Random Variables

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October 2nd 2009, 01:21 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!
October 4th 2009, 02:20 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 $E[X](=E[X 1_{A^c}])=1$ ).
October 4th 2009, 05:31 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?
October 7th 2009, 12:19 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 $E[X 1_A]$ ?
October 13th 2009, 06:35 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.
October 13th 2009, 06:58 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 $E(X1_A)=0$ . )