Sample space of sequence of random variables

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Say that I have sequence of iid bernoulli random variables $ (X_n)$ taking values in 0,1 with equal probability 0.5. Then I know by the weak law of large numbers that for every $epsilon>0, lim P(<epsilon)= 1$ where $S_n= X_1+...+X_n$
Would I be correct to assume that the sample space $Omega$ would be the set of infinite sequences in 0,1? So one possible w might be (0,1,0,1,....).? Thanks.







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    Say that I have sequence of iid bernoulli random variables $ (X_n)$ taking values in 0,1 with equal probability 0.5. Then I know by the weak law of large numbers that for every $epsilon>0, lim P(<epsilon)= 1$ where $S_n= X_1+...+X_n$
    Would I be correct to assume that the sample space $Omega$ would be the set of infinite sequences in 0,1? So one possible w might be (0,1,0,1,....).? Thanks.







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      Say that I have sequence of iid bernoulli random variables $ (X_n)$ taking values in 0,1 with equal probability 0.5. Then I know by the weak law of large numbers that for every $epsilon>0, lim P(<epsilon)= 1$ where $S_n= X_1+...+X_n$
      Would I be correct to assume that the sample space $Omega$ would be the set of infinite sequences in 0,1? So one possible w might be (0,1,0,1,....).? Thanks.







      share|cite|improve this question














      Say that I have sequence of iid bernoulli random variables $ (X_n)$ taking values in 0,1 with equal probability 0.5. Then I know by the weak law of large numbers that for every $epsilon>0, lim P(<epsilon)= 1$ where $S_n= X_1+...+X_n$
      Would I be correct to assume that the sample space $Omega$ would be the set of infinite sequences in 0,1? So one possible w might be (0,1,0,1,....).? Thanks.









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      edited Aug 9 at 0:05









      Mwatha

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      asked Aug 8 at 19:26









      jerry

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          Yes, that is the sample space and $(0,1,0,1,ldots)$ is an example of an outcome.



          Not sure what the WLLN has to do with that, but I'll note you have a typo. It is $S_n/n$ that converges in probability to $1/2,$ not $S_n.$






          share|cite|improve this answer




















          • Thanks! I forgot to divide by n. My confusion was because my book never discussed the sample space for the sum of RV as being the product of each sample space.. Those details are all left out.
            – jerry
            Aug 8 at 19:41











          • @jerry I should add that this is merely one 'natural' choice of sample space for this problem. Since the basic building blocks here are the binary random variables $X_i,$ it makes sense to characterize the outcomes as just the values of these random variables. However, there are other sample spaces that these same random variables can be defined on. For instance, we could let the sample space be $[0,1]$ then let the random variables $X_i$ be the digits of the number in $[0,1]$ expressed in binary.
            – spaceisdarkgreen
            Aug 8 at 19:49










          • @jerry Also it's best not to think about a sample space 'for the sum'. The sum is a RV defined on your sample space. But the sample space has all the information for all the RVs in the scenario.
            – spaceisdarkgreen
            Aug 8 at 19:53










          • Thanks! I think I see what you mean. I wanted a sample space for the sum to try to make sense of the LLN result in this simplest situation as that says something about the limit of the probability of outcomes w in the sample space such that bla bla so I was wondering what sample space the LLN was refering to.
            – jerry
            Aug 8 at 20:39










          • @jerry ahh I got it, like 'sample space for the sum' means what does the $omega$ mean in $S_n(omega)$? Yeah I think you've got it. Each $omega$ corresponds (somehow) to a possible infinite sequence $(X_1(omega),X_2(omega), ldots)$ and then $$S_n(omega) = X_1(omega)+X_2(omega)+ldots +X_n(omega) $$
            – spaceisdarkgreen
            Aug 9 at 0:52











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          1 Answer
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          1 Answer
          1






          active

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          active

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          up vote
          0
          down vote













          Yes, that is the sample space and $(0,1,0,1,ldots)$ is an example of an outcome.



          Not sure what the WLLN has to do with that, but I'll note you have a typo. It is $S_n/n$ that converges in probability to $1/2,$ not $S_n.$






          share|cite|improve this answer




















          • Thanks! I forgot to divide by n. My confusion was because my book never discussed the sample space for the sum of RV as being the product of each sample space.. Those details are all left out.
            – jerry
            Aug 8 at 19:41











          • @jerry I should add that this is merely one 'natural' choice of sample space for this problem. Since the basic building blocks here are the binary random variables $X_i,$ it makes sense to characterize the outcomes as just the values of these random variables. However, there are other sample spaces that these same random variables can be defined on. For instance, we could let the sample space be $[0,1]$ then let the random variables $X_i$ be the digits of the number in $[0,1]$ expressed in binary.
            – spaceisdarkgreen
            Aug 8 at 19:49










          • @jerry Also it's best not to think about a sample space 'for the sum'. The sum is a RV defined on your sample space. But the sample space has all the information for all the RVs in the scenario.
            – spaceisdarkgreen
            Aug 8 at 19:53










