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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?

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  • $\begingroup$ I was wondering about that question in the descrete case. Maybe, you can add this to your question as well. $\endgroup$
    – Antoine
    Apr 26, 2017 at 9:17
  • $\begingroup$ @Antoine I think the proof for discrete case is easy, but intuition is still hard to obtain. $\endgroup$
    – Matics
    Apr 26, 2017 at 9:33
  • $\begingroup$ Yeah, I would like to have some intuitive explanation. (The proof can be the same as in the continuous case: exchange the order of summation/integration.) $\endgroup$
    – Antoine
    Apr 26, 2017 at 10:09

2 Answers 2

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This answer gives an intuitive view in a backwards sort of way. Assume that you have a good handle on the mathematical proof that for nonnegative random variables, $$E[X] = \int_{0}^\infty P\{X > x\} \,\mathrm dx = \int_{0}^\infty [1-F(x)] \,\mathrm dx \tag{1}$$ and are aware that the world will end if you open that square bracket in the second integral in $(1)$ and write that integral as the difference of integrals of $1$ and $F(x)$ over the positive real line. I will describe a way of viewing this integral in a way that makes it intuitively obvious (at least to me, ymmv) why its value is the same as the value of the integral $$\int_{0}^\infty xf(x) \,\mathrm dx\tag{2}$$ which is the standard formula for the expected value of a nonnegative continuous random variable with density $f(x)$.

In the usual sense of Riemann integral, the second integral in $(1)$ is the area of the region (in the half-plane $\{(x,y)\colon x > 0\}$ bounded below by $F(x)$ and above by the line $y=1$. The Riemann integral calculates this area by dividing it into thin vertical (almost rectangular) strips of width $\Delta$ and height $\approx[1-F(x)]$ (and thus area $\approx [1-F(x)] \cdot \Delta$), adding up such areas, and then taking the limit of the sum as $\Delta \to 0$. This gives us the usual interpretation of $\int_{0}^\infty [1-F(x)] \,\mathrm dx$.

Another way of calculating the area of the region under consideration is to divide the area into thin horizontal strips of height $\Delta$. The bottom edge of the strip at height $y_0$ above the $x$ axis extends from $(0,y_0)$ to $(F^{-1}(y_0),y_0) = (x_0,y_0)$ where $x_0$ is the number such that $F(x_0) = y_0$, while the upper edge extends from $(0,y_0+\Delta)$ to $(F^{-1}(y_0+\Delta),y_0+\Delta) = (x_0+\delta, y_0+\Delta)$ where $\delta$ is such that $$F(x_0+\delta) = y_0+\Delta = F(x_0)+\Delta \implies \Delta \approx \left.\frac{\mathrm dF(x)}{\mathrm dx} \right|_{x = x_0} \cdot \delta = f(x_0)\cdot \delta.$$ Thus, the area of this thin horizontal strip is approximately $F^{-1}(y_0)\cdot \Delta = x_0\cdot f(x_0)\cdot \delta $. Adding up all such areas $F^{-1}(y_0)\cdot \Delta$ as $y_0$ varies from $0$ to $1$ and taking the limit as $\Delta \to 0$ is the same as adding up areas $x_0\cdot f(x_0)\cdot \delta $ as $x_0$ varies from $0$ to $\infty$ and taking the limit as $\delta \to 0$ which gives us $$\int_{0}^\infty xf(x) \,\mathrm dx,$$ the standard formula for the expected value of a nonnegative random variable with density $f(x)$. And that is why the integral on the right side of $(1)$ has the same value as the integral in $(2)$; they are just two different ways of computing the area of a specific region in the positive half-plane.

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Intuitive interpretation to non-statistician by mean survival time:

(1) Suppose there are 100 new-born monkeys, you follow them up until all of them die and record their survival times. Add them together and divide by 100, you get the mean survival time. (It is for the second equation).

(2) But I have another method. I count the # of alive monkeys every day, and add the #/100 to the previous results to get the accumulative sum. When the last monkey dies, the sum is the mean survival time. (It is for the first equation)

The benefit of (2) is I can have some information such as the mean survival time is larger than observed truncated mean before the last monkey dies. So if I die before the last monkey, I still have some contribution; but if I use (1), my contribution is 0.

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    $\begingroup$ You can directly type dollar symbol to include an equation. $\endgroup$
    – Matics
    Apr 27, 2017 at 5:10

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