Mixing
Probability disjunction: Sum probabilities greater than actual probability
$begingroup$
Is it true that $$P(A lor B lor C lor ...) leq P(A) + P(B) + P(C) + ...$$?
I saw the general form of the probability disjunction here: https://stats.stackexchange.com/questions/87533/whats-the-general-disjunction-rule-for-n-events , but the right-hand side of the formula contains positive terms beyond simply $P(A)$, $P(B)$, etc.
My attempt by induction:
$$P(A lor B) = P(A) + P(B) - P(A land B) leq P(A) + P(B)$$
Assume now $P(bigvee_{i=1}^n C_i) leq sum_{i=1}^nP(C_i)$
Then $$P(C_{n+1}lor bigvee_{i=1}^n C_i) = P(bigvee_{i=1}^n C_i) + P(C_{n+1}) - P(C_{n+1}land bigvee_{i=1}^n C_i) leq sum_{i=1}^nP(C_i) + P(C_{n+1}) - P(C_{n+1}land bigvee_{i=1}^n C_i) leq sum_{i=1}^nP(C_i) + P(C_{n+1})$$
probability
asked Jan 11 at 22:26
HintonHinton
1163
$endgroup$
add a comment |
$begingroup$
Is it true that $$P(A lor B lor C lor ...) leq P(A) + P(B) + P(C) + ...$$?
I saw the general form of the probability disjunction here: https://stats.stackexchange.com/questions/87533/whats-the-general-disjunction-rule-for-n-events , but the right-hand side of the formula contains positive terms beyond simply $P(A)$, $P(B)$, etc.
My attempt by induction:
$$P(A lor B) = P(A) + P(B) - P(A land B) leq P(A) + P(B)$$
Assume now $P(bigvee_{i=1}^n C_i) leq sum_{i=1}^nP(C_i)$
Then $$P(C_{n+1}lor bigvee_{i=1}^n C_i) = P(bigvee_{i=1}^n C_i) + P(C_{n+1}) - P(C_{n+1}land bigvee_{i=1}^n C_i) leq sum_{i=1}^nP(C_i) + P(C_{n+1}) - P(C_{n+1}land bigvee_{i=1}^n C_i) leq sum_{i=1}^nP(C_i) + P(C_{n+1})$$
probability
asked Jan 11 at 22:26
HintonHinton
1163
$endgroup$
$begingroup$
Does anyone have comments on the downvote? I'm not really trying to break rules here, would appreciate if someone could let me know what I'm doing wrong
$endgroup$
– Hinton
Jan 11 at 22:40
$begingroup$
Is the notation $vee$ common for unions $cup$ in certain regions of the world? This is a rarity, to me.
$endgroup$
– LoveTooNap29
Jan 12 at 1:27
$begingroup$
I have way more education in logic than probability so I used the notation that seemed most familiar to me
$endgroup$
– Hinton
Jan 12 at 22:08
add a comment |
$begingroup$
Is it true that $$P(A lor B lor C lor ...) leq P(A) + P(B) + P(C) + ...$$?
I saw the general form of the probability disjunction here: https://stats.stackexchange.com/questions/87533/whats-the-general-disjunction-rule-for-n-events , but the right-hand side of the formula contains positive terms beyond simply $P(A)$, $P(B)$, etc.
My attempt by induction:
$$P(A lor B) = P(A) + P(B) - P(A land B) leq P(A) + P(B)$$
Assume now $P(bigvee_{i=1}^n C_i) leq sum_{i=1}^nP(C_i)$
Then $$P(C_{n+1}lor bigvee_{i=1}^n C_i) = P(bigvee_{i=1}^n C_i) + P(C_{n+1}) - P(C_{n+1}land bigvee_{i=1}^n C_i) leq sum_{i=1}^nP(C_i) + P(C_{n+1}) - P(C_{n+1}land bigvee_{i=1}^n C_i) leq sum_{i=1}^nP(C_i) + P(C_{n+1})$$
probability
asked Jan 11 at 22:26
HintonHinton
1163
$endgroup$
Is it true that $$P(A lor B lor C lor ...) leq P(A) + P(B) + P(C) + ...$$?
I saw the general form of the probability disjunction here: https://stats.stackexchange.com/questions/87533/whats-the-general-disjunction-rule-for-n-events , but the right-hand side of the formula contains positive terms beyond simply $P(A)$, $P(B)$, etc.
