Intuitive explanation for $\mathbb{E}X= \int_0^\infty 1-F(x) \, dx$

Solution 1:

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.

Solution 2:

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 $t=\Phi^{-1}(X)$.