Mixing
Struggling with event spaces in probability
$begingroup$
My textbook gives the following definition of an event space (see attached). Omega here is the sample space (i.e. all possible outcomes of an experiment)
Based on this definition I have been trying to solve the following exercises, but haven't really gotten anywhere and would like some advice on how to continue: (see attached)
For 1.11: I tried to define a set $K$ where exactly $k$ of the points in $Omega$ belong to $K$, however I don't know how to do this concisely/what notation to use.
$K = (A_1setminus A_2... A_k)cup (A_2setminus A_1...A_k)cup...(A_ksetminus A_1...A_{k-1})$ does't work I think.
For 1.12: Isn't this just the power set hence the answer is $2^n$. I feel like this isn't right though. Won't an event space always be just the power set of omega?
probability probability-theory
edited Jan 14 at 20:11
Andrés E. Caicedo
65.5k8158249
asked Jan 14 at 19:59
xAlyxAly
404
$endgroup$
add a comment |
$begingroup$
My textbook gives the following definition of an event space (see attached). Omega here is the sample space (i.e. all possible outcomes of an experiment)
Based on this definition I have been trying to solve the following exercises, but haven't really gotten anywhere and would like some advice on how to continue: (see attached)
For 1.11: I tried to define a set $K$ where exactly $k$ of the points in $Omega$ belong to $K$, however I don't know how to do this concisely/what notation to use.
$K = (A_1setminus A_2... A_k)cup (A_2setminus A_1...A_k)cup...(A_ksetminus A_1...A_{k-1})$ does't work I think.
For 1.12: Isn't this just the power set hence the answer is $2^n$. I feel like this isn't right though. Won't an event space always be just the power set of omega?
probability probability-theory
edited Jan 14 at 20:11
Andrés E. Caicedo
65.5k8158249
asked Jan 14 at 19:59
xAlyxAly
404
$endgroup$
$begingroup$
I think you're right about 1.12
$endgroup$
– Zubin Mukerjee
Jan 14 at 20:04
$begingroup$
for 1.12 yes it's true for the power set and trivial set but you can construct the smallest sigma algebra containing any $A_i$.
$endgroup$
– user29418
Jan 14 at 20:13
add a comment |
$begingroup$
My textbook gives the following definition of an event space (see attached). Omega here is the sample space (i.e. all possible outcomes of an experiment)
Based on this definition I have been trying to solve the following exercises, but haven't really gotten anywhere and would like some advice on how to continue: (see attached)
For 1.11: I tried to define a set $K$ where exactly $k$ of the points in $Omega$ belong to $K$, however I don't know how to do this concisely/what notation to use.
$K = (A_1setminus A_2... A_k)cup (A_2setminus A_1...A_k)cup...(A_ksetminus A_1...A_{k-1})$ does't work I think.
For 1.12: Isn't this just the power set hence the answer is $2^n$. I feel like this isn't right though. Won't an event space always be just the power set of omega?
probability probability-theory
edited Jan 14 at 20:11
Andrés E. Caicedo
65.5k8158249
asked Jan 14 at 19:59
xAlyxAly
404
$endgroup$
My textbook gives the following definition of an event space (see attached). Omega here is the sample space (i.e. all possible outcomes of an experiment)
Based on this definition I have been trying to solve the following exercises, but haven't really gotten anywhere and would like some advice on how to continue: (see attached)
For 1.11: I tried to define a set $K$ where exactly $k$ of the points in $Omega$ belong to $K$, however I don't know how to do this concisely/what notation to use.
$K = (A_1setminus A_2... A_k)cup (A_2setminus A_1...A_k)cup...(A_ksetminus A_1...A_{k-1})$ does't work I think.
