Mixing
![What is Modern Analysis and Abstract Analysis? [closed]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZCTB4-bb-s-32SAm6ytjzFOU06cH9o8q-a_wq_UnH6lcd8hvws23CmBWHM6g256LzTX9yK1EiCCzTC1NSgtxlNk_J9hdr9bA3tD0Nqntbl521j4bLAm_uXPANuWLBhcCKTKzns2gaUodx/s72-c/1.jpg)
Correct interpretation of a conditional probability constraint
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Let $A$, $B$, $C$ be three events such that $P(Acup Bcup C)=1$.
Suppose that their event-intersection $I=Acap Bcap C$ is such that it holds the constraint
$$
P(I)=fracP(IA)+P(I.
$$
From this assumption, I would conclude that the knowledge about the occurrence (success/failure) of the event $A$ and of the event $B$ is enough to determine the probability that the event $I$ takes place or not, independently on what is the occurrence of the event $C$.
Is this interpretation correct?
Otherwise, what kind of information gives us such imposition (e.g. about their mutual exclusivity)?
So far, I was able to prove that such statement corresponds to $P(A)+P(B)=1$, which however seems not to involve the same interpretation (at least, not at first sight).
I am not an expert of conditional probability, therefore this question may be obvious for an expert, and I apologize in that case.
However, many thanks for your suggestions!
A similar problem is discussed here A problem of conditional probability, where the focus was on whether or not $P(I)=0$, whereas here I am more interested on the correct interpretation of the constraint. Sorry for some overlap!
probability probability-theory bayesian bayes-theorem
asked Aug 1 at 6:32
Andrea Prunotto
569114
 |Â
show 3 more comments
up vote
1
down vote
favorite
Let $A$, $B$, $C$ be three events such that $P(Acup Bcup C)=1$.
Suppose that their event-intersection $I=Acap Bcap C$ is such that it holds the constraint
$$
P(I)=fracP(IA)+P(I.
$$
From this assumption, I would conclude that the knowledge about the occurrence (success/failure) of the event $A$ and of the event $B$ is enough to determine the probability that the event $I$ takes place or not, independently on what is the occurrence of the event $C$.
Is this interpretation correct?
Otherwise, what kind of information gives us such imposition (e.g. about their mutual exclusivity)?
So far, I was able to prove that such statement corresponds to $P(A)+P(B)=1$, which however seems not to involve the same interpretation (at least, not at first sight).
I am not an expert of conditional probability, therefore this question may be obvious for an expert, and I apologize in that case.
However, many thanks for your suggestions!
A similar problem is discussed here A problem of conditional probability, where the focus was on whether or not $P(I)=0$, whereas here I am more interested on the correct interpretation of the constraint. Sorry for some overlap!
probability probability-theory bayesian bayes-theorem
asked Aug 1 at 6:32
Andrea Prunotto
569114
Could you perhaps say something about how you arrived at this interpretation of the constraint?
– joriki
Aug 1 at 6:52
Sure: $P(A)+P(B)=1$, $1=frac1P(A)+P(B)$, $P(I)=fracP^2(I)P(I)P(A)+P(I)P(B)$, $P(I)=fracP^2(I)P(I$, $P(I)=fracP(IA)+P(IfracP(A)P(B)P(A)P(B)$, $P(I)=fracP(IA)+P(I$.
– Andrea Prunotto
Aug 1 at 7:01
I extensively used the fact that $P(I|A)P(A)=P(A|I)P(I)$, and $P(A|I)=1$. Same for $B$.
– Andrea Prunotto
Aug 1 at 7:12
That was a misunderstanding. I meant not how you showed that it's equivalent to $P(A)+P(B)=1$ (though that's also interesting) but how you arrived at the interpretation about which you're asking whether it's correct.
– joriki
Aug 1 at 7:14
1
I see. Then consider $P(I)=P(Imid A)$. You could apply the same argument to that. But that's clearly not a statement that $I$ is independent of $C$ -- in fact, it's a statement that $I$ is independent of $A$. Whether an expression involves only conditioning on $A$ and $B$ has nothing to do with other events. In fact, even the trivial statement $P(I)=P(I)$ doesn't contain anything other than conditioning on $A$ and $B$, and here it's even more obvious that this tells us nothing about $C$.
