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Basic Probability-Measurable function, could anyone help me?

  1. Apr 5, 2008 #1
    I know this is basic maths, Though I just could get my head round it, so confused about the definition of measurable, could someone give me some idea please? Thanks so much!

    Exercise 1. Let = [0; 1] and F be the sigma-algebra generated by the dyadic intervals
    [k/4,(k+1)/4]; k = 0; 1; 2; 3 :
    (a) List all events in F.
    (b) Let X be function defined on omega (I don't know how to type the greek letter) as follows:
    X(w)={k/4 if k/4=< w < (k+1)/4 and w<1/2
    1/2 if w>=1/2 }
    where k = 0; 1; 2; 3. Show that X is F-measurable.
    (c) Give an example of a function Z : omega-->R such that Z takes only finite number of values and is not F-measurable.

    I guess once I've got the idea sorted..then I'll be able to understand more...
  2. jcsd
  3. Apr 6, 2008 #2
    someone help plz
  4. Apr 6, 2008 #3
    1. I believe the sigma algebra generated by a finite set is the power set. 2. This follows from the definition of a measurable function. Show that {f<a} is a set in the sigma algebra found in 1. 3. Again, this is just from the definition of measurable. Find a function such that {f<a} is not a set in your sigma-algebra.
  5. Apr 6, 2008 #4
    thanks Zentil!
    though, coul dyou please explian it a bit further that what's {f<a}..and wha'ts a here
  6. Apr 6, 2008 #5

    I believe this is in general not true.
  7. Apr 6, 2008 #6

    {f<a} means the set of all w in Omega such that f(w) < a. Since this is some subset of Omega you can ask it is in the sigma algebra or not. If this happens to be the case for all a, then f is measurable.
  8. Apr 6, 2008 #7
    Thanks so much for your reply!
    actually in one of my notes, the def is {w∈omega: X(w)<a}∈F, a∈R
    then X is F-measurable.

    This is very similar to the explaination that you were given, however I still don't get what's 'a' here represents.

    I understand that I've got to test all w to see weather they're in F or not. They're. Do I need to see if X(w) (ie k/4 and 1/2) are also in F or not?

    Since I'm not so clear about the definition here, so I couldn't given an example of NOT F-Measure. could you help please?
    Thanks so much for your help!
  9. Apr 6, 2008 #8
    'a' represents some value in the codomain of the random variable.

    This is not right. F is a sigma algebra and as such it contains SETS not elements.
    You have to check if {X<a} (whose meaning you now undestand) lies in F.

    I suggest you first try and figure out what {X<1/2} is in your specific example.

    It is the set of all w which, when plugged into X, give a result less than 1/2. What is this set?
  10. Apr 6, 2008 #9
    Sorry Pere...what does the 'a' specifically represent please? Can we use the question as the example? is it 1/2? Thanks!
  11. Apr 6, 2008 #10
    Yes, let's take your question as an example and consider the set {X<1/2} that is a=1/2 in the terminology used earlier.

    It might help to graph the function X(w). Then you can see where its values are less than 1/2.
  12. Apr 6, 2008 #11
    Thanks Pere...does the a=1/2 comes from X(w)=1/2 when w>=1/2 the first 1/2 or the second please?
    if it's from the first, then we have a=k/4 for the other condition do we?
    Sorry I just have'nt got my head around it....
  13. Apr 6, 2008 #12
    the graph does help thanks. If I graph X(w) all values of X(w) are either 0 or 1/2(though 1/2 isn't less than a, it's euqal to a, is it all right?)
  14. Apr 6, 2008 #13
    I just picked a=1/2 as an arbitrary example. You will eventually have to show it for all real values of a:smile:

    So again I ask you: what are the values of w such that X(w) is less than 1/2 ? It is a SET not a single value !

    EDIT: You function takes on the values 0, 1/4 and 1/2 so your graph is not quite right.
    Last edited: Apr 7, 2008
  15. Apr 6, 2008 #14
    sorry, I'm a bit confused here....so what's a exactly?
    is a a set value?
    or just any real value? or something else?

    what's the value for 'a' in my example please?
  16. Apr 7, 2008 #15
    Re-read what I wrote above and try to answer my very specific questions (not involving any a).

  17. Apr 7, 2008 #16
    Thanks for getting back to me...
    If just less than 1/2 (not including =1/2)
    {0, 1/4}
    Is this correct?
    May I know the connection between value of X(w) and F measurable? Could you please give me an example of function Z being not F-measurable?
  18. Apr 7, 2008 #17
    No, neither of those events are even in F. Events in F are closed intervals, not singletons. Maybe it's a good idea to revisit part a, where you list all the elements of the sigma algebra?
  19. Apr 7, 2008 #18
    No this is not correct. Give it another try.

