Probability negative information
I there, about this exercise:
An event F is said to carry negative information
about an event E, and we write $Frightarrow E$ if $P(E|F)leq P(E)$
Prove or give a counter example for the following claim:
If $Frightarrow E$ then $E rightarrow F $
Well, my attempt was:
$$
P(E|F) = frac{P(E cap F)}{P(F)} leq P(E) => P(E cap F)geq P(E)*P(F)
$$
$$
P(F|E) = frac{P(F cap E)}{P(E)} leq P(F) => P(F cap E)geq P(F)*P(E)
$$
And since $ P(F cap E) = P(E cap F) $
Than it seems pretty correct to me! (but the solution manual says I'm wrong).
What is wrong in my way of thinking?
Thanks!
probability conditional-probability
asked Dec 8 at 9:26
superuser123
1686
add a comment |
I there, about this exercise:
An event F is said to carry negative information
about an event E, and we write $Frightarrow E$ if $P(E|F)leq P(E)$
Prove or give a counter example for the following claim:
If $Frightarrow E$ then $E rightarrow F $
Well, my attempt was:
$$
P(E|F) = frac{P(E cap F)}{P(F)} leq P(E) => P(E cap F)geq P(E)*P(F)
$$
$$
P(F|E) = frac{P(F cap E)}{P(E)} leq P(F) => P(F cap E)geq P(F)*P(E)
$$
And since $ P(F cap E) = P(E cap F) $
Than it seems pretty correct to me! (but the solution manual says I'm wrong).
What is wrong in my way of thinking?
Thanks!
probability conditional-probability
asked Dec 8 at 9:26
superuser123
1686
Related: math.stackexchange.com/questions/1138373/…
– Henry
Dec 8 at 9:40
add a comment |
I there, about this exercise:
An event F is said to carry negative information
about an event E, and we write $Frightarrow E$ if $P(E|F)leq P(E)$
Prove or give a counter example for the following claim:
If $Frightarrow E$ then $E rightarrow F $
Well, my attempt was:
$$
P(E|F) = frac{P(E cap F)}{P(F)} leq P(E) => P(E cap F)geq P(E)*P(F)
$$
$$
P(F|E) = frac{P(F cap E)}{P(E)} leq P(F) => P(F cap E)geq P(F)*P(E)
$$
And since $ P(F cap E) = P(E cap F) $
Than it seems pretty correct to me! (but the solution manual says I'm wrong).
What is wrong in my way of thinking?
Thanks!
probability conditional-probability
asked Dec 8 at 9:26
superuser123
1686
I there, about this exercise:
An event F is said to carry negative information
about an event E, and we write $Frightarrow E$ if $P(E|F)leq P(E)$
Prove or give a counter example for the following claim:
If $Frightarrow E$ then $E rightarrow F $
Well, my attempt was:
$$
P(E|F) = frac{P(E cap F)}{P(F)} leq P(E) => P(E cap F)geq P(E)*P(F)
$$
$$
P(F|E) = frac{P(F cap E)}{P(E)} leq P(F) => P(F cap E)geq P(F)*P(E)
$$
And since $ P(F cap E) = P(E cap F) $
Than it seems pretty correct to me! (but the solution manual says I'm wrong).
What is wrong in my way of thinking?
Thanks!
probability conditional-probability
probability conditional-probability
asked Dec 8 at 9:26
superuser123
1686
asked Dec 8 at 9:26
superuser123
1686
asked Dec 8 at 9:26
superuser123
1686
asked Dec 8 at 9:26
superuser123
1686
asked Dec 8 at 9:26
superuser123
1686
1686
Related: math.stackexchange.com/questions/1138373/…
– Henry
Dec 8 at 9:40
add a comment |
Related: math.stackexchange.com/questions/1138373/…
– Henry
Dec 8 at 9:40
Related: math.stackexchange.com/questions/1138373/…
– Henry
Dec 8 at 9:40
Related: math.stackexchange.com/questions/1138373/…
– Henry
Dec 8 at 9:40
add a comment |
1 Answer
1
active
oldest
votes
The $geq$ signs in your attempt should be $leq$ signs.
That's all I can find.
Further the idea is okay.
If both sides are multiplied by factor $P(F)$ then $P(Emid F)leq P(E)$ changes into: $$P(Ecap F)leq P(E)P(F)$$and it is immediate then that the relation is symmetric.
answered Dec 8 at 9:32
drhab
97.4k544128
Thanks but the fact that $P(F) leq 1$ doesn't flip the sign when multiplying?
– superuser123
Dec 8 at 9:34
1
No, positive numbers do not flip the sign. Negative numbers do.
– drhab
Dec 8 at 9:36
Yes, foolish me :) Anyway, @drhab do you think it's a mistake in the manual?
– superuser123
Dec 8 at 9:36
1
If the manual is rejecting this proof then: yes, the manual is wrong.
