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
122k741116
asked Jan 11 at 23:43
LauraLaura
3368
$endgroup$
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
122k741116
asked Jan 11 at 23:43
LauraLaura
3368
$endgroup$
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
122k741116
asked Jan 11 at 23:43
LauraLaura
3368
$endgroup$
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
122k741116
asked Jan 11 at 23:43
LauraLaura
3368
edited Jan 11 at 23:43
Bernard
122k741116
asked Jan 11 at 23:43
LauraLaura
3368
edited Jan 11 at 23:43
Bernard
122k741116
edited Jan 11 at 23:43
Bernard
122k741116
edited Jan 11 at 23:43
Bernard
122k741116
122k741116
asked Jan 11 at 23:43
LauraLaura
3368
asked Jan 11 at 23:43
LauraLaura
3368
asked Jan 11 at 23:43
LauraLaura
3368
3368
add a comment |
add a comment |
1 Answer
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$begingroup$
We must have $P(B)>0$ for $P(A|B)$ to be defined (your textbook may vary).
Axiom 1: $P(A|B) := frac{P(A cap B)}{P(B)} ge 0$. There's no problem here.
Axiom 2: $P(Omega) = P(A | Omega) = frac{P(A)}{P(Omega)} = P(A) ne 1$. There is a problem here if $P(A) ne 1$. Of course there is no problem here if $P(A) = 1$ (even if $A ne Omega$, which is indeed possible).
Axiom 3: $P(A|cup_{text{countable} i} B_i) = sum_{text{countable} i} P(A|B_i)$. I think we don't have to check anymore because Axiom 2 is violated, but I think Axiom 3 is violated as well.
The correct probability function involving conditional probability should be $R(A)=P(A|B)$, I think:
Axiom 1: $P(A|B) := frac{P(A cap B)}{P(B)} ge 0$. There's no problem here.
Axiom 2: $P(Omega) = P(Omega | B) = frac{P(B)}{P(B)} = 1$. Now, there's no problem here.
Axiom 3: $P(cup_{text{countable} i} A_i | B) = sum_{text{countable} i} P(A_I |B)$. I think Axiom 3 is satisfied here. I'll leave it to you.
answered Feb 28 at 7:24
MitjacksonMitjackson
7011
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$begingroup$
We must have $P(B)>0$ for $P(A|B)$ to be defined (your textbook may vary).
Axiom 1: $P(A|B) := frac{P(A cap B)}{P(B)} ge 0$. There's no problem here.
Axiom 2: $P(Omega) = P(A | Omega) = frac{P(A)}{P(Omega)} = P(A) ne 1$. There is a problem here if $P(A) ne 1$. Of course there is no problem here if $P(A) = 1$ (even if $A ne Omega$, which is indeed possible).
Axiom 3: $P(A|cup_{text{countable} i} B_i) = sum_{text{countable} i} P(A|B_i)$. I think we don't have to check anymore because Axiom 2 is violated, but I think Axiom 3 is violated as well.
The correct probability function involving conditional probability should be $R(A)=P(A|B)$, I think:
Axiom 1: $P(A|B) := frac{P(A cap B)}{P(B)} ge 0$. There's no problem here.
Axiom 2: $P(Omega) = P(Omega | B) = frac{P(B)}{P(B)} = 1$. Now, there's no problem here.
Axiom 3: $P(cup_{text{countable} i} A_i | B) = sum_{text{countable} i} P(A_I |B)$. I think Axiom 3 is satisfied here. I'll leave it to you.
answered Feb 28 at 7:24
MitjacksonMitjackson
7011
$endgroup$
add a comment |
$begingroup$
We must have $P(B)>0$ for $P(A|B)$ to be defined (your textbook may vary).
Axiom 1: $P(A|B) := frac{P(A cap B)}{P(B)} ge 0$. There's no problem here.
