Basic Probability-Measurable function, could anyone help me?

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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...
 

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someone help plz
 
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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.
 
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thanks Zentil!
though, coul dyou please explian it a bit further that what's {f<a}..and wha'ts a here
 
1. I believe the sigma algebra generated by a finite set is the power set.

I believe this is in general not true.
 
though, coul dyou please explian it a bit further that what's {f<a}..and wha'ts a here

{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.
 
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{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.
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!
 
'a' represents some value in the codomain of the random variable.

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?
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?
 
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Sorry Pere...what does the 'a' specifically represent please? Can we use the question as the example? is it 1/2? Thanks!
 
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.
 
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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....
 
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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?)
 
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.
 
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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?
 
Re-read what I wrote above and try to answer my very specific questions (not involving any a).

So again I ask you: what are the values of w such that X(w) is less than 1/2 ?
 
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Thanks for getting back to me...
If just less than 1/2 (not including =1/2)
They're
{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?
 
quadraphonics
Thanks for getting back to me...
If just less than 1/2 (not including =1/2)
They're
{0, 1/4}
Is this correct?
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?
 
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!
 
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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
{[0,1/4),[1/4,1/2)}

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!
 
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
{[0,1/4),[1/4,1/2)}
!
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

{X<1/2}=[0,1/2)

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.
 
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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!
 
Th
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?
{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.
 
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Opps..sorry, I do, I do agree....
this time...
{X<3/4}=[0,1/2]
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?
 
{X<3/4}=[0,1/2]?
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].

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
This is a very good idea.
 
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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].
I'm so sorry, you must get tired by now.....sure, it should be [0,1]...I was keep thinking about the value of X(w), but when you ask {X<a} you are acutally after w such that...{w:X(w)<a}....I've been silly

Thanks so much for your help!!!
 

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