# [SOLVED] question on expectations and cumulative distribution functions

• Oct 26th 2008, 08:00 PM
joop
[SOLVED] question on expectations and cumulative distribution functions
I want to prove that $\displaystyle E(X) = \int_{0}^{\infty} (1-F_X(X))\, dx$ is true with the only information given being that X is a continuous RV such that P(X > 0) = 1.

The hint that I've received from my teacher is to replace $\displaystyle F_X(X)$ with the integral of the pdf, to get $\displaystyle E(X) = \int_{0}^{\infty} (1- \int_{0}^{\infty} f_T(t)\, dt )\, dx$ (using dummy variable t) and then switch the order I do the integrals.

Also, with $\displaystyle E(X) = \int_{-\infty}^{\infty} xf(x)\, dx$ (the definition of E(X)) I'm not sure how I would integrate this without knowing exactly what f(x) is.

Thanks in advance for any help :)
• Oct 27th 2008, 12:12 AM
joop
Ok, I realized that I should be working with
$\displaystyle E(X) = \int_{0}^{\infty} (1- \int_{0}^{x} f_T(t)\, dt )\, dx$

but I'm still trying to figure out reversing the order of integration.
• Oct 27th 2008, 12:23 AM
mr fantastic
Quote:

Originally Posted by joop
Ok, I realized that I should be working with
$\displaystyle E(X) = \int_{0}^{\infty} (1- \int_{0}^{x} f_T(t)\, dt )\, dx$

but I'm still trying to figure out reversing the order of integration.

I was going to point that out but had no time.

$\displaystyle \int_{0}^{\infty} (1- \int_{0}^{x} f_T(t)\, dt )\, dx$

$\displaystyle = \int_{0}^{\infty} \int_{x}^{+\infty} f_T(t)\, dt \, dx$

It should be smooth from here.
• Oct 27th 2008, 12:35 AM
joop
Would you please explain to me how you got
$\displaystyle \int_{0}^{\infty} (1- \int_{0}^{x} f_T(t)\, dt )\, dx$
$\displaystyle = \int_{0}^{\infty} \int_{x}^{\infty} f_T(t)\, dt \, dx$

To the best of my understanding,
$\displaystyle \int_{0}^{\infty} (1- \int_{0}^{x} f_T(t)\, dt )\, dx$
$\displaystyle = \int_{0}^{\infty} \, dx - \int_{0}^{\infty} \int_{0}^{x} f_T(t)\,dt \,dx$

and reversing the order,
$\displaystyle = \int_{0}^{\infty} \, dx - \int_{0}^{\infty} \int_{t}^{\infty} f_T(t)\,dx \,dt$

• Oct 27th 2008, 02:42 AM
mr fantastic
Quote:

Originally Posted by joop
Would you please explain to me how you got
$\displaystyle \int_{0}^{\infty} (1- \int_{0}^{x} f_T(t)\, dt )\, dx$
$\displaystyle = \int_{0}^{\infty} \int_{x}^{\infty} f_T(t)\, dt \, dx$

To the best of my understanding,
$\displaystyle \int_{0}^{\infty} (1- \int_{0}^{x} f_T(t)\, dt )\, dx$
$\displaystyle = \int_{0}^{\infty} \, dx - \int_{0}^{\infty} \int_{0}^{x} f_T(t)\,dt \,dx$

and reversing the order,
$\displaystyle = \int_{0}^{\infty} \, dx - \int_{0}^{\infty} \int_{t}^{\infty} f_T(t)\,dx \,dt$ Mr F says: I'd be interested to see how you do the first integral ......

You should have seen that the second line follows from:

$\displaystyle 1 = \int_{0}^{x} f_T(t)\, dt + \int_{x}^{+\infty} f_T(t)\, dt$

$\displaystyle \Rightarrow 1 - \int_{0}^{x} f_T(t)\, dt = \int_{x}^{+\infty} f_T(t)\, dt$.
• Oct 27th 2008, 08:10 AM
joop
Thanks. Here is what I got (which I'm sure is correct):

Starting from Mr. F's help, and reversing the order of integration:
$\displaystyle \int_{0}^{\infty} \int_{0}^{t} f(t) \, dt$
$\displaystyle = \int_{0}^{\infty} t f(t) \, dt$
rename t to x
$\displaystyle \int_{0}^{\infty} x f(x) \, dx = \int_{0}^{\infty} x f(x) \, dx$
(Defn of E(X) = what I just got)

Thanks again.
• Oct 27th 2008, 02:18 PM
mr fantastic
Quote:

Originally Posted by joop
Thanks. Here is what I got (which I'm sure is correct):

Starting from Mr. F's help, and reversing the order of integration:
$\displaystyle \int_{0}^{\infty} \int_{0}^{t} f(t) \, {\color{red}dx} \, dt$
$\displaystyle = \int_{0}^{\infty} t f(t) \, dt$
rename t to x
$\displaystyle \int_{0}^{\infty} x f(x) \, dx = \int_{0}^{\infty} x f(x) \, dx$
(Defn of E(X) = what I just got)

Thanks again.

You're welcome. I've corrected (in red) a small omission.