Mixing
How to define a random variable of an indicator function and find the smallest sigma algebra such that X is...
$begingroup$
Let $Omega$ = ${1,2,3}$
Define the random variable $x(omega) = omegacdot1_{{omegainOmega:omegageq2)}}$
That is :
$$1_{{omegainOmega:omegageq2)}} = begin{cases}
1& omegageq2,\
0& text{otherwise}.
end{cases}$$
Then I need to find the smallest sigma- algebra such that X is measurable, or induced by X.
Here is my attempt:
I used the inverse mapping of omega as follows:
$$x(omega)^{-1} = begin{cases}
phi& text{if } 1 notin omega land 2 notin omega land 3 notin omega \
0& text{if } 1 in omega land 2 notin omega land 3 notin omega \
2& text{if } 1 notin omega land 2 in omega land 3 notin omega \
3& text{if } 1 notin omega land 2 notin omega land 3 in omega \
Omega& text{if } 1 in omega land 2 in omega land 3 in omega \
end{cases}$$
Let the smallest sigma algebra be $lambda$
Then I get that $$lambda = { {1}, {2}, {3}, {2,3}, {1,3}, {1,2} ,phi ,lambda } $$
Which would inherently be the power set as I understand it.
I feel like I may be missing something here as I am seeing the smallest sigma algebra be the power set, however that may be the point of the exercise.
Can anyone shed some light on my intuition here? I appreciate it.
measure-theory elementary-set-theory lebesgue-measure
edited Jan 29 at 16:25
Andrés E. Caicedo
65.8k8160251
asked Jan 28 at 23:08
Victor NogueiraVictor Nogueira
1
$endgroup$
add a comment |
$begingroup$
Let $Omega$ = ${1,2,3}$
Define the random variable $x(omega) = omegacdot1_{{omegainOmega:omegageq2)}}$
That is :
$$1_{{omegainOmega:omegageq2)}} = begin{cases}
1& omegageq2,\
0& text{otherwise}.
end{cases}$$
Then I need to find the smallest sigma- algebra such that X is measurable, or induced by X.
Here is my attempt:
I used the inverse mapping of omega as follows:
$$x(omega)^{-1} = begin{cases}
phi& text{if } 1 notin omega land 2 notin omega land 3 notin omega \
0& text{if } 1 in omega land 2 notin omega land 3 notin omega \
2& text{if } 1 notin omega land 2 in omega land 3 notin omega \
3& text{if } 1 notin omega land 2 notin omega land 3 in omega \
Omega& text{if } 1 in omega land 2 in omega land 3 in omega \
end{cases}$$
Let the smallest sigma algebra be $lambda$
Then I get that $$lambda = { {1}, {2}, {3}, {2,3}, {1,3}, {1,2} ,phi ,lambda } $$
Which would inherently be the power set as I understand it.
I feel like I may be missing something here as I am seeing the smallest sigma algebra be the power set, however that may be the point of the exercise.
Can anyone shed some light on my intuition here? I appreciate it.
measure-theory elementary-set-theory lebesgue-measure
edited Jan 29 at 16:25
Andrés E. Caicedo
65.8k8160251
asked Jan 28 at 23:08
Victor NogueiraVictor Nogueira
1
$endgroup$
add a comment |
$begingroup$
Let $Omega$ = ${1,2,3}$
Define the random variable $x(omega) = omegacdot1_{{omegainOmega:omegageq2)}}$
That is :
$$1_{{omegainOmega:omegageq2)}} = begin{cases}
1& omegageq2,\
0& text{otherwise}.
end{cases}$$
Then I need to find the smallest sigma- algebra such that X is measurable, or induced by X.
