Intuitive explanation for $\mathbb{E}X= \int_0^\infty 1-F(x) \, dx$ I can see by manipulating the expression why $\mathbb{E}X$ works out to be $\int_0^\infty 1-F(x)\,dx$, where $F$ is the distribution function of $X$, but what is an intuitive explanation for why that is true? If at each point we sum the probability $\mathbb{P}(X>x)$, why should we end up with the expectation?
Thanks
 A: If you are looking for intuition, the discrete case is your best bet.
Look at $\sum_0^\infty P[X > n]$ and count how many times you count the set $\{X = k\}$. You don't count $\{X=0\}$ at all. The only one which includes $\{X=1\}$ is $P[X>0]$ so it gets counted once. You will count $\{X=2\}$ twice, when $n=0$ and $n=1$. And so on, you will count $\{X=n\}$ exactly $n$ times.
Thus we must have $\sum_0^\infty P[X>n] = \sum_0^\infty n P[X=n] = EX$. To make the argument rigorous and also extend to the continuous case, we simply apply Fubini's Theorem (Tonelli's Theorem) to say
$$
\int_0^\infty P[X>t]dt = \int_0^\infty\int1_{X>t}dPdt = \int\int_0^\infty 1_{X>t}dtdP = \int XdP = EX.
$$
Edit: As Evan mentions, of course we require $X$ to be nonnegative.
A: This is really just another way of stating nullUser's intuitive explanation, particularly focusing on the second half (Tonelli/Fubini).
Suppose for now that our variable has the specific form $X=f(t)$, where $t$ is chosen uniformly between $0$ and $1$ and $f$ is an increasing function of $t$.    In that case $E(X)$ has a natural interpretation: It just corresponds to the average value of $f$, which is just the area under $f$.  
There's two ways of thinking about this area.  One is "vertically", as $\int_0^1 f(t) dt$.  The other is "horizontally": integrate the width of the horizontal rectangle instead of the height of the vertical one.  And the width of a horizontal rectangle here just corresponds to the part of the $f(t)$ which is above $t$, i.e. the probability $X$ is at least $t$.   

This picture was for when $X$ had this particular $f(t)$ form.  For the general case, you should think of $t$ as representing a sort of percentile for $X$ (e.g. $t=0.5$ represents the median value of $X$, and $t=0.9$ a value which is only reached $10\%$ of the time).  This can also be thought of in terms of a change of variables where $t=\Phi^{-1}(X)$.  
