Sigma Algebra - Help please

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  • #2
haruspex
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The subsets in G all have the property that {x, T} is an element if and only if {x, H} is an element. So, when considering whether an actual result xy is a member of such subset, the result of the second coin toss is irrelevant.
(3/4, 6) is just an interval on the real line, i.e. (0.75, 6). In two coin tosses, the number of heads that may result is 0, 1 or 2. Of those, only 1 and 2 lie in B. The set of outcomes of the coin tossing that each lead to X lying in B are HT, TH, HH. But this set is not in G.
 
  • #3
The subsets in G all have the property that {x, T} is an element if and only if {x, H} is an element. So, when considering whether an actual result xy is a member of such subset, the result of the second coin toss is irrelevant.
(3/4, 6) is just an interval on the real line, i.e. (0.75, 6). In two coin tosses, the number of heads that may result is 0, 1 or 2. Of those, only 1 and 2 lie in B. The set of outcomes of the coin tossing that each lead to X lying in B are HT, TH, HH. But this set is not in G.

thank you very much for your reply but tbh I don't understand the first part of your answer that why the result of second coin toss is irrelevant?
I have managed to understand the half part of your second answer that ht, th, hh doesn't belong to G but I still don't understand that what is the logic behind telling about some interval of real line here i.e. (3/4, 6) where has it come from?
I will really appreciate if you can explain this to me.
 
  • #4
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I assume you followed the link to http://www.theponytail.net/wiki/pmwiki.php/Main/RandomVariable, which shows how Ω and F are defined.
G consists of the subsets {0, {HH,HT}, {TH,TT}, {HH,HT,TH,TT}}. All of these subsets have the property that xH is a member if and only if xT is a member. So in order to decide whether a given result xy is an element of the subset it is only necessary to check x.
It might help to put the subsets into words. {HH, HT} is the event that the outcome is either HT or HH. That is equivalent to specifying only that the first toss outcome is a Head, and so on for the other combinations. So none of the events care about the result of the second coin toss.

The interval (3/4, 6) is fairly arbitrary. The author could have chosen any interval (a, b) where 0 <= a < 1 and 5 < b <= 6.
 
  • #5
I assume you followed the link to http://www.theponytail.net/wiki/pmwiki.php/Main/RandomVariable, which shows how Ω and F are defined.
G consists of the subsets {0, {HH,HT}, {TH,TT}, {HH,HT,TH,TT}}. All of these subsets have the property that xH is a member if and only if xT is a member. So in order to decide whether a given result xy is an element of the subset it is only necessary to check x.
It might help to put the subsets into words. {HH, HT} is the event that the outcome is either HT or HH. That is equivalent to specifying only that the first toss outcome is a Head, and so on for the other combinations. So none of the events care about the result of the second coin toss.

The interval (3/4, 6) is fairly arbitrary. The author could have chosen any interval (a, b) where 0 <= a < 1 and 5 < b <= 6.

Thanks again for your response.
I am looking at:
http://www.theponytail.net/wiki/pmwiki.php/Main/SigmaAlgebra
the first answer is now crystal clear, OK (3/4, 6) is some arbitrary interval but I still don't understand that why have you used the 0 & 6 in the range? & what is the connection of this arbitrary interval?
 
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  • #6
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Yes, but that page doesn't define everything it uses. You need to follow the link from there to http://www.theponytail.net/wiki/pmwiki.php/Main/RandomVariable.
OK (3/4, 6) is some arbitrary interval but I still don't understand that why have you used the 0 & 6 in the range? & what is the connection of this arbitrary interval?
I may have added to the confusion here. I should have said:
The author could have chosen any interval (a, b) where 0 <= a < 1 and 2 < b.
I.e. any interval that includes 1 and 2 but not 0. The event that X maps into that interval is {HT, TH, HH}, which is not in G. So the random variable 'number of heads in two tosses' is not measurable in G.
 
  • #7
Yes, but that page doesn't define everything it uses. You need to follow the link from there to http://www.theponytail.net/wiki/pmwiki.php/Main/RandomVariable.

I may have added to the confusion here. I should have said:
The author could have chosen any interval (a, b) where 0 <= a < 1 and 2 < b.
I.e. any interval that includes 1 and 2 but not 0. The event that X maps into that interval is {HT, TH, HH}, which is not in G. So the random variable 'number of heads in two tosses' is not measurable in G.

why 0, 1 & 2 in interval (a, b) where 0 <= a < 1 and 2 < b
is it because 0 for null event, 1 for one head , and 2 for two heads?
 
