$|Corr(X,Y)|=1$ implies linear relationship a.e. or everywhere?
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The only proof of this implication is using the schwarz inequality, but that only proves that linear relationship for almost everywhere (a.e.), not for all events...
However, in most undergraduate textbooks, they all state that the relationship is as if for all events.
Is there another proof for the case of all events?
probability
asked yesterday
An old man in the sea.
1,59911031
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up vote
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The only proof of this implication is using the schwarz inequality, but that only proves that linear relationship for almost everywhere (a.e.), not for all events...
However, in most undergraduate textbooks, they all state that the relationship is as if for all events.
Is there another proof for the case of all events?
probability
asked yesterday
An old man in the sea.
1,59911031
1
The undergrad texts where you see this might not talk about probability in a measure theoretic context, hence why a.s. results might not be discussed.
– rubikscube09
yesterday
@rubikscube09 you're probably right. Thanks for the comment. ;)
– An old man in the sea.
yesterday
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The only proof of this implication is using the schwarz inequality, but that only proves that linear relationship for almost everywhere (a.e.), not for all events...
However, in most undergraduate textbooks, they all state that the relationship is as if for all events.
Is there another proof for the case of all events?
probability
asked yesterday
An old man in the sea.
1,59911031
The only proof of this implication is using the schwarz inequality, but that only proves that linear relationship for almost everywhere (a.e.), not for all events...
However, in most undergraduate textbooks, they all state that the relationship is as if for all events.
Is there another proof for the case of all events?
probability
probability
asked yesterday
An old man in the sea.
1,59911031
asked yesterday
An old man in the sea.
1,59911031
asked yesterday
An old man in the sea.
1,59911031
asked yesterday
An old man in the sea.
1,59911031
asked yesterday
An old man in the sea.
1,59911031
1,59911031
1
The undergrad texts where you see this might not talk about probability in a measure theoretic context, hence why a.s. results might not be discussed.
– rubikscube09
yesterday
@rubikscube09 you're probably right. Thanks for the comment. ;)
– An old man in the sea.
yesterday
add a comment |
1
The undergrad texts where you see this might not talk about probability in a measure theoretic context, hence why a.s. results might not be discussed.
– rubikscube09
yesterday
@rubikscube09 you're probably right. Thanks for the comment. ;)
– An old man in the sea.
yesterday
1
1
The undergrad texts where you see this might not talk about probability in a measure theoretic context, hence why a.s. results might not be discussed.
– rubikscube09
yesterday
The undergrad texts where you see this might not talk about probability in a measure theoretic context, hence why a.s. results might not be discussed.
– rubikscube09
yesterday
@rubikscube09 you're probably right. Thanks for the comment. ;)
– An old man in the sea.
yesterday
@rubikscube09 you're probably right. Thanks for the comment. ;)
– An old man in the sea.
yesterday
add a comment |
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The undergrad texts where you see this might not talk about probability in a measure theoretic context, hence why a.s. results might not be discussed.
– rubikscube09
yesterday
@rubikscube09 you're probably right. Thanks for the comment. ;)
– An old man in the sea.
yesterday