          • Thanks! I think I see what you mean. I wanted a sample space for the sum to try to make sense of the LLN result in this simplest situation as that says something about the limit of the probability of outcomes w in the sample space such that bla bla so I was wondering what sample space the LLN was refering to.
            – jerry
            Aug 8 at 20:39










          • @jerry ahh I got it, like 'sample space for the sum' means what does the $omega$ mean in $S_n(omega)$? Yeah I think you've got it. Each $omega$ corresponds (somehow) to a possible infinite sequence $(X_1(omega),X_2(omega), ldots)$ and then $$S_n(omega) = X_1(omega)+X_2(omega)+ldots +X_n(omega) $$
            – spaceisdarkgreen
            Aug 9 at 0:52















          up vote
          0
          down vote













          Yes, that is the sample space and $(0,1,0,1,ldots)$ is an example of an outcome.



          Not sure what the WLLN has to do with that, but I'll note you have a typo. It is $S_n/n$ that converges in probability to $1/2,$ not $S_n.$






          share|cite|improve this answer




















          • Thanks! I forgot to divide by n. My confusion was because my book never discussed the sample space for the sum of RV as being the product of each sample space.. Those details are all left out.
            – jerry
            Aug 8 at 19:41











          • @jerry I should add that this is merely one 'natural' choice of sample space for this problem. Since the basic building blocks here are the binary random variables $X_i,$ it makes sense to characterize the outcomes as just the values of these random variables. However, there are other sample spaces that these same random variables can be defined on. For instance, we could let the sample space be $[0,1]$ then let the random variables $X_i$ be the digits of the number in $[0,1]$ expressed in binary.
            – spaceisdarkgreen
            Aug 8 at 19:49










          • @jerry Also it's best not to think about a sample space 'for the sum'. The sum is a RV defined on your sample space. But the sample space has all the information for all the RVs in the scenario.
            – spaceisdarkgreen
            Aug 8 at 19:53










          • Thanks! I think I see what you mean. I wanted a sample space for the sum to try to make sense of the LLN result in this simplest situation as that says something about the limit of the probability of outcomes w in the sample space such that bla bla so I was wondering what sample space the LLN was refering to.
            – jerry
            Aug 8 at 20:39










          • @jerry ahh I got it, like 'sample space for the sum' means what does the $omega$ mean in $S_n(omega)$? Yeah I think you've got it. Each $omega$ corresponds (somehow) to a possible infinite sequence $(X_1(omega),X_2(omega), ldots)$ and then $$S_n(omega) = X_1(omega)+X_2(omega)+ldots +X_n(omega) $$
            – spaceisdarkgreen
            Aug 9 at 0:52













          up vote
          0
          down vote










          up vote
          0
          down vote









          Yes, that is the sample space and $(0,1,0,1,ldots)$ is an example of an outcome.



          Not sure what the WLLN has to do with that, but I'll note you have a typo. It is $S_n/n$ that converges in probability to $1/2,$ not $S_n.$






          share|cite|improve this answer












          Yes, that is the sample space and $(0,1,0,1,ldots)$ is an example of an outcome.



          Not sure what the WLLN has to do with that, but I'll note you have a typo. It is $S_n/n$ that converges in probability to $1/2,$ not $S_n.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 8 at 19:32









          spaceisdarkgreen

          27.9k21548




          27.9k21548











          • Thanks! I forgot to divide by n. My confusion was because my book never discussed the sample space for the sum of RV as being the product of each sample space.. Those details are all left out.
            – jerry
            Aug 8 at 19:41











          • @jerry I should add that this is merely one 'natural' choice of sample space for this problem. Since the basic building blocks here are the binary random variables $X_i,$ it makes sense to characterize the outcomes as just the values of these random variables. However, there are other sample spaces that these same random variables can be defined on. For instance, we could let the sample space be $[0,1]$ then let the random variables $X_i$ be the digits of the number in $[0,1]$ expressed in binary.
            – spaceisdarkgreen
            Aug 8 at 19:49










          • @jerry Also it's best not to think about a sample space 'for the sum'. The sum is a RV defined on your sample space. But the sample space has all the information for all the RVs in the scenario.
            – spaceisdarkgreen
            Aug 8 at 19:53










          • Thanks! I think I see what you mean. I wanted a sample space for the sum to try to make sense of the LLN result in this simplest situation as that says something about the limit of the probability of outcomes w in the sample space such that bla bla so I was wondering what sample space the LLN was refering to.
            – jerry
            Aug 8 at 20:39










          • @jerry ahh I got it, like 'sample space for the sum' means what does the $omega$ mean in $S_n(omega)$? Yeah I think you've got it. Each $omega$ corresponds (somehow) to a possible infinite sequence $(X_1(omega),X_2(omega), ldots)$ and then $$S_n(omega) = X_1(omega)+X_2(omega)+ldots +X_n(omega) $$
            – spaceisdarkgreen
            Aug 9 at 0:52

