Here is my attempt:
I used the inverse mapping of omega as follows:
$$x(omega)^{-1} = begin{cases}
phi& text{if } 1 notin omega land 2 notin omega land 3 notin omega \
0& text{if } 1 in omega land 2 notin omega land 3 notin omega \
2& text{if } 1 notin omega land 2 in omega land 3 notin omega \
3& text{if } 1 notin omega land 2 notin omega land 3 in omega \
Omega& text{if } 1 in omega land 2 in omega land 3 in omega \
end{cases}$$
Let the smallest sigma algebra be $lambda$
Then I get that $$lambda = { {1}, {2}, {3}, {2,3}, {1,3}, {1,2} ,phi ,lambda } $$
Which would inherently be the power set as I understand it.
I feel like I may be missing something here as I am seeing the smallest sigma algebra be the power set, however that may be the point of the exercise.
Can anyone shed some light on my intuition here? I appreciate it.
measure-theory elementary-set-theory lebesgue-measure
edited Jan 29 at 16:25
Andrés E. Caicedo
65.8k8160251
asked Jan 28 at 23:08
Victor NogueiraVictor Nogueira
1
$endgroup$
Let $Omega$ = ${1,2,3}$
Define the random variable $x(omega) = omegacdot1_{{omegainOmega:omegageq2)}}$
That is :
$$1_{{omegainOmega:omegageq2)}} = begin{cases}
1& omegageq2,\
0& text{otherwise}.
end{cases}$$
Then I need to find the smallest sigma- algebra such that X is measurable, or induced by X.
Here is my attempt:
I used the inverse mapping of omega as follows:
$$x(omega)^{-1} = begin{cases}
phi& text{if } 1 notin omega land 2 notin omega land 3 notin omega \
0& text{if } 1 in omega land 2 notin omega land 3 notin omega \
2& text{if } 1 notin omega land 2 in omega land 3 notin omega \
3& text{if } 1 notin omega land 2 notin omega land 3 in omega \
Omega& text{if } 1 in omega land 2 in omega land 3 in omega \
end{cases}$$
Let the smallest sigma algebra be $lambda$
Then I get that $$lambda = { {1}, {2}, {3}, {2,3}, {1,3}, {1,2} ,phi ,lambda } $$
Which would inherently be the power set as I understand it.
I feel like I may be missing something here as I am seeing the smallest sigma algebra be the power set, however that may be the point of the exercise.
Can anyone shed some light on my intuition here? I appreciate it.
measure-theory elementary-set-theory lebesgue-measure
measure-theory elementary-set-theory lebesgue-measure
edited Jan 29 at 16:25
Andrés E. Caicedo
65.8k8160251
asked Jan 28 at 23:08
Victor NogueiraVictor Nogueira
1
edited Jan 29 at 16:25
Andrés E. Caicedo
65.8k8160251
asked Jan 28 at 23:08
Victor NogueiraVictor Nogueira
1
edited Jan 29 at 16:25
Andrés E. Caicedo
65.8k8160251
edited Jan 29 at 16:25
Andrés E. Caicedo
65.8k8160251
edited Jan 29 at 16:25
Andrés E. Caicedo
65.8k8160251
65.8k8160251
asked Jan 28 at 23:08
Victor NogueiraVictor Nogueira
1
asked Jan 28 at 23:08
Victor NogueiraVictor Nogueira
1
asked Jan 28 at 23:08
Victor NogueiraVictor Nogueira
1
1
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
The sigma algebra generated is ${emptyset, {1},{2,3},{1,2,3}}$. [ $sigma(x)={x^{-1} (A):A text {is a Borel set in} mathbb R}$].
answered Jan 28 at 23:12
Kavi Rama MurthyKavi Rama Murthy
71.1k53170
$endgroup$
$begingroup$
So, essentially, the element ${1}$ needs to be in its own set and the compliment of that element is the elements ${2, 3}$ which are by definition included in the sigma algebra. Can these elements be listed as simply as that? My understanding in this space is somewhat limited, but I thought that ${2}$ and ${3}$ would need to be input as their own elements in the sigma algebra - hence my result.
