A probability density family for $x\in\mathbb{R}$ is

$$f(x) = k(\theta)\left[1 + (x/\theta)^2\right]^{-1}$$

parameterized by $\theta \gt 0$.

I am supposed to find $k(\theta)$ and then both $E(X)$ and $\text{Var}(X)$. I have found that $k =1/(\theta\pi)$. When I try to evaluate $E(X)$ I get that the corresponding integral diverges. Is this possible and does it imply that the variance diverges too?


You can find the answer on Wikipedia http://en.wikipedia.org/wiki/Cauchy_distribution. However, you are right, the integral for $E(X)$ does not converge, hence it means that $E(X)$ is undefined. Whereas for $E(X^2)$ it evaluates to infinity, e.g. $E(X^2)=\infty$. Therefore, in this case $Var(X) = E(X^2)-E(X)^2$ is undefined as well, though the second moment exists and is infinite.

  • 5
    $\begingroup$ Some people make a distinction between an expectation being undefined and being infinite. For this distribution it is undefined and not infinite. The expectation of $X^2$, on the other hand, is infinite. $\endgroup$
    – whuber
    May 18 '15 at 20:24
  • $\begingroup$ yes that's right. $\endgroup$
    – utobi
    May 18 '15 at 20:33
  • $\begingroup$ I've edited the answer accordingly $\endgroup$
    – utobi
    May 18 '15 at 20:52

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