Mutually independent $P(A ∩ B ∩ C) ne P(A ∩ B∩ A ∩ C)$
1
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I have events $A$, $B$, $C$ which are mutually independent. Let's say $A∩B$ and $A∩C$ are also independent. Then $$P(A ∩ B ∩ C) = P(A)P(B)P(C)$$ but
$$P(A ∩ B ∩ C) = P((A ∩ B) ∩ (A ∩ C)) = P(A ∩ B)P(A ∩ C) = P(A)P(B)P(A)P(C)$$
$$P(A)P(B)P(C) = P(A)P(B)P(A)P(C)$$
Why is it so? I guess that $A = Omega$ or $emptyset$? How then this conclusion can be drawn from given properties? Would really appreciate any input.
probability elementary-set-theory
edited Dec 17 '18 at 15:57
Tianlalu
3,08421138
asked Dec 17 '18 at 15:42
PaprikaPaprika
113
$endgroup$
|
show 9 more comments
1
$begingroup$
I have events $A$, $B$, $C$ which are mutually independent. Let's say $A∩B$ and $A∩C$ are also independent. Then $$P(A ∩ B ∩ C) = P(A)P(B)P(C)$$ but
$$P(A ∩ B ∩ C) = P((A ∩ B) ∩ (A ∩ C)) = P(A ∩ B)P(A ∩ C) = P(A)P(B)P(A)P(C)$$
$$P(A)P(B)P(C) = P(A)P(B)P(A)P(C)$$
Why is it so? I guess that $A = Omega$ or $emptyset$? How then this conclusion can be drawn from given properties? Would really appreciate any input.
probability elementary-set-theory
edited Dec 17 '18 at 15:57
Tianlalu
3,08421138
asked Dec 17 '18 at 15:42
PaprikaPaprika
113
$endgroup$
1
$begingroup$
Who said $Acap B$ and $Acap C$ are independent?
$endgroup$
– Don Thousand
Dec 17 '18 at 15:58
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@RushabhMehta That's given.
$endgroup$
– Paprika
Dec 17 '18 at 16:01
1
$begingroup$
$P(A)=0$ or $P(A)=1$ does not imply that $A=varnothing$ or $A=Omega$.
$endgroup$
– user593746
Dec 17 '18 at 16:29
1
$begingroup$
Take $Omega=Bbb{R}$ for example, say, with the standard normal distribution. Then, $A={0}$ is not empty but $P(A)=0$. Similarly, $A=Bbb{R}setminus{0}$ is not $Omega$ but $P(A)=1$.
$endgroup$
– user593746
Dec 17 '18 at 16:31
1
$begingroup$
I don't know. What is $Omega$ here? What is the probability distribution on $Omega$? If we don't know these two pieces of information, I'm afraid you cannot conclude $A=varnothing $or $A=Omega$.
$endgroup$
– user593746
Dec 17 '18 at 16:38
|
show 9 more comments
1
1
1
0
$begingroup$
I have events $A$, $B$, $C$ which are mutually independent. Let's say $A∩B$ and $A∩C$ are also independent. Then $$P(A ∩ B ∩ C) = P(A)P(B)P(C)$$ but
$$P(A ∩ B ∩ C) = P((A ∩ B) ∩ (A ∩ C)) = P(A ∩ B)P(A ∩ C) = P(A)P(B)P(A)P(C)$$
$$P(A)P(B)P(C) = P(A)P(B)P(A)P(C)$$
Why is it so? I guess that $A = Omega$ or $emptyset$? How then this conclusion can be drawn from given properties? Would really appreciate any input.
probability elementary-set-theory
edited Dec 17 '18 at 15:57
Tianlalu
3,08421138
asked Dec 17 '18 at 15:42
PaprikaPaprika
113
$endgroup$
I have events $A$, $B$, $C$ which are mutually independent. Let's say $A∩B$ and $A∩C$ are also independent. Then $$P(A ∩ B ∩ C) = P(A)P(B)P(C)$$ but
$$P(A ∩ B ∩ C) = P((A ∩ B) ∩ (A ∩ C)) = P(A ∩ B)P(A ∩ C) = P(A)P(B)P(A)P(C)$$
$$P(A)P(B)P(C) = P(A)P(B)P(A)P(C)$$
Why is it so? I guess that $A = Omega$ or $emptyset$? How then this conclusion can be drawn from given properties? Would really appreciate any input.
probability elementary-set-theory
probability elementary-set-theory
edited Dec 17 '18 at 15:57
Tianlalu
3,08421138
asked Dec 17 '18 at 15:42
PaprikaPaprika
113
edited Dec 17 '18 at 15:57
Tianlalu
3,08421138
asked Dec 17 '18 at 15:42
PaprikaPaprika
113
edited Dec 17 '18 at 15:57
Tianlalu
3,08421138
edited Dec 17 '18 at 15:57
Tianlalu
3,08421138
edited Dec 17 '18 at 15:57
Tianlalu
3,08421138
3,08421138
asked Dec 17 '18 at 15:42
PaprikaPaprika
113
asked Dec 17 '18 at 15:42
PaprikaPaprika
113
asked Dec 17 '18 at 15:42
PaprikaPaprika
113
113
1
$begingroup$
Who said $Acap B$ and $Acap C$ are independent?