$endgroup$
– Victor Nogueira
Jan 28 at 23:20
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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votes
$begingroup$
The sigma algebra generated is ${emptyset, {1},{2,3},{1,2,3}}$. [ $sigma(x)={x^{-1} (A):A text {is a Borel set in} mathbb R}$].
answered Jan 28 at 23:12
Kavi Rama MurthyKavi Rama Murthy
71.1k53170
$endgroup$
$begingroup$
So, essentially, the element ${1}$ needs to be in its own set and the compliment of that element is the elements ${2, 3}$ which are by definition included in the sigma algebra. Can these elements be listed as simply as that? My understanding in this space is somewhat limited, but I thought that ${2}$ and ${3}$ would need to be input as their own elements in the sigma algebra - hence my result.
$endgroup$
– Victor Nogueira
Jan 28 at 23:20
add a comment |
$begingroup$
The sigma algebra generated is ${emptyset, {1},{2,3},{1,2,3}}$. [ $sigma(x)={x^{-1} (A):A text {is a Borel set in} mathbb R}$].
answered Jan 28 at 23:12
Kavi Rama MurthyKavi Rama Murthy
71.1k53170
$endgroup$
$begingroup$
So, essentially, the element ${1}$ needs to be in its own set and the compliment of that element is the elements ${2, 3}$ which are by definition included in the sigma algebra. Can these elements be listed as simply as that? My understanding in this space is somewhat limited, but I thought that ${2}$ and ${3}$ would need to be input as their own elements in the sigma algebra - hence my result.
$endgroup$
– Victor Nogueira
Jan 28 at 23:20
add a comment |
$begingroup$
The sigma algebra generated is ${emptyset, {1},{2,3},{1,2,3}}$. [ $sigma(x)={x^{-1} (A):A text {is a Borel set in} mathbb R}$].
answered Jan 28 at 23:12
Kavi Rama MurthyKavi Rama Murthy
71.1k53170
$endgroup$
The sigma algebra generated is ${emptyset, {1},{2,3},{1,2,3}}$. [ $sigma(x)={x^{-1} (A):A text {is a Borel set in} mathbb R}$].
answered Jan 28 at 23:12
Kavi Rama MurthyKavi Rama Murthy
71.1k53170
answered Jan 28 at 23:12
Kavi Rama MurthyKavi Rama Murthy
71.1k53170
answered Jan 28 at 23:12
Kavi Rama MurthyKavi Rama Murthy
71.1k53170
answered Jan 28 at 23:12
Kavi Rama MurthyKavi Rama Murthy
71.1k53170
71.1k53170
$begingroup$
So, essentially, the element ${1}$ needs to be in its own set and the compliment of that element is the elements ${2, 3}$ which are by definition included in the sigma algebra. Can these elements be listed as simply as that? My understanding in this space is somewhat limited, but I thought that ${2}$ and ${3}$ would need to be input as their own elements in the sigma algebra - hence my result.
$endgroup$
– Victor Nogueira
Jan 28 at 23:20
add a comment |
$begingroup$
So, essentially, the element ${1}$ needs to be in its own set and the compliment of that element is the elements ${2, 3}$ which are by definition included in the sigma algebra. Can these elements be listed as simply as that? My understanding in this space is somewhat limited, but I thought that ${2}$ and ${3}$ would need to be input as their own elements in the sigma algebra - hence my result.
$endgroup$
– Victor Nogueira
Jan 28 at 23:20
$begingroup$
So, essentially, the element ${1}$ needs to be in its own set and the compliment of that element is the elements ${2, 3}$ which are by definition included in the sigma algebra. Can these elements be listed as simply as that? My understanding in this space is somewhat limited, but I thought that ${2}$ and ${3}$ would need to be input as their own elements in the sigma algebra - hence my result.
$endgroup$
– Victor Nogueira
Jan 28 at 23:20
$begingroup$
So, essentially, the element ${1}$ needs to be in its own set and the compliment of that element is the elements ${2, 3}$ which are by definition included in the sigma algebra. Can these elements be listed as simply as that? My understanding in this space is somewhat limited, but I thought that ${2}$ and ${3}$ would need to be input as their own elements in the sigma algebra - hence my result.
$endgroup$
– Victor Nogueira
Jan 28 at 23:20
add a comment |
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