Is there a good way to visualize the formula:
$$ E(x) = \int_{0}^{\infty} 1 - F(X) \,\mathrm{d}x $$ ?
for positive continuous random variables? I understand the formula as far as basic calculus goes, in terms of integration by parts. I suppose my insight is rather trivial at the analytical level, but it seems to me that
$$ \int_{0}^x 1 - F(X) \, \mathrm{d}x $$
must converge to $E(X)$ as $x$ goes to the supremum of the domain of $X$. So I wonder if there is a "nice" picture/animation that that makes a "reason" for this clear.
First, define the one-minus-CDF function $Q(x) := 1 - F(x)$. If we take a uniform random variable $U$ on $[0,1]$, the inverse of $Q$ applied to $U$ has the same distribution as $X$, that is, $Q^{-1}(U) \sim X$.
Now try this: Draw a graph of $1-F(X)$ for some positive random variable $X$ of your choice. Shade the area under the curve; note that this is $\int_0^\infty 1-F(x) dx$. Now, turn the graph 90 degrees counterclockwise. If we interpret the now-horizontal (previously vertical) axis as values of $U$, then the vertical axis will represent values of $Q^{-1}(U)$, and the shaded area is $$ \int_0^1 Q^{-1}(u)du = E[Q^{-1}(U)] = E[X]. $$