          • Thanks! I forgot to divide by n. My confusion was because my book never discussed the sample space for the sum of RV as being the product of each sample space.. Those details are all left out.
            – jerry
            Aug 8 at 19:41











          • @jerry I should add that this is merely one 'natural' choice of sample space for this problem. Since the basic building blocks here are the binary random variables $X_i,$ it makes sense to characterize the outcomes as just the values of these random variables. However, there are other sample spaces that these same random variables can be defined on. For instance, we could let the sample space be $[0,1]$ then let the random variables $X_i$ be the digits of the number in $[0,1]$ expressed in binary.
            – spaceisdarkgreen
            Aug 8 at 19:49










          • @jerry Also it's best not to think about a sample space 'for the sum'. The sum is a RV defined on your sample space. But the sample space has all the information for all the RVs in the scenario.
            – spaceisdarkgreen
            Aug 8 at 19:53










          • Thanks! I think I see what you mean. I wanted a sample space for the sum to try to make sense of the LLN result in this simplest situation as that says something about the limit of the probability of outcomes w in the sample space such that bla bla so I was wondering what sample space the LLN was refering to.
            – jerry
            Aug 8 at 20:39










          • @jerry ahh I got it, like 'sample space for the sum' means what does the $omega$ mean in $S_n(omega)$? Yeah I think you've got it. Each $omega$ corresponds (somehow) to a possible infinite sequence $(X_1(omega),X_2(omega), ldots)$ and then $$S_n(omega) = X_1(omega)+X_2(omega)+ldots +X_n(omega) $$
            – spaceisdarkgreen
            Aug 9 at 0:52
















          Thanks! I forgot to divide by n. My confusion was because my book never discussed the sample space for the sum of RV as being the product of each sample space.. Those details are all left out.
          – jerry
          Aug 8 at 19:41





          Thanks! I forgot to divide by n. My confusion was because my book never discussed the sample space for the sum of RV as being the product of each sample space.. Those details are all left out.
          – jerry
          Aug 8 at 19:41













          @jerry I should add that this is merely one 'natural' choice of sample space for this problem. Since the basic building blocks here are the binary random variables $X_i,$ it makes sense to characterize the outcomes as just the values of these random variables. However, there are other sample spaces that these same random variables can be defined on. For instance, we could let the sample space be $[0,1]$ then let the random variables $X_i$ be the digits of the number in $[0,1]$ expressed in binary.
          – spaceisdarkgreen
          Aug 8 at 19:49




          @jerry I should add that this is merely one 'natural' choice of sample space for this problem. Since the basic building blocks here are the binary random variables $X_i,$ it makes sense to characterize the outcomes as just the values of these random variables. However, there are other sample spaces that these same random variables can be defined on. For instance, we could let the sample space be $[0,1]$ then let the random variables $X_i$ be the digits of the number in $[0,1]$ expressed in binary.
          – spaceisdarkgreen
          Aug 8 at 19:49












          @jerry Also it's best not to think about a sample space 'for the sum'. The sum is a RV defined on your sample space. But the sample space has all the information for all the RVs in the scenario.
          – spaceisdarkgreen
          Aug 8 at 19:53




          @jerry Also it's best not to think about a sample space 'for the sum'. The sum is a RV defined on your sample space. But the sample space has all the information for all the RVs in the scenario.
          – spaceisdarkgreen
          Aug 8 at 19:53












          Thanks! I think I see what you mean. I wanted a sample space for the sum to try to make sense of the LLN result in this simplest situation as that says something about the limit of the probability of outcomes w in the sample space such that bla bla so I was wondering what sample space the LLN was refering to.
          – jerry
          Aug 8 at 20:39




          Thanks! I think I see what you mean. I wanted a sample space for the sum to try to make sense of the LLN result in this simplest situation as that says something about the limit of the probability of outcomes w in the sample space such that bla bla so I was wondering what sample space the LLN was refering to.
          – jerry
          Aug 8 at 20:39












          @jerry ahh I got it, like 'sample space for the sum' means what does the $omega$ mean in $S_n(omega)$? Yeah I think you've got it. Each $omega$ corresponds (somehow) to a possible infinite sequence $(X_1(omega),X_2(omega), ldots)$ and then $$S_n(omega) = X_1(omega)+X_2(omega)+ldots +X_n(omega) $$
          – spaceisdarkgreen
          Aug 9 at 0:52





          @jerry ahh I got it, like 'sample space for the sum' means what does the $omega$ mean in $S_n(omega)$? Yeah I think you've got it. Each $omega$ corresponds (somehow) to a possible infinite sequence $(X_1(omega),X_2(omega), ldots)$ and then $$S_n(omega) = X_1(omega)+X_2(omega)+ldots +X_n(omega) $$
          – spaceisdarkgreen
          Aug 9 at 0:52













           

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