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Random Variable Measurability w.r.t. Sigma Fields

  1. Sep 14, 2011 #1
    Hello everyone,

    I'm having a little trouble with a probability problem with three parts; I think I'm having trouble wrapping my head around just what's going on here. If anyone could give me a starting point, I'd appreciate it.

    Here's the problem (Billingsley 5.1) (X a random variable)

    a. Show that X is measurable w.r.t. the sigma field J iff sigma(X) is a subset of J. Show that X is a measurable w.r.t. sigma(Y) iff sigma(x) is a subset of sigma(Y)

    b. Show that if J = {empty set, omega}, then X is measurable w.r.t. J iff X is constant.

    c. Suppose that P(A) is 0 or 1 for every A in J. This holds, for example, if J is the tail field of an independent sequence, or if J consists of the countable and cocountable sets on the unit interval with Lebesgue measure. Show that if X is measurable w.r.t. J, then P[X=c] = 1 for some constant c.

    Thanks for any and all help!

    Best regards
  2. jcsd
  3. Sep 14, 2011 #2
    Did you see topology already?? The techniques used here ressembles the techniques in topology.

    Anyway, for (a), you need to prove that if X is measurable, then [itex]\sigma(X)[/itex] is a subset of J.

    So, firstly, how is [itex]\sigma(X)[/itex] defined?
  4. Sep 14, 2011 #3

    I thought there might be some sort of topological argument, but the book is very analysis-oriented, so I was trying to stick to that line of thinking. σ(X) is defined as the smallest sigma field that X is measurable w.r.t., i.e. the intersection of all such fields.
  5. Sep 14, 2011 #4
    OK, so [itex]\sigma(X)[/itex] is the smallest sigma field such that X is measurable. Doesn't that make it easy to show that [itex]\sigma(X)\subseteq J[/itex]?? X is measurable w.r.t. J after all...
  6. Sep 14, 2011 #5
    Ah yes, of course it does. I'm not sure why I missed that. Thanks. As for the other parts, any suggestions?
  7. Sep 14, 2011 #6
    For the reverse, you know that [itex]\sigma(X)\subseteq J[/itex]and that [itex]\sigma(X)[/itex] makes X measurable. You need to show that J makes X measurable. So by going to a finer sigma field, you preserve measurability. That shouldn't be too difficult?
  8. Sep 14, 2011 #7
    No, it's not bad at all. It's more parts b and c that I remain a bit lost on
  9. Sep 14, 2011 #8
    OK take X to be measurable wrt the trivial sigma field. We wish to prove that X is constant.

    What is [itex]\{X=c\}[/itex] for each c in [itex]\mathbb{R}[/itex]?? A measurable set w.r.t. the trivial sigma field,right?? What can you deduce?
  10. Sep 14, 2011 #9
    Yes, definitely measurable. Oh! If X is not constant, then the inverse image maps to a set strictly smaller than the space, or am I completely confused now? My apologies, for some reason I'm having particular difficulty on this one.
  11. Sep 14, 2011 #10
    OK, [itex]\{X=c\}[/itex] is a measurable set wrt to the trivial sigma-algebra. But the trivial sigma-algebra only has two measurable sets: [itex]\emptyset[/itex] and [itex]\Omega[/itex].

    So what are the only possibilities for [itex]\{X=c\}[/itex]??
  12. Sep 14, 2011 #11
    Ah, either the empty set or the entire space. So, for {X=C} to be measurable it either has to be the empty set or omega.
  13. Sep 14, 2011 #12
    Yes! And what does it mean that [itex]\{X=c\}=\Omega[/itex]??
  14. Sep 14, 2011 #13
    It means the set of x in omega s.t. X(x) = c is the entire space, correct?
  15. Sep 14, 2011 #14
    So, if [itex]\{X=c\}=\Omega[/itex], then the function is constant, right??

    What you have proven is that either [itex]\{X=c\}[/itex] is empty or is omega. If there is a c such that [itex]\{X=c\}[/itex], then you're done.

    Now prove that there must exist such a c. Hint: what is [itex]\{X\in \mathbb{R}\}[/itex]??
  16. Sep 15, 2011 #15
    Yes, the function is assuredly constant. X in R is the set of omega such that X(w) is in R. However, X is a random variable, so it maps from the the space to R. So, I feel the connections starting to coalesce in my head... I think now, Since J is the empty set or the whole space and we've shown that either {X=c} is empty or is omega, then since X in R is the set of omega such that X(w) is in R, which should be the entire space. Then, since if a X(y) = d, whereas X(x) = c elsewhere, both these sets would map back to sets strictly smaller than the space, but strictly larger than the empty set by the way they're defined, yes?
  17. Sep 15, 2011 #16
    Seems alright!!
  18. Sep 15, 2011 #17
    Awesome, I can't thank you enough for both your help and patience!
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