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It's provably correct that for a nonnegative random variable denoted as $Z$. The expectation of $Z$ can be written as follows: $$\mathbb{E}[Z] = \int_{x=0}^{\infty}\Pr[Z\geq x]dx.$$

Well, it can be proved by methods "integration by part", however, I think there may exist a more intuitive interpretation and a direct connection with ordinary definition of expectation $$\mathbb{E}[Z] = \int_{x=0}^{\infty}f(x)xdx$$

Hope someone could give some hints, thx.

Also, for discrete case, is there also intuitive interpretation?

Matics
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1 Answers1

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I like to think of this in three dimensions, using Cartesian coordinates $x,y,z.$

Construct a surface $S$ defined by $z = f_Z(x).$ That is, take the graph of $z=f_Z(x)$ in the $x,z$ plane and "extrude" it in the direction of the $y$ axis.

Now take the region of space consisting of points $(x,y,z)$ such that $0 \leq y \leq x$ and $0 \leq z \leq f_Z(x).$ This is a region bounded below by the $x,y$ plane, bounded above by $S,$ lying between the vertical planes $y=x$ and $y=0.$

The volume of this region is $\mathbb E[Z],$ computed in one of the following two ways:

If we take slices perpendicular to the $x$ axis, the slice at $x=t$ is a rectangle with width $t,$ height $f_Z(t),$ and area $t f_Z(t).$ Integrate from $t=0$ to $t=\infty.$

If we take slices perpendicular to the $y$ axis, the slice at $y=t$ has area equal to a region of the $x,z$ plane bounded below by $z=0,$ bounded below by $z=f_Z(x),$ and bounded on the left by $x=t$; this region has area $P(Z > t).$ Integrate from $t=0$ to $t=\infty.$

David K
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