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.
probability probability-theory measure-theory probability-limit-theorems
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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.
probability probability-theory measure-theory probability-limit-theorems
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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.
probability probability-theory measure-theory probability-limit-theorems
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.
probability probability-theory measure-theory probability-limit-theorems
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 $S_n.$
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.
â 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.
â 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.
â 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.
â 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 $$S_n(omega) = X_1(omega)+X_2(omega)+ldots +X_n(omega) $$
â spaceisdarkgreen
Aug 9 at 0:52
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $S_n.$
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.
â 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.
â 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.
â 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.
â 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 $$S_n(omega) = X_1(omega)+X_2(omega)+ldots +X_n(omega) $$
â spaceisdarkgreen
Aug 9 at 0:52
add a comment |Â
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 $S_n.$
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.
â 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.
â 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.
â 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.
â 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 $$S_n(omega) = X_1(omega)+X_2(omega)+ldots +X_n(omega) $$
â spaceisdarkgreen
Aug 9 at 0:52
add a comment |Â
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 $S_n.$
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 $S_n.$
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.
â 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.
â 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.
â 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.
â 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 $$S_n(omega) = X_1(omega)+X_2(omega)+ldots +X_n(omega) $$
â spaceisdarkgreen
Aug 9 at 0:52
add a comment |Â
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.
â 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.
â 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.
â 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.
â 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 $$S_n(omega) = X_1(omega)+X_2(omega)+ldots +X_n(omega) $$
â 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.
â 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.
â 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.
â 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.
â 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.
â 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.
â 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.
â 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.
â 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 $$S_n(omega) = X_1(omega)+X_2(omega)+ldots +X_n(omega) $$
â 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 $$S_n(omega) = X_1(omega)+X_2(omega)+ldots +X_n(omega) $$
â spaceisdarkgreen
Aug 9 at 0:52
add a comment |Â
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