– joriki
Aug 1 at 7:33
 |Â
show 3 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $A$, $B$, $C$ be three events such that $P(Acup Bcup C)=1$.
Suppose that their event-intersection $I=Acap Bcap C$ is such that it holds the constraint
$$
P(I)=fracP(IA)+P(I.
$$
From this assumption, I would conclude that the knowledge about the occurrence (success/failure) of the event $A$ and of the event $B$ is enough to determine the probability that the event $I$ takes place or not, independently on what is the occurrence of the event $C$.
Is this interpretation correct?
Otherwise, what kind of information gives us such imposition (e.g. about their mutual exclusivity)?
So far, I was able to prove that such statement corresponds to $P(A)+P(B)=1$, which however seems not to involve the same interpretation (at least, not at first sight).
I am not an expert of conditional probability, therefore this question may be obvious for an expert, and I apologize in that case.
However, many thanks for your suggestions!
A similar problem is discussed here A problem of conditional probability, where the focus was on whether or not $P(I)=0$, whereas here I am more interested on the correct interpretation of the constraint. Sorry for some overlap!
probability probability-theory bayesian bayes-theorem
asked Aug 1 at 6:32
Andrea Prunotto
569114
Let $A$, $B$, $C$ be three events such that $P(Acup Bcup C)=1$.
Suppose that their event-intersection $I=Acap Bcap C$ is such that it holds the constraint
$$
P(I)=fracP(IA)+P(I.
$$
From this assumption, I would conclude that the knowledge about the occurrence (success/failure) of the event $A$ and of the event $B$ is enough to determine the probability that the event $I$ takes place or not, independently on what is the occurrence of the event $C$.
Is this interpretation correct?
Otherwise, what kind of information gives us such imposition (e.g. about their mutual exclusivity)?
So far, I was able to prove that such statement corresponds to $P(A)+P(B)=1$, which however seems not to involve the same interpretation (at least, not at first sight).
I am not an expert of conditional probability, therefore this question may be obvious for an expert, and I apologize in that case.
However, many thanks for your suggestions!
A similar problem is discussed here A problem of conditional probability, where the focus was on whether or not $P(I)=0$, whereas here I am more interested on the correct interpretation of the constraint. Sorry for some overlap!
probability probability-theory bayesian bayes-theorem
asked Aug 1 at 6:32
Andrea Prunotto
569114
edited Aug 1 at 7:06
asked Aug 1 at 6:32
Andrea Prunotto
569114
asked Aug 1 at 6:32
Andrea Prunotto
569114
asked Aug 1 at 6:32
Andrea Prunotto
569114
569114
Could you perhaps say something about how you arrived at this interpretation of the constraint?
– joriki
Aug 1 at 6:52
Sure: $P(A)+P(B)=1$, $1=frac1P(A)+P(B)$, $P(I)=fracP^2(I)P(I)P(A)+P(I)P(B)$, $P(I)=fracP^2(I)P(I$, $P(I)=fracP(IA)+P(IfracP(A)P(B)P(A)P(B)$, $P(I)=fracP(IA)+P(I$.
– Andrea Prunotto
Aug 1 at 7:01
I extensively used the fact that $P(I|A)P(A)=P(A|I)P(I)$, and $P(A|I)=1$. Same for $B$.
– Andrea Prunotto
Aug 1 at 7:12
That was a misunderstanding. I meant not how you showed that it's equivalent to $P(A)+P(B)=1$ (though that's also interesting) but how you arrived at the interpretation about which you're asking whether it's correct.
– joriki
Aug 1 at 7:14
1
I see. Then consider $P(I)=P(Imid A)$. You could apply the same argument to that. But that's clearly not a statement that $I$ is independent of $C$ -- in fact, it's a statement that $I$ is independent of $A$. Whether an expression involves only conditioning on $A$ and $B$ has nothing to do with other events. In fact, even the trivial statement $P(I)=P(I)$ doesn't contain anything other than conditioning on $A$ and $B$, and here it's even more obvious that this tells us nothing about $C$.
– joriki
Aug 1 at 7:33
 |Â
show 3 more comments
Could you perhaps say something about how you arrived at this interpretation of the constraint?