Axiom 2: $P(Omega) = P(A | Omega) = frac{P(A)}{P(Omega)} = P(A) ne 1$. There is a problem here if $P(A) ne 1$. Of course there is no problem here if $P(A) = 1$ (even if $A ne Omega$, which is indeed possible).
Axiom 3: $P(A|cup_{text{countable} i} B_i) = sum_{text{countable} i} P(A|B_i)$. I think we don't have to check anymore because Axiom 2 is violated, but I think Axiom 3 is violated as well.
The correct probability function involving conditional probability should be $R(A)=P(A|B)$, I think:
Axiom 1: $P(A|B) := frac{P(A cap B)}{P(B)} ge 0$. There's no problem here.
Axiom 2: $P(Omega) = P(Omega | B) = frac{P(B)}{P(B)} = 1$. Now, there's no problem here.
Axiom 3: $P(cup_{text{countable} i} A_i | B) = sum_{text{countable} i} P(A_I |B)$. I think Axiom 3 is satisfied here. I'll leave it to you.
answered Feb 28 at 7:24
MitjacksonMitjackson
7011
$endgroup$
add a comment |
$begingroup$
We must have $P(B)>0$ for $P(A|B)$ to be defined (your textbook may vary).
Axiom 1: $P(A|B) := frac{P(A cap B)}{P(B)} ge 0$. There's no problem here.
Axiom 2: $P(Omega) = P(A | Omega) = frac{P(A)}{P(Omega)} = P(A) ne 1$. There is a problem here if $P(A) ne 1$. Of course there is no problem here if $P(A) = 1$ (even if $A ne Omega$, which is indeed possible).
Axiom 3: $P(A|cup_{text{countable} i} B_i) = sum_{text{countable} i} P(A|B_i)$. I think we don't have to check anymore because Axiom 2 is violated, but I think Axiom 3 is violated as well.
The correct probability function involving conditional probability should be $R(A)=P(A|B)$, I think:
Axiom 1: $P(A|B) := frac{P(A cap B)}{P(B)} ge 0$. There's no problem here.
Axiom 2: $P(Omega) = P(Omega | B) = frac{P(B)}{P(B)} = 1$. Now, there's no problem here.
Axiom 3: $P(cup_{text{countable} i} A_i | B) = sum_{text{countable} i} P(A_I |B)$. I think Axiom 3 is satisfied here. I'll leave it to you.
answered Feb 28 at 7:24
MitjacksonMitjackson
7011
$endgroup$
We must have $P(B)>0$ for $P(A|B)$ to be defined (your textbook may vary).
Axiom 1: $P(A|B) := frac{P(A cap B)}{P(B)} ge 0$. There's no problem here.
Axiom 2: $P(Omega) = P(A | Omega) = frac{P(A)}{P(Omega)} = P(A) ne 1$. There is a problem here if $P(A) ne 1$. Of course there is no problem here if $P(A) = 1$ (even if $A ne Omega$, which is indeed possible).
Axiom 3: $P(A|cup_{text{countable} i} B_i) = sum_{text{countable} i} P(A|B_i)$. I think we don't have to check anymore because Axiom 2 is violated, but I think Axiom 3 is violated as well.
The correct probability function involving conditional probability should be $R(A)=P(A|B)$, I think:
Axiom 1: $P(A|B) := frac{P(A cap B)}{P(B)} ge 0$. There's no problem here.
Axiom 2: $P(Omega) = P(Omega | B) = frac{P(B)}{P(B)} = 1$. Now, there's no problem here.
Axiom 3: $P(cup_{text{countable} i} A_i | B) = sum_{text{countable} i} P(A_I |B)$. I think Axiom 3 is satisfied here. I'll leave it to you.
answered Feb 28 at 7:24
MitjacksonMitjackson
7011
answered Feb 28 at 7:24
MitjacksonMitjackson
7011
answered Feb 28 at 7:24
MitjacksonMitjackson
7011
answered Feb 28 at 7:24
MitjacksonMitjackson
7011
7011
add a comment |
add a comment |
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