My attempt by induction:
$$P(A lor B) = P(A) + P(B) - P(A land B) leq P(A) + P(B)$$
Assume now $P(bigvee_{i=1}^n C_i) leq sum_{i=1}^nP(C_i)$
Then $$P(C_{n+1}lor bigvee_{i=1}^n C_i) = P(bigvee_{i=1}^n C_i) + P(C_{n+1}) - P(C_{n+1}land bigvee_{i=1}^n C_i) leq sum_{i=1}^nP(C_i) + P(C_{n+1}) - P(C_{n+1}land bigvee_{i=1}^n C_i) leq sum_{i=1}^nP(C_i) + P(C_{n+1})$$
probability
probability
asked Jan 11 at 22:26
HintonHinton
1163
asked Jan 11 at 22:26
HintonHinton
1163
asked Jan 11 at 22:26
HintonHinton
1163
asked Jan 11 at 22:26
HintonHinton
1163
asked Jan 11 at 22:26
HintonHinton
1163
1163
$begingroup$
Does anyone have comments on the downvote? I'm not really trying to break rules here, would appreciate if someone could let me know what I'm doing wrong
$endgroup$
– Hinton
Jan 11 at 22:40
$begingroup$
Is the notation $vee$ common for unions $cup$ in certain regions of the world? This is a rarity, to me.
$endgroup$
– LoveTooNap29
Jan 12 at 1:27
$begingroup$
I have way more education in logic than probability so I used the notation that seemed most familiar to me
$endgroup$
– Hinton
Jan 12 at 22:08
add a comment |
$begingroup$
Does anyone have comments on the downvote? I'm not really trying to break rules here, would appreciate if someone could let me know what I'm doing wrong
$endgroup$
– Hinton
Jan 11 at 22:40
$begingroup$
Is the notation $vee$ common for unions $cup$ in certain regions of the world? This is a rarity, to me.
$endgroup$
– LoveTooNap29
Jan 12 at 1:27
$begingroup$
I have way more education in logic than probability so I used the notation that seemed most familiar to me
$endgroup$
– Hinton
Jan 12 at 22:08
$begingroup$
Does anyone have comments on the downvote? I'm not really trying to break rules here, would appreciate if someone could let me know what I'm doing wrong
$endgroup$
– Hinton
Jan 11 at 22:40
$begingroup$
Does anyone have comments on the downvote? I'm not really trying to break rules here, would appreciate if someone could let me know what I'm doing wrong
$endgroup$
– Hinton
Jan 11 at 22:40
$begingroup$
Is the notation $vee$ common for unions $cup$ in certain regions of the world? This is a rarity, to me.
$endgroup$
– LoveTooNap29
Jan 12 at 1:27
$begingroup$
Is the notation $vee$ common for unions $cup$ in certain regions of the world? This is a rarity, to me.
$endgroup$
– LoveTooNap29
Jan 12 at 1:27
$begingroup$
I have way more education in logic than probability so I used the notation that seemed most familiar to me
$endgroup$
– Hinton
Jan 12 at 22:08
$begingroup$
I have way more education in logic than probability so I used the notation that seemed most familiar to me
$endgroup$
– Hinton
Jan 12 at 22:08
add a comment |
3 Answers
3
active
oldest
votes
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Yes the statement is true. Let $(A_i)_{igeq 1}$ be a collection of events. Then note that
$$
begin{align}
Pleft(bigcup_{i=1}^infty A_iright)
&=P(A_1)+P(A_2bar{A_1})+P(A_3bar{A_2}bar{A_1})+P(A_4bar{A_3}bar{A_2}bar{A_1})+dotsb\
&leq P(A_1)+P(A_2)+P(A_3)+P(A_4)+dotsb
end{align}
$$
where the first equality is by countable additivity and the second by the fact that $Asubseteq Bimplies P(A)leq P(B)$.
answered Jan 11 at 22:38
Foobaz JohnFoobaz John
22.1k41352
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add a comment |
$begingroup$
Your work is correct.