– joriki
Aug 1 at 6:52
Sure: $P(A)+P(B)=1$, $1=frac1P(A)+P(B)$, $P(I)=fracP^2(I)P(I)P(A)+P(I)P(B)$, $P(I)=fracP^2(I)P(I$, $P(I)=fracP(IA)+P(IfracP(A)P(B)P(A)P(B)$, $P(I)=fracP(IA)+P(I$.
– Andrea Prunotto
Aug 1 at 7:01
I extensively used the fact that $P(I|A)P(A)=P(A|I)P(I)$, and $P(A|I)=1$. Same for $B$.
– Andrea Prunotto
Aug 1 at 7:12
That was a misunderstanding. I meant not how you showed that it's equivalent to $P(A)+P(B)=1$ (though that's also interesting) but how you arrived at the interpretation about which you're asking whether it's correct.
– joriki
Aug 1 at 7:14
1
I see. Then consider $P(I)=P(Imid A)$. You could apply the same argument to that. But that's clearly not a statement that $I$ is independent of $C$ -- in fact, it's a statement that $I$ is independent of $A$. Whether an expression involves only conditioning on $A$ and $B$ has nothing to do with other events. In fact, even the trivial statement $P(I)=P(I)$ doesn't contain anything other than conditioning on $A$ and $B$, and here it's even more obvious that this tells us nothing about $C$.
– joriki
Aug 1 at 7:33
Could you perhaps say something about how you arrived at this interpretation of the constraint?
– joriki
Aug 1 at 6:52
Could you perhaps say something about how you arrived at this interpretation of the constraint?
– joriki
Aug 1 at 6:52
Sure: $P(A)+P(B)=1$, $1=frac1P(A)+P(B)$, $P(I)=fracP^2(I)P(I)P(A)+P(I)P(B)$, $P(I)=fracP^2(I)P(I$, $P(I)=fracP(IA)+P(IfracP(A)P(B)P(A)P(B)$, $P(I)=fracP(IA)+P(I$.
– Andrea Prunotto
Aug 1 at 7:01
Sure: $P(A)+P(B)=1$, $1=frac1P(A)+P(B)$, $P(I)=fracP^2(I)P(I)P(A)+P(I)P(B)$, $P(I)=fracP^2(I)P(I$, $P(I)=fracP(IA)+P(IfracP(A)P(B)P(A)P(B)$, $P(I)=fracP(IA)+P(I$.
– Andrea Prunotto
Aug 1 at 7:01
I extensively used the fact that $P(I|A)P(A)=P(A|I)P(I)$, and $P(A|I)=1$. Same for $B$.
– Andrea Prunotto
Aug 1 at 7:12
I extensively used the fact that $P(I|A)P(A)=P(A|I)P(I)$, and $P(A|I)=1$. Same for $B$.
– Andrea Prunotto
Aug 1 at 7:12
That was a misunderstanding. I meant not how you showed that it's equivalent to $P(A)+P(B)=1$ (though that's also interesting) but how you arrived at the interpretation about which you're asking whether it's correct.
– joriki
Aug 1 at 7:14
That was a misunderstanding. I meant not how you showed that it's equivalent to $P(A)+P(B)=1$ (though that's also interesting) but how you arrived at the interpretation about which you're asking whether it's correct.
– joriki
Aug 1 at 7:14
1
1
I see. Then consider $P(I)=P(Imid A)$. You could apply the same argument to that. But that's clearly not a statement that $I$ is independent of $C$ -- in fact, it's a statement that $I$ is independent of $A$. Whether an expression involves only conditioning on $A$ and $B$ has nothing to do with other events. In fact, even the trivial statement $P(I)=P(I)$ doesn't contain anything other than conditioning on $A$ and $B$, and here it's even more obvious that this tells us nothing about $C$.
– joriki
Aug 1 at 7:33
I see. Then consider $P(I)=P(Imid A)$. You could apply the same argument to that. But that's clearly not a statement that $I$ is independent of $C$ -- in fact, it's a statement that $I$ is independent of $A$. Whether an expression involves only conditioning on $A$ and $B$ has nothing to do with other events. In fact, even the trivial statement $P(I)=P(I)$ doesn't contain anything other than conditioning on $A$ and $B$, and here it's even more obvious that this tells us nothing about $C$.
