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

[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."

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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!

 

[Hurkyl]

"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...

 

[NoMoreExams]

This works for discrete, cont., and mixed. Though the derivative statement applies for cont. only. Use integration by parts for cont. case.

 

[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?

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!

 

[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.

 

[Hurkyl]

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.

 

[Hurkyl]

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...

 

[kingwinner]

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!

 

[HallsofIvy]

Hurkyl, you don't "need" measure theory to write a sum as an integral. The Riemann-Stieljes integral will do that.

 

[Adeimantus]

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.

 

[Preno]

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.

 

[lorta230]

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!

 

[jason1995]

$$\begin{align*} E[X] &= E\bigg[\int_0^X 1\,dx\bigg]\\ &= E\bigg[\int_0^\infty 1_{\{X>x\}}\,dx\bigg]\\ &= \int_0^\infty E[1_{\{X>x\}}]\,dx\\ &= \int_0^\infty P(X > x)\,dx\\ &= \int_0^\infty (1 - F(x))\,dx \end{align*}$$

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.