– drhab
Dec 8 at 9:38
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The $geq$ signs in your attempt should be $leq$ signs.
That's all I can find.
Further the idea is okay.
If both sides are multiplied by factor $P(F)$ then $P(Emid F)leq P(E)$ changes into: $$P(Ecap F)leq P(E)P(F)$$and it is immediate then that the relation is symmetric.
answered Dec 8 at 9:32
drhab
97.4k544128
Thanks but the fact that $P(F) leq 1$ doesn't flip the sign when multiplying?
– superuser123
Dec 8 at 9:34
1
No, positive numbers do not flip the sign. Negative numbers do.
– drhab
Dec 8 at 9:36
Yes, foolish me :) Anyway, @drhab do you think it's a mistake in the manual?
– superuser123
Dec 8 at 9:36
1
If the manual is rejecting this proof then: yes, the manual is wrong.
– drhab
Dec 8 at 9:38
add a comment |
The $geq$ signs in your attempt should be $leq$ signs.
That's all I can find.
Further the idea is okay.
If both sides are multiplied by factor $P(F)$ then $P(Emid F)leq P(E)$ changes into: $$P(Ecap F)leq P(E)P(F)$$and it is immediate then that the relation is symmetric.
answered Dec 8 at 9:32
drhab
97.4k544128
Thanks but the fact that $P(F) leq 1$ doesn't flip the sign when multiplying?
– superuser123
Dec 8 at 9:34
1
No, positive numbers do not flip the sign. Negative numbers do.
– drhab
Dec 8 at 9:36
Yes, foolish me :) Anyway, @drhab do you think it's a mistake in the manual?
– superuser123
Dec 8 at 9:36
1
If the manual is rejecting this proof then: yes, the manual is wrong.
– drhab
Dec 8 at 9:38
add a comment |
The $geq$ signs in your attempt should be $leq$ signs.
That's all I can find.
Further the idea is okay.
If both sides are multiplied by factor $P(F)$ then $P(Emid F)leq P(E)$ changes into: $$P(Ecap F)leq P(E)P(F)$$and it is immediate then that the relation is symmetric.
answered Dec 8 at 9:32
drhab
97.4k544128
The $geq$ signs in your attempt should be $leq$ signs.
That's all I can find.
Further the idea is okay.
If both sides are multiplied by factor $P(F)$ then $P(Emid F)leq P(E)$ changes into: $$P(Ecap F)leq P(E)P(F)$$and it is immediate then that the relation is symmetric.
answered Dec 8 at 9:32
drhab
97.4k544128
edited Dec 8 at 9:35
answered Dec 8 at 9:32
drhab
97.4k544128
answered Dec 8 at 9:32
drhab
97.4k544128
answered Dec 8 at 9:32
drhab
97.4k544128
97.4k544128
Thanks but the fact that $P(F) leq 1$ doesn't flip the sign when multiplying?
– superuser123
Dec 8 at 9:34
1
No, positive numbers do not flip the sign. Negative numbers do.
– drhab
Dec 8 at 9:36
Yes, foolish me :) Anyway, @drhab do you think it's a mistake in the manual?
– superuser123
Dec 8 at 9:36
1
If the manual is rejecting this proof then: yes, the manual is wrong.
– drhab
Dec 8 at 9:38
add a comment |
Thanks but the fact that $P(F) leq 1$ doesn't flip the sign when multiplying?
– superuser123
Dec 8 at 9:34
1
No, positive numbers do not flip the sign. Negative numbers do.
– drhab
Dec 8 at 9:36
Yes, foolish me :) Anyway, @drhab do you think it's a mistake in the manual?
– superuser123
Dec 8 at 9:36
1
If the manual is rejecting this proof then: yes, the manual is wrong.
– drhab
Dec 8 at 9:38
Thanks but the fact that $P(F) leq 1$ doesn't flip the sign when multiplying?
– superuser123
Dec 8 at 9:34
Thanks but the fact that $P(F) leq 1$ doesn't flip the sign when multiplying?
– superuser123
Dec 8 at 9:34
1
1
No, positive numbers do not flip the sign. Negative numbers do.
– drhab
Dec 8 at 9:36
No, positive numbers do not flip the sign. Negative numbers do.
– drhab
Dec 8 at 9:36
Yes, foolish me :) Anyway, @drhab do you think it's a mistake in the manual?
– superuser123
Dec 8 at 9:36
Yes, foolish me :) Anyway, @drhab do you think it's a mistake in the manual?
– superuser123
Dec 8 at 9:36
1
1
If the manual is rejecting this proof then: yes, the manual is wrong.
– drhab
Dec 8 at 9:38
If the manual is rejecting this proof then: yes, the manual is wrong.
– drhab
Dec 8 at 9:38
add a comment |
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Related: math.stackexchange.com/questions/1138373/…
– Henry
Dec 8 at 9:40