expect otherwise, without a finite and well-defined mean and variance. No $mu$, no $(X-mu^2)$, no variance. The characteristic function provides another way to look at it: we can't very well write $mu=varphi'(0)$ because the modulus has undefined derivative at $0$ (the one-sided limits $lim_{tto 0^pm}frac{|t|}{t}$ differ).


• It can be shown, however, that the median of $n$ samples is asymptotically Normal for large $n$. (The proof is a bit more involved than a standard CLT argument for means; you can get an overview here.) By contrast, if you compute the mean of a gradually growing sample, it'll bounce around like crazy, because (as shown above) it's Cauchy-distributed.