– joriki
Aug 1 at 7:33
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
With reference to your comments under the question:
Consider $P(I)=P(Imid A)$. You could apply the same argument to that. But that's clearly not a statement that $I$ is independent of $C$ – in fact, it's a statement that $I$ is independent of $A$. Whether an expression involves only conditioning on $A$ and $B$ has nothing to do with other events. In fact, even the trivial statement $P(I)=P(I)$ doesn't contain anything other than conditioning on $A$ and $B$, and here it's even more obvious that this tells us nothing about $C$.
answered Aug 1 at 7:36
joriki
164k10179328
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
With reference to your comments under the question:
Consider $P(I)=P(Imid A)$. You could apply the same argument to that. But that's clearly not a statement that $I$ is independent of $C$ – in fact, it's a statement that $I$ is independent of $A$. Whether an expression involves only conditioning on $A$ and $B$ has nothing to do with other events. In fact, even the trivial statement $P(I)=P(I)$ doesn't contain anything other than conditioning on $A$ and $B$, and here it's even more obvious that this tells us nothing about $C$.
answered Aug 1 at 7:36
joriki
164k10179328
add a comment |Â
up vote
2
down vote
accepted
With reference to your comments under the question:
Consider $P(I)=P(Imid A)$. You could apply the same argument to that. But that's clearly not a statement that $I$ is independent of $C$ – in fact, it's a statement that $I$ is independent of $A$. Whether an expression involves only conditioning on $A$ and $B$ has nothing to do with other events. In fact, even the trivial statement $P(I)=P(I)$ doesn't contain anything other than conditioning on $A$ and $B$, and here it's even more obvious that this tells us nothing about $C$.
answered Aug 1 at 7:36
joriki
164k10179328
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
With reference to your comments under the question:
Consider $P(I)=P(Imid A)$. You could apply the same argument to that. But that's clearly not a statement that $I$ is independent of $C$ – in fact, it's a statement that $I$ is independent of $A$. Whether an expression involves only conditioning on $A$ and $B$ has nothing to do with other events. In fact, even the trivial statement $P(I)=P(I)$ doesn't contain anything other than conditioning on $A$ and $B$, and here it's even more obvious that this tells us nothing about $C$.
answered Aug 1 at 7:36
joriki
164k10179328
With reference to your comments under the question:
Consider $P(I)=P(Imid A)$. You could apply the same argument to that. But that's clearly not a statement that $I$ is independent of $C$ – in fact, it's a statement that $I$ is independent of $A$. Whether an expression involves only conditioning on $A$ and $B$ has nothing to do with other events. In fact, even the trivial statement $P(I)=P(I)$ doesn't contain anything other than conditioning on $A$ and $B$, and here it's even more obvious that this tells us nothing about $C$.
answered Aug 1 at 7:36
joriki
164k10179328
answered Aug 1 at 7:36
joriki
164k10179328
answered Aug 1 at 7:36
joriki
164k10179328
answered Aug 1 at 7:36
joriki
164k10179328
164k10179328
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2868782%2fcorrect-interpretation-of-a-conditional-probability-constraint%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Could you perhaps say something about how you arrived at this interpretation of the constraint?
– joriki
Aug 1 at 6:52
Sure: $P(A)+P(B)=1$, $1=frac1P(A)+P(B)$, $P(I)=fracP^2(I)P(I)P(A)+P(I)P(B)$, $P(I)=fracP^2(I)P(I$, $P(I)=fracP(IA)+P(IfracP(A)P(B)P(A)P(B)$, $P(I)=fracP(IA)+P(I$.
– Andrea Prunotto
Aug 1 at 7:01
I extensively used the fact that $P(I|A)P(A)=P(A|I)P(I)$, and $P(A|I)=1$. Same for $B$.
– Andrea Prunotto
Aug 1 at 7:12
That was a misunderstanding. I meant not how you showed that it's equivalent to $P(A)+P(B)=1$ (though that's also interesting) but how you arrived at the interpretation about which you're asking whether it's correct.
– joriki
Aug 1 at 7:14
1
I see. Then consider $P(I)=P(Imid A)$. You could apply the same argument to that. But that's clearly not a statement that $I$ is independent of $C$ -- in fact, it's a statement that $I$ is independent of $A$. Whether an expression involves only conditioning on $A$ and $B$ has nothing to do with other events. In fact, even the trivial statement $P(I)=P(I)$ doesn't contain anything other than conditioning on $A$ and $B$, and here it's even more obvious that this tells us nothing about $C$.
– joriki
Aug 1 at 7:33