  • #8
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why 0, 1 & 2 in interval (a, b) where 0 <= a < 1 and 2 < b
is it because 0 for null event, 1 for one head , and 2 for two heads?
For the purpose of the article, the author wanted a r.v. X which would not be measurable in G. The author chose 'number of heads' as the r.v. Next, the author needed a set of values of X which did not, as a set, correspond to an event in G. The author chose the set {1, 2} ("either one or two heads"). Finally, the author needed a measurable set in F that included that set but no other possible values of X. Any interval containing 1 and 2 but not 0 would have done. The author chose (3/4, 6).
So there are several arbitrary choices in there. The author could have chosen the X values {0, 1} instead, and to go with them maybe the interval (-100, π/2).
 
  • #9
For the purpose of the article, the author wanted a r.v. X which would not be measurable in G. The author chose 'number of heads' as the r.v. Next, the author needed a set of values of X which did not, as a set, correspond to an event in G. The author chose the set {1, 2} ("either one or two heads"). Finally, the author needed a measurable set in F that included that set but no other possible values of X. Any interval containing 1 and 2 but not 0 would have done. The author chose (3/4, 6).
So there are several arbitrary choices in there. The author could have chosen the X values {0, 1} instead, and to go with them maybe the interval (-100, π/2).

So, does it mean that B is a Borel set which belongs to Borel-algebra and a random variable is some function that gives some value which belongs to B (Borel set) but that Borel set is not in G ? Please correct me if I am wrong.
My second question is now that should I believe that given G (σ-algebra) is a subset of Borel-algebra but converse is not true therefore as Borel-algebra is the smallest algebra then G is more smaller than Borel-algebra, if it is true then why Borel-algebra is called smallest algebra as there exist a σ-algebra such as G which is smaller than Borel-algebra.
Sorry if my questions are absurd but I am trying to understand these concepts since last night & now it's morning I am still confused.
thanks in advance
 
  • #10
Also, is it true if Ω = {HH,HT,TH,TT}
then Borel-sigma-algebra = {empty set, {HH,HT,TH,TT}} ?
And the Borel set B=(3/4,6) has come from this Ω ? if yes, then how? I mean contains character values such as hh, th etc whereas B contains an interval with number values. Also, what will be the maximum value of b in 0 <= a < 1 and 2 < b ?
What does X^-1 stands here? is it inverse transformation sampling or is it the one given here in abstract of:
http://users.uoa.gr/~npapadat/papers/Selfinverse.pdf
 
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  • #11
Yes, but that page doesn't define everything it uses. You need to follow the link from there to http://www.theponytail.net/wiki/pmwiki.php/Main/RandomVariable.

I may have added to the confusion here. I should have said:
The author could have chosen any interval (a, b) where 0 <= a < 1 and 2 < b.
I.e. any interval that includes 1 and 2 but not 0. The event that X maps into that interval is {HT, TH, HH}, which is not in G. So the random variable 'number of heads in two tosses' is not measurable in G.
why not 0 is included in the interval? And isn't {HH, HT}the event that X maps into that interval? as 1=HT & 2=HH ?
 
  • #12
For the purpose of the article, the author wanted a r.v. X which would not be measurable in G. The author chose 'number of heads' as the r.v. Next, the author needed a set of values of X which did not, as a set, correspond to an event in G. The author chose the set {1, 2} ("either one or two heads"). Finally, the author needed a measurable set in F that included that set but no other possible values of X. Any interval containing 1 and 2 but not 0 would have done. The author chose (3/4, 6).
So there are several arbitrary choices in there. The author could have chosen the X values {0, 1} instead, and to go with them maybe the interval (-100, π/2).

Is F the power set of Ω ?
 
  • #13
For the purpose of the article, the author wanted a r.v. X which would not be measurable in G. The author chose 'number of heads' as the r.v. Next, the author needed a set of values of X which did not, as a set, correspond to an event in G. The author chose the set {1, 2} ("either one or two heads"). Finally, the author needed a measurable set in F that included that set but no other possible values of X. Any interval containing 1 and 2 but not 0 would have done. The author chose (3/4, 6).
So there are several arbitrary choices in there. The author could have chosen the X values {0, 1} instead, and to go with them maybe the interval (-100, π/2).

So the author has picked (3/4, 6) because as 0=TT is not included (because we are counting the numbers of heads & TT doesn't contain it) therefore it's probability 1/4 has been subtracted from 1 which makes it 1, and then as the probabilities of HH=1, HT=1/2, and TH=1/2 and the total sum of these three ω1ω2ω3 ε Ω is equal to 1 and as ω1ω2ω3 can also be arranged as ω1ω3ω2, ω2ω1ω3, ω2ω3ω1, ω3ω1ω2, ω3ω2ω1 (the permutation of 3 is 6) hence 6*1=6 .
Please correct me if I am wrong.
 