    Maybe it helps to first rephrase my question in your own words.
    You're looking for ALL values w in [0,1] (=Omega) such that the function X evaluated at w is less than 1/2.

    What is X(1/5)? Is this less then 1/2? If so then 1/5 is an element of the set {X<1/2}.

    What is X(1/3)? Is it less than 1/2? If so, 1/3 is in {X<1/2}.

    1/3 and 1/5 are some of the w values we are looking for. Try to find all!
  20. Apr 7, 2008 #19
    Thanks for your patience!

    Is it
    for all w<1/2 their X(w) is less than 1/2

    but for all w>=1/2, their X(w) equals 1/2
    Therefore, the set is

    Would this be right?
    if we consider {X<=1/2}
    then it's {[0,1/4),[1/4,1/2),[1/2,3/4),[3/4,1)}

    Since the events in F contains:
    {emptyset, [0,1/4],[1/4,1/2],[1/2,3/4],[3/4,1],[0,1],[0,1/2],[1/4,3/4],[1/4,1],[1/2,1]} (could you please help me to check it this? I still feel a bit redundant to use [] instead of [) )

    so to see whether X is F measurable, do I need to see whether all values of w is in F? sorry I still can't see what does the real number 'a' here do...........
    Could you please give me an example of function Z being not F-measurable?
    Thanks heaps!
  21. Apr 8, 2008 #20
    X(w) is less than 1/2 for all w strictly less than 1/2. This is correct. Therefore the set {X<1/2} which by defintion consists of those w for which X(w) is less than 1/2 is


    As quadrophonics indicated earlier, this is an interval, not a union of singletons. You should be aware of whether the intervals are open or closed, sometimes this turns out to be crucial.

    This was for a=1/2.
    We must however check {X<a} for all real numbers a. So let's try a=1/3: What is {X<1/3}, i.e. what is the set of w such that X(w) is less than 1/3 ?

    Once you have understood what the sets {X<a} are and how you find them in your example, you have to check whether or not they're elements of your sigma algebra. Recall that a sigma algebra is a set of sets, so altough {X<a} is most likely to be an interval it can well be an element of F.

    I'll check later if what you found for F is correct, first try and understand the sets {X<1/3} and {X<a} for any 0<a<1.
  22. Apr 8, 2008 #21
    Thanks so much for your answer and leading my way!
    so, {X<1/3}=[0,1/4), which is in F. Do we just pick a value for 'a' and test whether the interval is in F please? and then we conclude, if this is ture for all real number between 0 and 1, then X is in F. I guess I get it....so a is not a fix number, it's any number within the range?

    as of for the subsets in F, I guess, I'm missing a whole lot of values, and I forgot to include compliments and intersections in, I should, shouldn't I?
    Many thanks!
  23. Apr 8, 2008 #22
    {X,1/3}=[0,1/4] is not right.

    Your function X is zero on [0, 1/4), it is 1/4 on [1/4, 1/2) and it is 1/2 on [1/2, 1], do you agree?

    So the interval where it is less than 1/3 is [0, 1/2) the same as {X < 1/2} :smile:
    Another try, what is {X < 3/4}?

    For the sigma algebra F ... yes you have to find a collection of sets which contains the dyadic intervals and is closed under taking the complement or unions / intersections, so you basically start with the dyadic intervals and combine / intersect them until you don't get anything new.

    Then if for each real number a, the set {X<a} is in F, X is measurable.
  24. Apr 8, 2008 #23
    Opps..sorry, I do, I do agree....
    this time...
    Sorry, I'm such a hard student to teach, am I?
    One more question please, for how to choose a, do I need to separate a into the following intervals for convinience? ie 0<a<=1/4, 1/4<a<1/2,a>1/2 so that a∈R, or, do I just randomly select a few values, eg, 1/2, 1/3,and say X are all in F after a few trials? which way is better please?
  25. Apr 8, 2008 #24
    Ehm, didn't you say you agree that the function X is 1/2 on [1/2, 1]..... that is for any w in [1/2, 1] X(w)=1/2, which is less than 3/4 .... in fact X is ALWAYS less then 3/4 so {X<3/4}=[0, 1].

    This is a very good idea.
  26. Apr 8, 2008 #25
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