For 1.12: Isn't this just the power set hence the answer is $2^n$. I feel like this isn't right though. Won't an event space always be just the power set of omega?
probability probability-theory
probability probability-theory
edited Jan 14 at 20:11
Andrés E. Caicedo
65.5k8158249
asked Jan 14 at 19:59
xAlyxAly
404
edited Jan 14 at 20:11
Andrés E. Caicedo
65.5k8158249
asked Jan 14 at 19:59
xAlyxAly
404
edited Jan 14 at 20:11
Andrés E. Caicedo
65.5k8158249
edited Jan 14 at 20:11
Andrés E. Caicedo
65.5k8158249
edited Jan 14 at 20:11
Andrés E. Caicedo
65.5k8158249
65.5k8158249
asked Jan 14 at 19:59
xAlyxAly
404
asked Jan 14 at 19:59
xAlyxAly
404
asked Jan 14 at 19:59
xAlyxAly
404
404
$begingroup$
I think you're right about 1.12
$endgroup$
– Zubin Mukerjee
Jan 14 at 20:04
$begingroup$
for 1.12 yes it's true for the power set and trivial set but you can construct the smallest sigma algebra containing any $A_i$.
$endgroup$
– user29418
Jan 14 at 20:13
add a comment |
$begingroup$
I think you're right about 1.12
$endgroup$
– Zubin Mukerjee
Jan 14 at 20:04
$begingroup$
for 1.12 yes it's true for the power set and trivial set but you can construct the smallest sigma algebra containing any $A_i$.
$endgroup$
– user29418
Jan 14 at 20:13
$begingroup$
I think you're right about 1.12
$endgroup$
– Zubin Mukerjee
Jan 14 at 20:04
$begingroup$
I think you're right about 1.12
$endgroup$
– Zubin Mukerjee
Jan 14 at 20:04
$begingroup$
for 1.12 yes it's true for the power set and trivial set but you can construct the smallest sigma algebra containing any $A_i$.
$endgroup$
– user29418
Jan 14 at 20:13
$begingroup$
for 1.12 yes it's true for the power set and trivial set but you can construct the smallest sigma algebra containing any $A_i$.
$endgroup$
– user29418
Jan 14 at 20:13
add a comment |
1 Answer
1
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$begingroup$
Hint for 1.11: one way to write "points that belong to only $A_1, A_3, A_5$ and no other $A_i$" is $$A_1 cap A_2^c cap A_3 cap A_4^c cap A_5 cap A_6^c cap cdots cap A_m^c,$$ (if you are unfamiliar with this complement notation, I denote the complement of $B$ by $B^c := Omega setminus B$).
Using something like the above as a building block can help you finish this question.
1.12: The English is a little misleading in the definition. $mathcal{F}$ is some collection of subsets of $Omega$ that satisfies the three conditions. It is not necessarily the power set (all subsets). Note that the power set always satisfies the three conditions.
Hint: is there a natural way to group the elements of $mathcal{F}$ into pairs?
answered Jan 14 at 20:06
angryavianangryavian
41.1k23380
$endgroup$
$begingroup$
What is the best way to choose the $A_i$'s that contain the set of points? I defined a set $K = {A'_1,A'_2,...,A'_k}$ with $k$ members as well as the set $M = {A''_1,A''_2,...,A''_{m-k}}$ with $m-k$ members. Then I used your hint to complete the question.
$endgroup$
– xAly
Jan 14 at 21:52
$begingroup$
For 1.12: Let $mathcal{F} = {A_1,A_2,...,A_m}$, so $mathcal{F}$ has $m$ members. Now since $A_1 in mathcal{F}$,$A^c_1 in mathcal{F}$, and so we can pair each $A_i$ with $A^c_i$, so there must be an even number of elements in $mathcal{F}$.
$endgroup$
– xAly
Jan 14 at 22:03
add a comment |
Your Answer
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1 Answer
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$begingroup$
Hint for 1.11: one way to write "points that belong to only $A_1, A_3, A_5$ and no other $A_i$" is $$A_1 cap A_2^c cap A_3 cap A_4^c cap A_5 cap A_6^c cap cdots cap A_m^c,$$ (if you are unfamiliar with this complement notation, I denote the complement of $B$ by $B^c := Omega setminus B$).
Using something like the above as a building block can help you finish this question.
