# Random Variables

• Oct 2nd 2009, 02:21 PM
azdang
Random Variables
Given $(\Omega, \mathcal{A}, P)$, suppose X is a random variable with $X\geq0$ and E{X}=1. Definite $Q:\mathcal{A}->R$ by Q(A)= $E{(X1_A)}$. Show that if P(A) = 0, then Q(A)=0. Give an example that shows that Q(A)=0 does not in general imply P(A)=0.

I'm pretty sure that $E{(1_A)}=P(A)$, so I was thinking that maybe we could break Q(A) up into $E{(X)}E{(1_A)}$. However, this would be assuming that X and 1_A are independent random variables? Would this be true? If so, then it's very obvious why, if P(A)=0, then Q(A)=0. Anyone have any hints? Thanks!
• Oct 4th 2009, 03:20 AM
Laurent
Quote:

Originally Posted by azdang
Given $(\Omega, \mathcal{A}, P)$, suppose X is a random variable with $X\geq0$ and E{X}=1. Definite $Q:\mathcal{A}->R$ by Q(A)= $E{(X1_A)}$. Show that if P(A) = 0, then Q(A)=0. Give an example that shows that Q(A)=0 does not in general imply P(A)=0.

I'm pretty sure that $E{(1_A)}=P(A)$, so I was thinking that maybe we could break Q(A) up into $E{(X)}E{(1_A)}$. However, this would be assuming that X and 1_A are independent random variables? Would this be true? If so, then it's very obvious why, if P(A)=0, then Q(A)=0. Anyone have any hints? Thanks!

It is false (without further assumption) that $Q(A)=E{(X)}E{(1_A)}$. However, if $P(A)=0$, then $X 1_A=0$ almost-everywhere (it is 0 outside $A$, and $A$ is negligible), hence $Q(A)=0$.

As for the example, you must choose $X$ in such a way that $X 1_A$ is zero everywhere, i.e. $X$ is 0 on $A$, and it is chosen outside A in such a way that $E[X](=E[X 1_{A^c}])=1$).
• Oct 4th 2009, 06:31 PM
azdang
Ooh, the first part seems so obvious. Thank you, Laurent. :)

As for the second part, I'm still a little lost. My first question is: Is it correct that $X1_A$ = X if x is in A and 0 if x is not in A?
• Oct 7th 2009, 01:19 PM
Laurent
Quote:

Originally Posted by azdang
Ooh, the first part seems so obvious. Thank you, Laurent. :)

As for the second part, I'm still a little lost. My first question is: Is it correct that $X1_A$ = X if x is in A and 0 if x is not in A?

No. What is true is that $X(\omega) 1_A(\omega)=X(\omega)$ if $\omega\in A$ and $X(\omega) 1_A(\omega)=0$ else. Remember $A$ is an event, not a subset of $\mathbb{R}$.

For instance (you should write this more formally), suppose we toss a fair coin, and define the event $A=\{\text{the outcome is heads}\}$ and $X$ is defined to be 0 in case heads shows up, and to be 2 if tails shows up. Then what are $P(A)$, $E[X]$ and $E[X 1_A]$?
• Oct 13th 2009, 07:35 AM
azdang
Well, P(A) should be .5, correct? So, wouldn't $EX1_A$ be 0 in either case (if we roll a heads or a tales)? So in this case, Q(A) = 0, but P(A) does not equal 0.
• Oct 13th 2009, 07:58 AM
Laurent
Quote:

Originally Posted by azdang
Well, P(A) should be .5, correct? So, wouldn't $EX1_A$ be 0 in either case (if we roll a heads or a tales)? So in this case, Q(A) = 0, but P(A) does not equal 0.

Yes. (More correctly, $X1_A$ is 0 in both cases, hence $E(X1_A)=0$. )