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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?

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    $\begingroup$ Maybe it's because if you're unconscious, you don't have any expectations (in the usual, psychological sense). $\endgroup$ – celtschk Oct 27 '15 at 21:06
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    $\begingroup$ Well, I'd have said the law of the unconscious statistician was "$E[g(X)]=\int_{\mathbb R} g(X)f_X(x)dx$". Never been clear why it was called that...always thought it was because it let you do the computation without ever thinking through the distribution of $g(X)$. Don't hear the term much any more. Don't see why it might be either amusing or otherwise. $\endgroup$ – lulu Oct 27 '15 at 21:08
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    $\begingroup$ Note: I have at least some support for my use of the term en.wikipedia.org/wiki/Law_of_the_unconscious_statistician $\endgroup$ – lulu Oct 27 '15 at 21:10
  • $\begingroup$ @lulu: The talk of the wikipedia page you've liked offers some insight on the naming of this theorem: see DeGroot and Schervish, p. 214, first paragraph. "Theorem 4.1.1 is called the law of the unconscious statistician because many people treat equations (4.1.9) and (4.1.10) as the definition of E[r(X)] and forget that the definition of the mean of Y=r(X) is given in Definitions 4.1.2 and 4.1.4." While I don't have the precise citation here, it's save to assume that $r=g$ and that the definitions refer to the definition of the mean of a random variable Y in this case and a R.V. given by Y=r(X) $\endgroup$ – Roland Jan 4 '16 at 14:49
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    $\begingroup$ I've always wondered this myself. Thanks for asking it. $\endgroup$ – CommonerG Jan 5 '16 at 20:33
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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:

$$E(g(x)) = \int_{-\infty}^{\infty} g(x) f_x(x) dx$$

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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.

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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.

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  • $\begingroup$ I disagree. This is an answer to why one call this the "law of the unconscious statistician". $\endgroup$ – Diego Jan 11 '16 at 3:32
  • $\begingroup$ I do not understand this answer. Why would finite or infinite be connected with the name? And what factual support is there for this opinion? $\endgroup$ – Xi'an Oct 16 at 7:35

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