The law of the unconscious statistician In Casella and Berger's Statistical Inference (2nd edition) it says at the start of section 2.2 (page 55) when defining expectations that

If $ \mathrm{E} \,|g(X)| = \infty $ we say that $ \mathrm{E} \,g(X) $ does not exists. (Ross 1988 refers to this as the "law of the unconscious statistician." We do not find this amusing.)

Why


*

*would one call this the "law of the unconscious statistician"? Perhaps it is that I'm not a native speaker of English, but I have really no idea what being "unconscious" has to do with defining existence of expectations.

*can this be (or not be) considered amusing? 
 A: The "law of the unconscious statistician" refers to the theorem :
$$ E[g(X)] = \int\limits_R g(X)f_X(x) dx $$
According to  this forum, the theorem name comes from the fact that some statisticians present this as the definition of the expected value rather than a theorem. It seems that some statisticians did not like the name (including Casella and Berger's, I guess) and it was removed in later editions of the book. 
A: This may be total nuts but "infinite" in Russian is "бес-конечный", its opposite "конечный" [kɐˈnʲet͡ɕnɨj] and it sounds as "conscious"  /kŏnʹshəs/. Infinite thus can be related to unconscious. So may be this is an obscure way to make fun of a fellow Russian. Some have made great contribution to the Statistics like Komogorow or Smirnov of Kolmogorov–Smirnov test fame 
I just want to add this here as even if that's not the true explanation it could very well be. Some may find it not amusing, however.
A: In his lectures on probability, Blitzstein gives the following explanation:
Say you've computed $E(X)$ for some continuous distribution $X$:
$$E(X) = \int_{-\infty}^{\infty} x f_x(x) dx$$
where $f_x(x)$ is the PDF for $X$. Now you're looking to compute the variance: 
 $E(X^2) - [E(X)]^2$.
Now you need to compute $E(X^2)$, which is the expected value of a new distribution, $Y = X^2$: 
$$E(X^2) = E(Y) = \int_{-\infty}^{\infty} y f_y(y) dx$$
Well the unconscious statistician doesn't feel like computing another PDF $f_y$... so instead just reasons by analogy that if $E(X) = \int_{-\infty}^{\infty} x f_x(x) dx$ then surely he can simply replace the x with an $x^2$:
$$E(X^2) = \int_{-\infty}^{\infty} x^2 f_x(x) dx$$
Well that doesn't sound very legitimate! It looks like something he'd derive if he were half asleep or even drunk, but in general, this laziness turns out to be true:
$$E(g(x)) = \int_{-\infty}^{\infty} g(x) f_x(x) dx$$
