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In a question about the conditions that are necessary for a probably distribution to have no defined mean, R.M made the remark:

"Take the Cauchy distribution and chop off the tails - even arbitrarily far out, like at plus/minus the xkcd number - and (once re-normalized) you suddenly get something ... (that) ... has a defined mean"

Is that statement true not just for a "chopped" Cauchy distribution, but for all bounded probability distributions?

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  • $\begingroup$ @whuber, I selected the specific part of the statement that was relevant to the question that I posed. $\endgroup$
    – user120911
    Commented Dec 26, 2019 at 15:59

2 Answers 2

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Note that the definition of bounded you're using in your question is non-standard. I would say that your distributions have compact support. In any case...

Here's a proof that the integral defining the mean exists.

Suppose that $X$ is a random variable with chopped off tails, like you specify. Take $f$ to be the density function of $X$ (we could work with the CDF instead if we wished to, which would give a slightly more general proof). Then by your assumption, there is some interval $[-A, A]$ outside of which, the function $f$ is identically zero. Within this interval, the density function is non-negative, by its usual properties.

The integral $\int_{-A}^{A} f(x) dx$ exists and is finite, it is equal to one. Therefore, we can bound:

$$ \int_{-A}^{A} \left| x f(x) \right| dx \leq \int_{-A}^{A} A f(x) dx = A \int_{-A}^{A} f(x) dx \leq A $$

So, the function $x f(x)$ is dominated by the integrable function $A f(x)$ on the interval $[-A, A]$. From the Dominated Convergence Theorem, it follows immediately that $x f(x)$ is integrable on $[-A, A]$, and the integral is finite (being bounded by the integral of $A f(x)$, which is bounded by $A$).

Finally, since $f$ is zero outside of the specified interval, it's enough for us to observe that:

$$ \int_{-A}^A x f(x) dx = \int_{-\infty}^{\infty} x f(x) dx $$

to finish things off.

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    $\begingroup$ +1. You can also make this result fully general for any CDF $F$ by expressing the mean as $\int_0^A(1-F(x)+F(-x))\,\mathrm{d}x \le 2A.$ $\endgroup$
    – whuber
    Commented Dec 26, 2019 at 17:38
  • $\begingroup$ I would probably have said compact support as well, but what makes you say "bounded" is non-standard in this usage? I wouldn't have blinked an eye at the use of bounded here, and in fact your "compact support" link says "a function has compact support if and only if it has bounded support" in its second sentence. $\endgroup$
    – user88719
    Commented Dec 29, 2019 at 19:35
  • $\begingroup$ Just that it conflicts with my training that bounded always applies to the co-domain of a mathematical object. A bounded function satisfies f(x) < M for some M and all x in the domain. So a random variable being bounded would mean the same to me, X(w) < M for all w in the sample space. It may be common in non-mathematical domains, but I suspect most mathematicians will be a bit confused by this use of bounded before clarification. Bounded support may be a good middle ground. $\endgroup$ Commented Dec 30, 2019 at 0:22
  • $\begingroup$ Isn't "X(w) < M for all w in the sample space" exactly what is meant here? Or maybe for all w outside a set of probability zero? But I would call that "essentially bounded." It's the support of the density that we're talking about, but the codomain of the random variable, right? $\endgroup$
    – user88719
    Commented Dec 30, 2019 at 3:34
  • $\begingroup$ I believe the OP means that $X(\omega) = 0$ for all $\omega$ outside of a bounded set in the sample space. Under mild assumptions, that does imply that the random variable is bounded, for sure. $\endgroup$ Commented Dec 30, 2019 at 18:39
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It is true that all bounded random variables have a well-defined expectation. See https://kurser.math.su.se/pluginfile.php/9291/mod_resource/content/1/lecture-5e.pdf (pages $4$ and $5$).


With regard to unbounded random variables, the problem is a matter of improper integrals. It is indeed the behaviour of the density as the argument goes to $\pm \infty$ that causes the non-integrability of certain distributions (e.g. Cauchy distribution), as explained in the Explanation of undefined moments section of this article: https://en.wikipedia.org/wiki/Cauchy_distribution.

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  • $\begingroup$ The first link is down. $\endgroup$ Commented Mar 19 at 8:13

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