1.12: The English is a little misleading in the definition. $mathcal{F}$ is some collection of subsets of $Omega$ that satisfies the three conditions. It is not necessarily the power set (all subsets). Note that the power set always satisfies the three conditions.
Hint: is there a natural way to group the elements of $mathcal{F}$ into pairs?
answered Jan 14 at 20:06
angryavianangryavian
41.1k23380
$endgroup$
$begingroup$
What is the best way to choose the $A_i$'s that contain the set of points? I defined a set $K = {A'_1,A'_2,...,A'_k}$ with $k$ members as well as the set $M = {A''_1,A''_2,...,A''_{m-k}}$ with $m-k$ members. Then I used your hint to complete the question.
$endgroup$
– xAly
Jan 14 at 21:52
$begingroup$
For 1.12: Let $mathcal{F} = {A_1,A_2,...,A_m}$, so $mathcal{F}$ has $m$ members. Now since $A_1 in mathcal{F}$,$A^c_1 in mathcal{F}$, and so we can pair each $A_i$ with $A^c_i$, so there must be an even number of elements in $mathcal{F}$.
$endgroup$
– xAly
Jan 14 at 22:03
add a comment |
$begingroup$
Hint for 1.11: one way to write "points that belong to only $A_1, A_3, A_5$ and no other $A_i$" is $$A_1 cap A_2^c cap A_3 cap A_4^c cap A_5 cap A_6^c cap cdots cap A_m^c,$$ (if you are unfamiliar with this complement notation, I denote the complement of $B$ by $B^c := Omega setminus B$).
Using something like the above as a building block can help you finish this question.
1.12: The English is a little misleading in the definition. $mathcal{F}$ is some collection of subsets of $Omega$ that satisfies the three conditions. It is not necessarily the power set (all subsets). Note that the power set always satisfies the three conditions.
Hint: is there a natural way to group the elements of $mathcal{F}$ into pairs?
answered Jan 14 at 20:06
angryavianangryavian
41.1k23380
$endgroup$
$begingroup$
What is the best way to choose the $A_i$'s that contain the set of points? I defined a set $K = {A'_1,A'_2,...,A'_k}$ with $k$ members as well as the set $M = {A''_1,A''_2,...,A''_{m-k}}$ with $m-k$ members. Then I used your hint to complete the question.
$endgroup$
– xAly
Jan 14 at 21:52
$begingroup$
For 1.12: Let $mathcal{F} = {A_1,A_2,...,A_m}$, so $mathcal{F}$ has $m$ members. Now since $A_1 in mathcal{F}$,$A^c_1 in mathcal{F}$, and so we can pair each $A_i$ with $A^c_i$, so there must be an even number of elements in $mathcal{F}$.
$endgroup$
– xAly
Jan 14 at 22:03
add a comment |
$begingroup$
Hint for 1.11: one way to write "points that belong to only $A_1, A_3, A_5$ and no other $A_i$" is $$A_1 cap A_2^c cap A_3 cap A_4^c cap A_5 cap A_6^c cap cdots cap A_m^c,$$ (if you are unfamiliar with this complement notation, I denote the complement of $B$ by $B^c := Omega setminus B$).
Using something like the above as a building block can help you finish this question.
1.12: The English is a little misleading in the definition. $mathcal{F}$ is some collection of subsets of $Omega$ that satisfies the three conditions. It is not necessarily the power set (all subsets). Note that the power set always satisfies the three conditions.
Hint: is there a natural way to group the elements of $mathcal{F}$ into pairs?
answered Jan 14 at 20:06
angryavianangryavian
41.1k23380
$endgroup$
Hint for 1.11: one way to write "points that belong to only $A_1, A_3, A_5$ and no other $A_i$" is $$A_1 cap A_2^c cap A_3 cap A_4^c cap A_5 cap A_6^c cap cdots cap A_m^c,$$ (if you are unfamiliar with this complement notation, I denote the complement of $B$ by $B^c := Omega setminus B$).
Using something like the above as a building block can help you finish this question.
