Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given that $X\ge 0, q>0 $ and $P$ is the probability measure, I need to prove the following in a probability theory perspective:

  1. $E X = \int_{0}^\infty P(X>x) \,dx$
  2. $EX = \int_{0}^\infty xf(x) \,dx$
  3. $EX^q = \int_{0}^\infty qx^{q-1}P(X>x) \,dx$

In proving 1, I want to start with the definition of expected value which is: $EX = \int_{\Omega} X \,dP$, but not sure how to proceed. I checked with the other examples on this site describing this, but they use the existence of density function to prove 1. Can we prove 1 without using it? Also I think if I can prove 3 first, I can get 1 by setting $q=1$. Please help

share|improve this question
    
Someone down-voted this. Could they explain why? –  Michael Hardy Nov 28 '12 at 13:58
    
OK, my answer has now been updated to include a sketch of a proof of 2. It also shows how 2 entails 1, and how 1 entails 3. –  Michael Hardy Nov 30 '12 at 2:33

2 Answers 2

Assertion 2 is often taken as the definition of the expectation of a random variable with density, hence if you do not take it as a definition you might want to explain what is.

Assertions 1 and 3 are valid in the wider context of nonnegative random variables, with density or not. Assertion 1 is a special case of Assertion 3, which follows from Tonelli theorem applied to the almost sure identity $$ X^q=\int_0^Xqx^{q-1}\,\mathrm dx=\int_0^{+\infty}qx^{q-1}\mathbf 1_{X\geqslant x}\,\mathrm dx. $$

share|improve this answer

One of my first thoughts is to try to derive $(1)$ from $(2)$ by integrating by parts: $$ \int_0^\infty xf(x)\,dx = \int x\,dv = xv-\int v\,dx = \left.xF(x)\right|_0^\infty - \int_0^\infty F(x)\,dx\ldots \text{?} $$ This won't work, for reasons you'll see if you think about what that last thing says. So go from $0$ to $a$ and then take $\lim\limits_{a\to\infty}$: $$ aF(a) - \int_0^a F(x)\,dx = \int_0^a F(a)-F(x)\,dx = \int_{(0,\infty)} 1_{(0,a]}(x) (F(a)-F(x)) \,dx. $$ You can then use the monotone convergence theorem to get $$ \int_0^\infty 1-F(x)\,dx. $$ as the limit.

(Of course then you'd want to show $(2)$ follows from $\mathbb EX = \int_\Omega X\,dP$.)

Later addendum: Let $Y=X^q$. Then if you've established $(1)$, then you have $$ \mathbb EY = \int_0^\infty P(Y>y)\,dy. $$ Let $y=x^q$ so that $dy=qx^{q-1}\,dx$. And notice that $Y>y$ iff $X>x$, so that $P(Y>y)=P(X>x)$. Then the integral above becomes $$ \int_0^\infty P(X>x)\left(qx^{q-1}\,dx\right). $$

Still later addendum: How to prove $(2)$ is something I'd forgotten the details of.

We have a probability measure $P$ on measurable subsets of a space $\Omega$, and a measurable function $X:\Omega\to\mathbb R^+$. "Measurable" will mean that inverse-images of Borel subsets of $\mathbb R$ are measurable subsets of $\Omega$. The "probability distribution" of $X$ is a probability measure $Q$ on Borel subsets of $\mathbb R^+$, defined by $Q(A) = P(\{\omega\in\Omega : X(\omega)\in A\})$. So the problem is to prove that $$ \int_\Omega X(\omega)\,P(d\omega) = \int_{\mathbb R^+} x \, Q(dx).\tag{i} $$ The integral on the left in $(i)$ is the supremum of integrals of simple function (nonnegative functions with finite images) that are $\le X$. The one on the right is the supremum of integrals of simple functions of $x$ that are $\le x$. It remains to show that there is a bijection between simple functions of $X$ that are $\le X$ and simple functions of $x$ that are $\le x$, and that that bijection preserves the values of the integrals. You therefore get the same set of values of integrals of simple functions in both cases, hence the same sup in both cases.

share|improve this answer
    
Thank you Michael, Do you have any idea about 3? –  Kamini Nov 28 '12 at 5:26
    
I've added something about how to derive $(3)$ from $(1)$. –  Michael Hardy Nov 28 '12 at 13:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.