Let $X$ be a random variable with support on $S \subseteq [0, \infty)$. Let $f$ be the pdf of $X$ and $F$ be the cdf.
I'm trying to get some intuition behind this identity for random variables with support on the nonnegative reals. I have a proof (given below).
$$ E[X] := \int_0^{\infty} f(t)tdt = \int_0^\infty 1 - F(s)ds $$
I've managed to convince myself that it is true, but I don't have the faintest idea why the integral of the complement of the cdf should mean anything in particular, let alone something nice like the expected value.
I got the proof mostly by pushing symbols around and remembering that you can sometimes do nifty things with indicator functions and reordering sums.
Proof
Start with LHS
$$ \int_{t \in [0,\infty)} f(t) t \; dt $$
Express $t$ as integral of indicator function.
$$ \int_{t \in [0,\infty)} f(t) \left( \int_{s \in [0, \infty)} \text{I}[t \ge s] ds \right) dt $$
move constant inside integral.
$$ \int_{t \in [0,\infty)} \left( \int_{s \in [0, \infty)} f(t) \text{I}[t \ge s] ds \right) dt $$
reorder integrals
$$ \int_{s \in [0,\infty)} \left( \int_{t \in [0, \infty)} f(t) \text{I}[t \ge s] dt \right) ds $$
This indicator function is equivalent to evaluating the inner integral on the interval $[s, \infty)$
$$ \int_{s \in [0,\infty)} \left( \int_{t \in [s, \infty)} f(t) dt \right) ds $$
difference of integrals over $[0,\infty)$ and $[0, s)$
$$ \int_{s \in [0,\infty)} \left( \int_{t \in [0, \infty)} f(t) dt - \int_{t \in [0,s)} f(t) dt \right) ds $$
Use the fact that $f$ is a pdf and integrates to one and definition of $F$
$$ \int_{ s \in [0, \infty) } 1 - F(s) ds $$
Thus
$$ \int_{t \in [0,\infty)} f(t)tdt = \int_{s \in [0,\infty)} 1 - F(s)ds $$
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