1.12: The English is a little misleading in the definition. $mathcal{F}$ is some collection of subsets of $Omega$ that satisfies the three conditions. It is not necessarily the power set (all subsets). Note that the power set always satisfies the three conditions.
Hint: is there a natural way to group the elements of $mathcal{F}$ into pairs?
answered Jan 14 at 20:06
angryavianangryavian
41.1k23380
answered Jan 14 at 20:06
angryavianangryavian
41.1k23380
answered Jan 14 at 20:06
angryavianangryavian
41.1k23380
answered Jan 14 at 20:06
angryavianangryavian
41.1k23380
41.1k23380
$begingroup$
What is the best way to choose the $A_i$'s that contain the set of points? I defined a set $K = {A'_1,A'_2,...,A'_k}$ with $k$ members as well as the set $M = {A''_1,A''_2,...,A''_{m-k}}$ with $m-k$ members. Then I used your hint to complete the question.
$endgroup$
– xAly
Jan 14 at 21:52
$begingroup$
For 1.12: Let $mathcal{F} = {A_1,A_2,...,A_m}$, so $mathcal{F}$ has $m$ members. Now since $A_1 in mathcal{F}$,$A^c_1 in mathcal{F}$, and so we can pair each $A_i$ with $A^c_i$, so there must be an even number of elements in $mathcal{F}$.
$endgroup$
– xAly
Jan 14 at 22:03
add a comment |
$begingroup$
What is the best way to choose the $A_i$'s that contain the set of points? I defined a set $K = {A'_1,A'_2,...,A'_k}$ with $k$ members as well as the set $M = {A''_1,A''_2,...,A''_{m-k}}$ with $m-k$ members. Then I used your hint to complete the question.
$endgroup$
– xAly
Jan 14 at 21:52
$begingroup$
For 1.12: Let $mathcal{F} = {A_1,A_2,...,A_m}$, so $mathcal{F}$ has $m$ members. Now since $A_1 in mathcal{F}$,$A^c_1 in mathcal{F}$, and so we can pair each $A_i$ with $A^c_i$, so there must be an even number of elements in $mathcal{F}$.
$endgroup$
– xAly
Jan 14 at 22:03
$begingroup$
What is the best way to choose the $A_i$'s that contain the set of points? I defined a set $K = {A'_1,A'_2,...,A'_k}$ with $k$ members as well as the set $M = {A''_1,A''_2,...,A''_{m-k}}$ with $m-k$ members. Then I used your hint to complete the question.
$endgroup$
– xAly
Jan 14 at 21:52
$begingroup$
What is the best way to choose the $A_i$'s that contain the set of points? I defined a set $K = {A'_1,A'_2,...,A'_k}$ with $k$ members as well as the set $M = {A''_1,A''_2,...,A''_{m-k}}$ with $m-k$ members. Then I used your hint to complete the question.
$endgroup$
– xAly
Jan 14 at 21:52
$begingroup$
For 1.12: Let $mathcal{F} = {A_1,A_2,...,A_m}$, so $mathcal{F}$ has $m$ members. Now since $A_1 in mathcal{F}$,$A^c_1 in mathcal{F}$, and so we can pair each $A_i$ with $A^c_i$, so there must be an even number of elements in $mathcal{F}$.
$endgroup$
– xAly
Jan 14 at 22:03
$begingroup$
For 1.12: Let $mathcal{F} = {A_1,A_2,...,A_m}$, so $mathcal{F}$ has $m$ members. Now since $A_1 in mathcal{F}$,$A^c_1 in mathcal{F}$, and so we can pair each $A_i$ with $A^c_i$, so there must be an even number of elements in $mathcal{F}$.
$endgroup$
– xAly
Jan 14 at 22:03
add a comment |
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$begingroup$
I think you're right about 1.12
$endgroup$
– Zubin Mukerjee
Jan 14 at 20:04
$begingroup$
for 1.12 yes it's true for the power set and trivial set but you can construct the smallest sigma algebra containing any $A_i$.
$endgroup$
– user29418
Jan 14 at 20:13