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.


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 $F$

$$ \int_{ s \in [0, \infty) } 1 - F(s) ds $$


$$ \int_{t \in [0,\infty)} f(t)tdt = \int_{s \in [0,\infty)} 1 - F(s)ds $$


marked as duplicate by joriki probability Jun 23 '16 at 5:55

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  • 2
    $\begingroup$ It's probably easier to note that $1-F(s) = 1 - P(X \leq s) = P(X >s)$, and the relationship between expectation and the latter expression is much easier to derive the relationship with expectation: math.stackexchange.com/questions/64186/… $\endgroup$ – Brenton Jun 23 '16 at 5:29
  • $\begingroup$ @Brenton Your link pretty much exactly answers the question I had. I think my question should be marked as a duplicate. $\endgroup$ – Gregory Nisbet Jun 23 '16 at 5:35
  • $\begingroup$ @GregoryNisbet: Done :-) $\endgroup$ – joriki Jun 23 '16 at 5:55

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