  • #14
haruspex
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You seem to be very confused still about the relationships.
Ω is the set of all possible atomic events, i.e. outcomes from two coin tosses.
X is a function from Ω to [itex]\Re[/itex], namely, the number of heads observed.
B is a Borel set in [itex]\Re[/itex].
So, does it mean that B is a Borel set which belongs to Borel-algebra and a random variable is some function that gives some value which belongs to B (Borel set) but that Borel set is not in G ?
Yes, except for the last bit. It's the inverse image of B that's not in G.
In the web page you link to, the author refers to the concept of a random variable being 'measurable'. Unfortunately, the link there provided only discusses measurability of subsets of Ω. For a definition of measurability of a r.v. see http://en.wikipedia.org/wiki/Measurable_function (a r.v. is a function).
My second question is now that should I believe that given G (σ-algebra) is a subset of Borel-algebra but converse is not true therefore as Borel-algebra is the smallest algebra then G is more smaller than Borel-algebra, if it is true then why Borel-algebra is called smallest algebra as there exist a σ-algebra such as G which is smaller than Borel-algebra.
G is a subset of "F", but I haven't found a definition of F at that site. I assume it's the powerset of Ω. Is that a Borel algebra? AFAIK, Borel algebras are only defined for topological spaces. If you mean the Borel algebra on [itex]\Re[/itex] then certainly G is not a subset of that, since it is not a subset of [itex]\Re[/itex].
The Borel algebra on [itex]\Re[/itex] is the not the smallest. It's the smallest that contains all open intervals. You could make a smaller one by applying the discrete topology to [itex]\Re[/itex].
Also, is it true if Ω = {HH,HT,TH,TT} then Borel-sigma-algebra = {empty set, {HH,HT,TH,TT}} ?
As above, you have to specify the topology. Different topologies on Ω will give different Borel algebras. Is there a natural topology for a given measure space? I doubt it.
And the Borel set B=(3/4,6) has come from this Ω ?
No. B comes from the Borel algebra on [itex]\Re[/itex].
What does X^-1 stands here?
It's the inverse image of the argument set under the function X. I.e. for A [itex]\subset\Re[/itex], X-1(A) is the set of elements ω of Ω for which X(ω) [itex]\in[/itex] A.
why not 0 is included in the interval?
The author excluded 0 because he/she does not want X-1(B) to include TT.
So the author has picked (3/4, 6) because as 0=TT is not included (because we are counting the numbers of heads & TT doesn't contain it)
I don't understand what you meant by the bit in parentheses. The author wants to exclude TT in order to show that X is not G-measurable.
therefore its probability 1/4 has been subtracted from 1 which makes it 1, and then as the probabilities of HH=1, HT=1/2, and TH=1/2 and the total sum of these three ω1ω2ω3 ε Ω is equal to 1 and as ω1ω2ω3 can also be arranged as ω1ω3ω2, ω2ω1ω3, ω2ω3ω1, ω3ω1ω2, ω3ω2ω1 (the permutation of 3 is 6) hence 6*1=6 .
Please correct me if I am wrong.
I struggle to make any sense of that. How does subtracting 1/4 from 1 make 1? Where do you get HH=1 from? My guess is that you're thinking of conditional probabilities here, which has nothing to do with what we're discussing.
 
  • #15
You seem to be very confused still about the relationships.
Ω is the set of all possible atomic events, i.e. outcomes from two coin tosses.
X is a function from Ω to [itex]\Re[/itex], namely, the number of heads observed.
B is a Borel set in [itex]\Re[/itex].

Thanks again.
Yes, I understand this part but what I am asking that how does X^-1 (0.75,6)={HH,HT,TH} how does the borel set B=(0.75,6) corresponds to {HH,HT,TH} ?
 
  • #16
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Thanks again.
Yes, I understand this part but what I am asking that how does X^-1 (0.75,6)={HH,HT,TH} how does the borel set B=(0.75,6) corresponds to {HH,HT,TH} ?
X only takes 3 values: 0, 1 and 2.
X-1(B) is the set of pairs of tosses that produce a head count in the interval B.
Of the values 0, 1 and 2, only 1 and 2 are in B.
So X-1(B) is the set of pairs of tosses that produce a head count of either 1 or 2.
These pairs are HH, HT, TH.
 
  • #17
X only takes 3 values: 0, 1 and 2.
X-1(B) is the set of pairs of tosses that produce a head count in the interval B.
Of the values 0, 1 and 2, only 1 and 2 are in B.
So X-1(B) is the set of pairs of tosses that produce a head count of either 1 or 2.
These pairs are HH, HT, TH.

