Is it possible for an expected value not to exist?

Question

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?

utobi · Accepted Answer · 2015-05-18T20:49:23.423

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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.

edited May 18 '15 at 20:49

answered May 18 '15 at 19:55

utobi

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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. – whuber May 18 '15 at 20:24
yes that's right. – utobi May 18 '15 at 20:33
I've edited the answer accordingly – utobi May 18 '15 at 20:52

Is it possible for an expected value not to exist?

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