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Let $X$ be a random variable with values in $[0,+\infty)$

Show that $E(X)=\int_{0}^\infty P(X \geq t) dt$

Hint: first show that the expected value of a random variable with values in $\mathbb{N}_{0}$ is given by $E(X)=\sum_{k=1}^\infty P(X \geq k)$.

I already proved the Hint, but I don't understand how to use it to prove the first statement. My approach would be the following:

  • from the definition we know that $E(X)=\int_{0}^\infty X dP$

  • now write $\int_{0}^\infty X dP=\int_{\mathbb{N}_{0}} X dP + \int_{\mathbb{R}_{\geq0}\setminus\mathbb{N}_{0}} X dP$

but here I get stuck because, even if we somehow success to use the hint to solve the first integral, i have no idea how to approach the second, but maybe my approach is totally wrong. I would be very grateful for any help you can give for solving this problem.

Lee David Chung Lin
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Axxxiom
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1 Answers1

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\begin{align} E[X] &= \int_{x = 0}^{\infty} f_X(x) \cdot x \,\mathrm{d}x \\ &= \int_{x = 0}^{\infty} f_X(x) \cdot \left( \int_{t = 0}^x 1 \,\mathrm{d}t \right) \,\mathrm{d}x \\ &= \iint_{\Omega} f_X(x) \,\mathrm{d}x \,\mathrm{d}t \tag*{Eq.(1)} \end{align} where $\Omega$ is the lower triangular region (below the diagonal $t = x$) in $x$-$t$ plane with $x$-axis being horizontal and $t$-axis being vertical.

Eq.$(1)$ above is doing the integration "vertical first". One can do this horizontal first such that

\begin{align} E[X] &= \iint_{\Omega} f_X(x) \,\mathrm{d}x \,\mathrm{d}t \\ &= \int_{t = 0}^{\infty} \left( \int_{x = t}^{\infty} f_X(x) \,\mathrm{d}x \right) \,\mathrm{d}t \\ &= \int_{t = 0}^{\infty} P(X \geq t) \,\mathrm{d}t \end{align} When you proved the discrete case earlier, you should have noticed the pattern that is often seen in triangular numbers. That pattern was supposed to be the hint for the continuous case, not splitting the reals into integers and non-integers.

For the discrete case, denote $P_k \equiv P( X = k )$, then \begin{align} E[X] &= \sum_{k = 1}^{\infty} k\cdot P_k \\ &= P_1 + P_2 + P_3 + P_4 + P_5 + \cdots + P_k + P_{k+1} + \cdots \\ &\hphantom{ {}={} } \hphantom{P_1} {}+ P_2 + P_3 + P_4 + P_5 + \cdots + P_k + P_{k+1} + \cdots \\ &\hphantom{ {}={} } \hphantom{P_1 + P_2 } {}+ P_3 + P_4 + P_5 + \cdots + P_k + P_{k+1} + \cdots \\ &\hphantom{ {}={} } \hphantom{P_1 + P_2 + P_3 } {}+ P_4 + P_5 + \cdots + P_k + P_{k+1} + \cdots \\ &\hspace{96pt}\vdots \\ &\text{at the $k$-th row} \hspace{69pt} {} + P_k + P_{k+1} + \cdots \\ &\hspace{139pt} {}+ P_{k+1} + \cdots \\ &\hspace{160pt} \vdots \end{align} Where each of the original summand $k \cdot P_k$ is now "aligned vertically" so that when looking at each row horizontally one gets $P( X \geq k)$. This mental picture is what you are supposed to have (and not just the algebraic manipulation).

Lee David Chung Lin
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  • Thank you for your fast answer! I have a few questions: you are using the definition of Exp.value with the prob. density. Why can you do that? Does there always exist a probability density? Can you explain how you get from the second last to the last equality of Eq.(1)? I don't understand what it means to do the integration "vertical" or "horizontal" first, can you tell me how i have to figure out the "geometry" behind that calculations? ty – Axxxiom Oct 21 '18 at 18:19
  • basically i do not understand how the integration limits x=0, infinity, x=t, infinity "melt" into one only simbol $\Omega$ – Axxxiom Oct 21 '18 at 18:22
  • First let me ask you this: what is the result of $\int_{t = 0}^x 1 ,\mathrm{d}t$? – Lee David Chung Lin Oct 21 '18 at 18:23
  • the result is x – Axxxiom Oct 21 '18 at 18:25
  • Yes, so we make the $x$ in the original integral into this auxiliary integration with respect to $t$. Now this becomes a 2-dim integral on the $x$-$t$ plane over a certain region. – Lee David Chung Lin Oct 21 '18 at 18:26
  • Look at the discrete case where I have aligned the terms for you. Doing the sum (integral) "vertical first" gives you each column of $k \cdot P_k$, while doing the sum (integral) "horizontal first" gives you $P( X \geq k)$. – Lee David Chung Lin Oct 21 '18 at 18:30
  • This is basic calculus. When doing a 2-dim integral over a region, you can always do it at least two ways: adding infinitesimally thin vertical columns, or adding infinitesimally thin horizontal "rows" (strips/stripes). – Lee David Chung Lin Oct 21 '18 at 18:31
  • Am I understating correctly: you do $\int_{x=0}^{\infty} \int_{t=0}^{x} f(x) dt dx$ . This way you are calculating the "area" between f and the positive x axis by adding tiny vertical lines? – Axxxiom Oct 21 '18 at 18:37
  • in other words this calculates the "area" of $\Omega$. But shouldn't we calculate the area under f? – Axxxiom Oct 21 '18 at 18:39
  • No, you are still thinking in terms of integration of a 1-dim function. A two-dimensional integration is $z = g(x,y)$ where $z$ the function value (the surface) is usually pictured as the vertical height and the $x$-$y$ plane is horizontal. Here we have $x$-$t$ plane as the horizontal plane and the surface is $f_X(x)$ which depends on only $x$ (but is still a surface). Now we are calculating the volume under the surface, just like under $z = g(x,y)$ the surface. – Lee David Chung Lin Oct 21 '18 at 18:43
  • When scanning over a region (for whatever surface it is), one can scan "one direction first" or "scan the other direction first". Yeah it's unfortunate that when sketching on a paper we can only do 2-dim so there's the confounding of phrases. – Lee David Chung Lin Oct 21 '18 at 18:47
  • This applies to the discrete case as well, only that the function value (surface height) is defined only over the grid points. At each grid (lattice) point indexed by TWO coordinates (since we have made the layout as a "2-dim" table), the function value are the individual $P_k$. – Lee David Chung Lin Oct 21 '18 at 18:50
  • Ok, this is much clearer now! I will go check out some 2-dim integration examples to get practice on this topic. Thank you very much for your help! – Axxxiom Oct 21 '18 at 18:52