$endgroup$
– Don Thousand
Dec 17 '18 at 15:58
$begingroup$
@RushabhMehta That's given.
$endgroup$
– Paprika
Dec 17 '18 at 16:01
1
$begingroup$
$P(A)=0$ or $P(A)=1$ does not imply that $A=varnothing$ or $A=Omega$.
$endgroup$
– user593746
Dec 17 '18 at 16:29
1
$begingroup$
Take $Omega=Bbb{R}$ for example, say, with the standard normal distribution. Then, $A={0}$ is not empty but $P(A)=0$. Similarly, $A=Bbb{R}setminus{0}$ is not $Omega$ but $P(A)=1$.
$endgroup$
– user593746
Dec 17 '18 at 16:31
1
$begingroup$
I don't know. What is $Omega$ here? What is the probability distribution on $Omega$? If we don't know these two pieces of information, I'm afraid you cannot conclude $A=varnothing $or $A=Omega$.
$endgroup$
– user593746
Dec 17 '18 at 16:38
|
show 9 more comments
1
$begingroup$
Who said $Acap B$ and $Acap C$ are independent?
$endgroup$
– Don Thousand
Dec 17 '18 at 15:58
$begingroup$
@RushabhMehta That's given.
$endgroup$
– Paprika
Dec 17 '18 at 16:01
1
$begingroup$
$P(A)=0$ or $P(A)=1$ does not imply that $A=varnothing$ or $A=Omega$.
$endgroup$
– user593746
Dec 17 '18 at 16:29
1
$begingroup$
Take $Omega=Bbb{R}$ for example, say, with the standard normal distribution. Then, $A={0}$ is not empty but $P(A)=0$. Similarly, $A=Bbb{R}setminus{0}$ is not $Omega$ but $P(A)=1$.
$endgroup$
– user593746
Dec 17 '18 at 16:31
1
$begingroup$
I don't know. What is $Omega$ here? What is the probability distribution on $Omega$? If we don't know these two pieces of information, I'm afraid you cannot conclude $A=varnothing $or $A=Omega$.
$endgroup$
– user593746
Dec 17 '18 at 16:38
1
1
$begingroup$
Who said $Acap B$ and $Acap C$ are independent?
$endgroup$
– Don Thousand
Dec 17 '18 at 15:58
$begingroup$
Who said $Acap B$ and $Acap C$ are independent?
$endgroup$
– Don Thousand
Dec 17 '18 at 15:58
$begingroup$
@RushabhMehta That's given.
$endgroup$
– Paprika
Dec 17 '18 at 16:01
$begingroup$
@RushabhMehta That's given.
$endgroup$
– Paprika
Dec 17 '18 at 16:01
1
1
$begingroup$
$P(A)=0$ or $P(A)=1$ does not imply that $A=varnothing$ or $A=Omega$.
$endgroup$
– user593746
Dec 17 '18 at 16:29
$begingroup$
$P(A)=0$ or $P(A)=1$ does not imply that $A=varnothing$ or $A=Omega$.
$endgroup$
– user593746
Dec 17 '18 at 16:29
1
1
$begingroup$
Take $Omega=Bbb{R}$ for example, say, with the standard normal distribution. Then, $A={0}$ is not empty but $P(A)=0$. Similarly, $A=Bbb{R}setminus{0}$ is not $Omega$ but $P(A)=1$.
$endgroup$
– user593746
Dec 17 '18 at 16:31
$begingroup$
Take $Omega=Bbb{R}$ for example, say, with the standard normal distribution. Then, $A={0}$ is not empty but $P(A)=0$. Similarly, $A=Bbb{R}setminus{0}$ is not $Omega$ but $P(A)=1$.
$endgroup$
– user593746
Dec 17 '18 at 16:31
1
1
$begingroup$
I don't know. What is $Omega$ here? What is the probability distribution on $Omega$? If we don't know these two pieces of information, I'm afraid you cannot conclude $A=varnothing $or $A=Omega$.
$endgroup$
– user593746
Dec 17 '18 at 16:38
$begingroup$
I don't know. What is $Omega$ here? What is the probability distribution on $Omega$? If we don't know these two pieces of information, I'm afraid you cannot conclude $A=varnothing $or $A=Omega$.
$endgroup$
– user593746
Dec 17 '18 at 16:38
|
show 9 more comments
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1
$begingroup$
Who said $Acap B$ and $Acap C$ are independent?
$endgroup$
– Don Thousand
Dec 17 '18 at 15:58
$begingroup$
@RushabhMehta That's given.
$endgroup$
– Paprika
Dec 17 '18 at 16:01
1
$begingroup$
$P(A)=0$ or $P(A)=1$ does not imply that $A=varnothing$ or $A=Omega$.
$endgroup$
– user593746
Dec 17 '18 at 16:29
1
$begingroup$
Take $Omega=Bbb{R}$ for example, say, with the standard normal distribution. Then, $A={0}$ is not empty but $P(A)=0$. Similarly, $A=Bbb{R}setminus{0}$ is not $Omega$ but $P(A)=1$.
$endgroup$
– user593746
Dec 17 '18 at 16:31
1
$begingroup$
I don't know. What is $Omega$ here? What is the probability distribution on $Omega$? If we don't know these two pieces of information, I'm afraid you cannot conclude $A=varnothing $or $A=Omega$.
$endgroup$
– user593746
Dec 17 '18 at 16:38