Simple Expected Value Of Continuous Variable Question

Question

The normal expected value that I am used to is the following with $f(x)$ as the probability density function:

$E[X] = \int_{-\infty}^\infty{xf(x) \, dx}$

My basic probability textbook is doing a basic proof with a nonnegative random variable $Y$ that begins:

$E[Y] = \int_0^\infty{P\{Y > y\} \, dy}$

Can someone explain this to me? Why does this expected value definition hold? Where does it come from? This doesn't make intuitive sense to me. I imagine that I'm missing something obvious?

http://math.stackexchange.com/questions/172841/integral-of-cdf-equals-expected-value — , Jun 28 '15 at 02:45
The link that d.k.o. posted shows that for nonnegative random variables, the second equation is another way to compute the expectation. — angryavian, Jun 28 '15 at 03:03
One can show it using Fubini Tonelli. Here is a similar question: http://math.stackexchange.com/q/1329112/27978. — copper.hat, Jun 28 '15 at 06:37

score 1 · Accepted Answer · answered Jun 28 '15 at 05:24

It's probably easier to see in the discrete case where we can write,

$$ \begin{align} \text{E}(Y) =& \sum_{k=1}^{\infty} k \, P(Y = k) \\ =& P(Y = 1) + \\ & P(Y = 2) + P(Y = 2) + \\ & P(Y = 3) + P(Y = 3) + P(Y = 3 + \\ & \ldots \\ =& \sum_{k=0}^{\infty} P(Y > k) , \end{align} $$

where in the last step we're simply summing by columns instead of rows. So all we're doing is changing the order of summation (which is permitted because the terms being added are nonnegative), and it's basically the same idea in the continuous case.

Simple Expected Value Of Continuous Variable Question

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