Mixing
Does this function define a probability function?
$begingroup$
Show that for a given random event $A$, the function $R: alpha rightarrow mathbb{R}$ defined as $R (B) = P (A | B)$ does not satisfy the axioms: https://en.wikipedia.org/wiki/Probability_axioms
$ alpha $ is the sigma-algebra
This is what I tried:
It is a simple counterexample: I have $A_{i}´s$ that belongs to $ alpha $. If I take the $emptyset$ set. So $R (B) = P (A | B)$ it is not well defined.
I am not sure. Any help? Am I right?
Also, If I define $P(B)>0$ of course my counterexmple doesnt work. How I could proof this?
probability probability-theory measure-theory probability-distributions
edited Jan 11 at 23:43
Bernard
121k740116
asked Jan 11 at 23:43
LauraLaura
2408
$endgroup$
add a comment |
$begingroup$
Show that for a given random event $A$, the function $R: alpha rightarrow mathbb{R}$ defined as $R (B) = P (A | B)$ does not satisfy the axioms: https://en.wikipedia.org/wiki/Probability_axioms
$ alpha $ is the sigma-algebra
This is what I tried:
It is a simple counterexample: I have $A_{i}´s$ that belongs to $ alpha $. If I take the $emptyset$ set. So $R (B) = P (A | B)$ it is not well defined.
I am not sure. Any help? Am I right?
Also, If I define $P(B)>0$ of course my counterexmple doesnt work. How I could proof this?
probability probability-theory measure-theory probability-distributions
edited Jan 11 at 23:43
Bernard
121k740116
asked Jan 11 at 23:43
LauraLaura
2408
$endgroup$
add a comment |
$begingroup$
Show that for a given random event $A$, the function $R: alpha rightarrow mathbb{R}$ defined as $R (B) = P (A | B)$ does not satisfy the axioms: https://en.wikipedia.org/wiki/Probability_axioms
$ alpha $ is the sigma-algebra
This is what I tried:
It is a simple counterexample: I have $A_{i}´s$ that belongs to $ alpha $. If I take the $emptyset$ set. So $R (B) = P (A | B)$ it is not well defined.
I am not sure. Any help? Am I right?
Also, If I define $P(B)>0$ of course my counterexmple doesnt work. How I could proof this?
probability probability-theory measure-theory probability-distributions
edited Jan 11 at 23:43
Bernard
121k740116
asked Jan 11 at 23:43
LauraLaura
2408
$endgroup$
Show that for a given random event $A$, the function $R: alpha rightarrow mathbb{R}$ defined as $R (B) = P (A | B)$ does not satisfy the axioms: https://en.wikipedia.org/wiki/Probability_axioms
$ alpha $ is the sigma-algebra
This is what I tried:
It is a simple counterexample: I have $A_{i}´s$ that belongs to $ alpha $. If I take the $emptyset$ set. So $R (B) = P (A | B)$ it is not well defined.
I am not sure. Any help? Am I right?
Also, If I define $P(B)>0$ of course my counterexmple doesnt work. How I could proof this?
probability probability-theory measure-theory probability-distributions
probability probability-theory measure-theory probability-distributions
edited Jan 11 at 23:43
Bernard
121k740116
asked Jan 11 at 23:43
LauraLaura
2408
edited Jan 11 at 23:43
Bernard
121k740116
asked Jan 11 at 23:43
LauraLaura
2408
edited Jan 11 at 23:43
Bernard
121k740116
edited Jan 11 at 23:43
Bernard
121k740116
edited Jan 11 at 23:43
Bernard
121k740116
121k740116
asked Jan 11 at 23:43
LauraLaura
2408
asked Jan 11 at 23:43
LauraLaura
2408
asked Jan 11 at 23:43
LauraLaura
2408
2408
add a comment |
add a comment |
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