# Statistics: E(X) = Integral(0 to infinity) of (1-F(x))dx

by kingwinner
Tags: 1fxdx, infinity, integral0, statistics
 P: 1,270 "If X is non-negative, then E(X) = Integral(0 to infinity) of (1-F(x))dx, where F(x) is the cumulative distribution function of X." ============================ First of all, does X have to be a continuous random variable here? Or will the above result hold for both continuous and discrete random variable X? Secondly, the source that states this result gives no proof of it. I searched the internet but was unable to find a proof of it. I know that by definition, since X is non-negative, we have E(X) = Integral(0 to infinity) of x f(x)dx where f(x) is the density function of X. What's next? Thanks for any help!
Emeritus
PF Gold
P: 16,091
 Quote by kingwinner "If X is non-negative, then E(X) = Integral(0 to infinity) of (1-F(x))dx, where F(x) is the cumulative distribution function of X." ... E(X) = Integral(0 to infinity) of x f(x)dx where f(x) is the density function of X. What's next?
Well, the thing you know has an x in it, but the thing you're trying to get to doesn't... and the thing you're trying to get to has an F in it, but the thing you know has the derivative of F in it....
 P: 626 This works for discrete, cont., and mixed. Though the derivative statement applies for cont. only. Use integration by parts for cont. case.
 P: 1,270 Statistics: E(X) = Integral(0 to infinity) of (1-F(x))dx But for discrete random variable X, would it still make sense to talk about "integration"? (i.e.INTEGRAL(0 to infinity) of (1-F(x))dx) Or should it be replaced by a (sigma) sum? Do you mean using integration by parts for the expression of the definition of E(X)? What should I let u and dv be? Thanks!
 P: 626 No, in the discrete case you would be using sums instead of integrals since expectation is defined in terms of a sum not integrals. Well if u = 1 - F(x) then du = -f'(x) dx and dv = dx then v = x. So now you have x*S(x) (evalulated between your limits) + integral(x*f(x) dx). Obviously the first part of your sum vanishes since at "infinity" S(x) -> 0 and at 0, x*S(x) = 0. And so now you are left with what your usual definition of E(X). Note: this is a very handwavy proof as you would really want to be rigorous when talking about the limits that make the first term vanish.
Emeritus
PF Gold
P: 16,091
 Quote by kingwinner But for discrete random variable X, would it still make sense to talk about "integration"? (i.e.INTEGRAL(0 to infinity) of (1-F(x))dx) Or should it be replaced by a (sigma) sum?
It does when you learn measure theory. Until then, just replace it with a sum without thinking about it.
Emeritus
PF Gold
P: 16,091
 Quote by kingwinner What should I let u and dv be?
Did you think about that question at all? I practically told you what u and dv should be in post #2....
P: 1,270
 Quote by NoMoreExams No, in the discrete case you would be using sums instead of integrals since expectation is defined in terms of a sum not integrals. Well if u = 1 - F(x) then du = -f'(x) dx and dv = dx then v = x. So now you have x*S(x) (evalulated between your limits) + integral(x*f(x) dx). Obviously the first part of your sum vanishes since at "infinity" S(x) -> 0 and at 0, x*S(x) = 0. And so now you are left with what your usual definition of E(X). Note: this is a very handwavy proof as you would really want to be rigorous when talking about the limits that make the first term vanish.
Just one point I am having troubles with: (in red)

lim x(1-F(x))
x->inf
This actually gives "infinity times 0" which is an indeterminate form and requires L'Hopital's Rule. I tried many different ways but was still unable to figure out what the limit is going to be...how can we prove that the limit is equal to 0?

Thanks!
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 Hurkyl, you don't "need" measure theory to write a sum as an integral. The Riemann-Stieljes integral will do that.
 P: 100 Another way to do it is to write the expected value as $$E[X]=\int_{0}^{\infty}sf(s)ds = \int_{s=0}^{\infty}\int_{x=0}^{s}f(s)dxds$$ and then change the order of the integrals to get your formula. To see what the new bounds on the integrals would be, draw a picture of the region of integration. You can use this same approach to find that $$E[X^2] = \int_{0}^{\infty}s^2f(s)ds = \int_{s=0}^{\infty}\int_{x=0}^{s}2xf(s)dxds= \int_{0}^{\infty}2x(1-F(x))dx$$ which is also valid for X nonnegative.
P: 159
 Quote by NoMoreExams No, in the discrete case you would be using sums instead of integrals since expectation is defined in terms of a sum not integrals.
Or, equivalently, write the probability distribution as a sum of delta functions.
 P: 1 Integration by parts: $$m_X=\int^{\infty}_{0} x f_X(x) dx = -\int^{\infty}_{0} x (-f_X(x) dx)$$ eq(1) Let $$u=x$$ and $$dv = -f_X(x) dx$$ Thus $$du=dx$$ and $$v = 1-F_X(x)$$ Chech that $$dv/dx = d/dx (1-F_X(x)) = d/dx(-F_X(x)) = -f_X(x)$$ o.k. Then substitute in (1) $$m_X=-[uv|^{\infty}_{0}-\int^{\infty}_{0}vdu]$$ $$m_X=-[x[1-F_X(x)]|^{\infty}_{0}]+\int^{\infty}_{0}[1-F_X(x)]dx$$ The first term is zero at x = 0. As $$x\rightarrow\infty, 1-F_X(x)$$ tends to zero faster than the increase of $$x$$ and thus $$x[1-F_X(x)]\rightarrow0$$ Therefore $$m_X=\int^{\infty}_{0}[1-F_X(x)]dx$$ QED Enjoy!
 P: 5 \begin{align*} E[X] &= E\bigg[\int_0^X 1\,dx\bigg]\\ &= E\bigg[\int_0^\infty 1_{\{X>x\}}\,dx\bigg]\\ &= \int_0^\infty E[1_{\{X>x\}}]\,dx\\ &= \int_0^\infty P(X > x)\,dx\\ &= \int_0^\infty (1 - F(x))\,dx \end{align*} By the way, this formula is true no matter what kind of random variable X is, and we do not need anything more than freshman calculus to understand the integral on the right-hand side. (We need neither measure theory nor Stieltjes integrals.) Even when X is discrete, the function 1 - F(x) is still at least piecewise continuous, so the integral makes perfectly good sense, even when understood as a good old-fashioned Riemann integral.

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