Perhaps an easier way to do it: choose $A=C_{n+1}$ and $B=bigvee_{i=1}^n C_i$. The induction step then follows from the case with only $A$ and $B$.
answered Jan 11 at 22:34
DashermanDasherman
1,075817
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add a comment |
$begingroup$
Hint: Let $A_1,A_2 ldots, A_r$ be a collection of events. Then
$$A_1 lor A_2 lor A_3 lor ldots lor A_r = A'_1 lor A'_2 ldots lor A'_r,$$,
where
$$A_1 doteq A_1; A'_l doteq A_l setminus (A_1 lor A_2 ldots lor A_{l-1}) text{for each } l=1,2,ldots,r $$
So
$$mathbb{P}[A_1 lor A_2 lor A_3 lor ldots lor A_r] = mathbb{P}[A'_1 lor A'_2 ldots lor A'_r].$$
And as the $A'_l$s are mutually disjoint,
$$ mathbb{P}[A'_1 lor A'_2 ldots lor A'_r] = sum_{l=1}^r mathbb{P}[A'_l]$$
Yielding
$$mathbb{P}[A_1 lor A_2 lor A_3 lor ldots lor A_r] = mathbb{P}[A'_1 lor A'_2 ldots lor A'_r] = sum_{l=1}^r mathbb{P}[A'_l].$$
But note that $mathbb{P}[A_l] ge mathbb{P}[A'_l]$ for each $l$, because $A'_l subseteq A_l$. Can you finish from here.
answered Jan 11 at 22:40
MikeMike
3,951412
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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$begingroup$
Yes the statement is true. Let $(A_i)_{igeq 1}$ be a collection of events. Then note that
$$
begin{align}
Pleft(bigcup_{i=1}^infty A_iright)
&=P(A_1)+P(A_2bar{A_1})+P(A_3bar{A_2}bar{A_1})+P(A_4bar{A_3}bar{A_2}bar{A_1})+dotsb\
&leq P(A_1)+P(A_2)+P(A_3)+P(A_4)+dotsb
end{align}
$$
where the first equality is by countable additivity and the second by the fact that $Asubseteq Bimplies P(A)leq P(B)$.
answered Jan 11 at 22:38
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
add a comment |
$begingroup$
Yes the statement is true. Let $(A_i)_{igeq 1}$ be a collection of events. Then note that
$$
begin{align}
Pleft(bigcup_{i=1}^infty A_iright)
&=P(A_1)+P(A_2bar{A_1})+P(A_3bar{A_2}bar{A_1})+P(A_4bar{A_3}bar{A_2}bar{A_1})+dotsb\
&leq P(A_1)+P(A_2)+P(A_3)+P(A_4)+dotsb
end{align}
$$
where the first equality is by countable additivity and the second by the fact that $Asubseteq Bimplies P(A)leq P(B)$.
answered Jan 11 at 22:38
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
add a comment |
$begingroup$
Yes the statement is true. Let $(A_i)_{igeq 1}$ be a collection of events. Then note that
$$
begin{align}
Pleft(bigcup_{i=1}^infty A_iright)
&=P(A_1)+P(A_2bar{A_1})+P(A_3bar{A_2}bar{A_1})+P(A_4bar{A_3}bar{A_2}bar{A_1})+dotsb\
&leq P(A_1)+P(A_2)+P(A_3)+P(A_4)+dotsb
end{align}
$$
where the first equality is by countable additivity and the second by the fact that $Asubseteq Bimplies P(A)leq P(B)$.
answered Jan 11 at 22:38
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
Yes the statement is true. Let $(A_i)_{igeq 1}$ be a collection of events. Then note that
$$
begin{align}
Pleft(bigcup_{i=1}^infty A_iright)
&=P(A_1)+P(A_2bar{A_1})+P(A_3bar{A_2}bar{A_1})+P(A_4bar{A_3}bar{A_2}bar{A_1})+dotsb\
&leq P(A_1)+P(A_2)+P(A_3)+P(A_4)+dotsb
end{align}
$$
where the first equality is by countable additivity and the second by the fact that $Asubseteq Bimplies P(A)leq P(B)$.
answered Jan 11 at 22:38
Foobaz JohnFoobaz John
22.1k41352
answered Jan 11 at 22:38
Foobaz JohnFoobaz John
22.1k41352
answered Jan 11 at 22:38
Foobaz JohnFoobaz John
22.1k41352
answered Jan 11 at 22:38
Foobaz JohnFoobaz John
22.1k41352
22.1k41352
add a comment |
add a comment |
$begingroup$
Your work is correct.
Perhaps an easier way to do it: choose $A=C_{n+1}$ and $B=bigvee_{i=1}^n C_i$. The induction step then follows from the case with only $A$ and $B$.
answered Jan 11 at 22:34
DashermanDasherman
1,075817
$endgroup$
add a comment |
$begingroup$
Your work is correct.
Perhaps an easier way to do it: choose $A=C_{n+1}$ and $B=bigvee_{i=1}^n C_i$. The induction step then follows from the case with only $A$ and $B$.
answered Jan 11 at 22:34
DashermanDasherman
1,075817
$endgroup$
add a comment |
$begingroup$
Your work is correct.
