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?
Best Answer
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.