OK now I understand it:

A coin is tossed twice, if Ω={HH,HT,TH,TT}
and Borel Sigma algebra = 2^3 = { {1}, {1, 2}, {1, 2, 0}, {2}, {1, 0}, ∅, {0}, {2, 0} } then {1, 2} corresponds to {HH, HT, TH} and {1, 2} so far I understand upto this level, but I am confused that how/why on real line it has the interval (0.75, 6)? is it because HH,HT, & TH are total 6 outcomes in {HH,HT,TH} (counting as HH=HT=TH=2) so that's why 6 is in (0.75, 6), and in Ω there are 8 outcomes (again counting as HH=HT=TH=TT=2, and total 8) & (chosen set of selected outcomes)/(total outcomes) i.e. 6/8 = 0.75 and now that's why 0.75 is in (0.75, 6). Similarly we can calculate it like as 3 outcomes have been chosen from Ω i.e. HH,HT,TH so 3/4 = 0.75 & that's why 0.75 is in (0.75, 6) and as each outcome consists of 2 possible values therefore 3*2=6 & that's why 6 is in (0.75, 6) ??????

Thanks for your patience for helping me, in fact teaching me.
 
  • #18
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I am confused that how/why on real line it has the interval (0.75, 6)? is it because HH,HT, & TH are total 6 outcomes in {HH,HT,TH} (counting as HH=HT=TH=2) so that's why 6 is in (0.75, 6), and in Ω there are 8 outcomes (again counting as HH=HT=TH=TT=2, and total 8) & (chosen set of selected outcomes)/(total outcomes) i.e. 6/8 = 0.75 and now that's why 0.75 is in (0.75, 6). Similarly we can calculate it like as 3 outcomes have been chosen from Ω i.e. HH,HT,TH so 3/4 = 0.75 & that's why 0.75 is in (0.75, 6) and as each outcome consists of 2 possible values therefore 3*2=6 & that's why 6 is in (0.75, 6) ??????
X({HT, TH, HH}) doesn't have the interval (0.75, 6). It takes the values 1 and 2. The author could have chosen, for the purposes of the demonstration, any Borel set on [itex]\Re[/itex] that included 1 and 2 but not 0. (0.75, 6) is a completely arbitrary choice from all those possibilities.
E.g. {HT, TH, HH} = X-1(0.9, 2.1) = X-1(0.01, 423) = X-1(0.75, 6) = X-1((0.9, 1.1) [itex]\cup[/itex] (1.9, 2.1)) = X-1([itex]\Re[/itex] - 0)
 
  • #19
X({HT, TH, HH}) doesn't have the interval (0.75, 6). It takes the values 1 and 2. The author could have chosen, for the purposes of the demonstration, any Borel set on [itex]\Re[/itex] that included 1 and 2 but not 0. (0.75, 6) is a completely arbitrary choice from all those possibilities.
E.g. {HT, TH, HH} = X-1(0.9, 2.1) = X-1(0.01, 423) = X-1(0.75, 6) = X-1((0.9, 1.1) [itex]\cup[/itex] (1.9, 2.1)) = X-1([itex]\Re[/itex] - 0)

OKKKKKKKKKKKKKKKKK....................now I got it.

Please accept my thanks for your help.

God bless you.
 
  • #20
1
0
I never heard of "Sigma Algebra"... but you might be referring to problems involving the sigma sign, Σ, that represents summation notation... a series or, rather, the sum of a sequence

5
∑ n = 1 + 2 + 3 + 4 + 5 = 15
n=1

See, we are creating a series... a sequence of values that are added together. The Sigma sign is followed by a function... in this case "n"

The variable used under the Sigma sign is the dependent variable of the function following the Sigma sign.

The function has an initial input value represented by the equation under the Sigma sign. The function has a terminating input value represented by the value above the Sigma sign. The dependent variable increments by one at each position in the sequence.

In the above example, the first time through n = 1, the second time, n = 2, so on and so forth until n = 5. The values determined by the function, in this case 1, 2 ,3, 4, and 5 are all added together.

Try your hand at this problem:
5
∑ 2n²x
n=3

It should equal 104x
 
  • #21
I never heard of "Sigma Algebra"... but you might be referring to problems involving the sigma sign, Σ, that represents summation notation... a series or, rather, the sum of a sequence

5
∑ n = 1 + 2 + 3 + 4 + 5 = 15
n=1

See, we are creating a series... a sequence of values that are added together. The Sigma sign is followed by a function... in this case "n"

The variable used under the Sigma sign is the dependent variable of the function following the Sigma sign.

The function has an initial input value represented by the equation under the Sigma sign. The function has a terminating input value represented by the value above the Sigma sign. The dependent variable increments by one at each position in the sequence.

In the above example, the first time through n = 1, the second time, n = 2, so on and so forth until n = 5. The values determined by the function, in this case 1, 2 ,3, 4, and 5 are all added together.

Try your hand at this problem:
5
∑ 2n²x
n=3

It should equal 104x

Sigma algebra is a measure of information. You are referring to summation notation Ʃ which is totally a different concept from sigma algebra. for more information check:
http://en.wikipedia.org/wiki/Sigma-algebra
 

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