Perhaps an easier way to do it: choose $A=C_{n+1}$ and $B=bigvee_{i=1}^n C_i$. The induction step then follows from the case with only $A$ and $B$.
answered Jan 11 at 22:34
DashermanDasherman
1,075817
$endgroup$
Your work is correct.
Perhaps an easier way to do it: choose $A=C_{n+1}$ and $B=bigvee_{i=1}^n C_i$. The induction step then follows from the case with only $A$ and $B$.
answered Jan 11 at 22:34
DashermanDasherman
1,075817
answered Jan 11 at 22:34
DashermanDasherman
1,075817
answered Jan 11 at 22:34
DashermanDasherman
1,075817
answered Jan 11 at 22:34
DashermanDasherman
1,075817
1,075817
add a comment |
add a comment |
$begingroup$
Hint: Let $A_1,A_2 ldots, A_r$ be a collection of events. Then
$$A_1 lor A_2 lor A_3 lor ldots lor A_r = A'_1 lor A'_2 ldots lor A'_r,$$,
where
$$A_1 doteq A_1; A'_l doteq A_l setminus (A_1 lor A_2 ldots lor A_{l-1}) text{for each } l=1,2,ldots,r $$
So
$$mathbb{P}[A_1 lor A_2 lor A_3 lor ldots lor A_r] = mathbb{P}[A'_1 lor A'_2 ldots lor A'_r].$$
And as the $A'_l$s are mutually disjoint,
$$ mathbb{P}[A'_1 lor A'_2 ldots lor A'_r] = sum_{l=1}^r mathbb{P}[A'_l]$$
Yielding
$$mathbb{P}[A_1 lor A_2 lor A_3 lor ldots lor A_r] = mathbb{P}[A'_1 lor A'_2 ldots lor A'_r] = sum_{l=1}^r mathbb{P}[A'_l].$$
But note that $mathbb{P}[A_l] ge mathbb{P}[A'_l]$ for each $l$, because $A'_l subseteq A_l$. Can you finish from here.
answered Jan 11 at 22:40
MikeMike
3,951412
$endgroup$
add a comment |
$begingroup$
Hint: Let $A_1,A_2 ldots, A_r$ be a collection of events. Then
$$A_1 lor A_2 lor A_3 lor ldots lor A_r = A'_1 lor A'_2 ldots lor A'_r,$$,
where
$$A_1 doteq A_1; A'_l doteq A_l setminus (A_1 lor A_2 ldots lor A_{l-1}) text{for each } l=1,2,ldots,r $$
So
$$mathbb{P}[A_1 lor A_2 lor A_3 lor ldots lor A_r] = mathbb{P}[A'_1 lor A'_2 ldots lor A'_r].$$
And as the $A'_l$s are mutually disjoint,
$$ mathbb{P}[A'_1 lor A'_2 ldots lor A'_r] = sum_{l=1}^r mathbb{P}[A'_l]$$
Yielding
$$mathbb{P}[A_1 lor A_2 lor A_3 lor ldots lor A_r] = mathbb{P}[A'_1 lor A'_2 ldots lor A'_r] = sum_{l=1}^r mathbb{P}[A'_l].$$
But note that $mathbb{P}[A_l] ge mathbb{P}[A'_l]$ for each $l$, because $A'_l subseteq A_l$. Can you finish from here.
answered Jan 11 at 22:40
MikeMike
3,951412
$endgroup$
add a comment |
$begingroup$
Hint: Let $A_1,A_2 ldots, A_r$ be a collection of events. Then
$$A_1 lor A_2 lor A_3 lor ldots lor A_r = A'_1 lor A'_2 ldots lor A'_r,$$,
where
$$A_1 doteq A_1; A'_l doteq A_l setminus (A_1 lor A_2 ldots lor A_{l-1}) text{for each } l=1,2,ldots,r $$
So
$$mathbb{P}[A_1 lor A_2 lor A_3 lor ldots lor A_r] = mathbb{P}[A'_1 lor A'_2 ldots lor A'_r].$$
And as the $A'_l$s are mutually disjoint,
$$ mathbb{P}[A'_1 lor A'_2 ldots lor A'_r] = sum_{l=1}^r mathbb{P}[A'_l]$$
Yielding
$$mathbb{P}[A_1 lor A_2 lor A_3 lor ldots lor A_r] = mathbb{P}[A'_1 lor A'_2 ldots lor A'_r] = sum_{l=1}^r mathbb{P}[A'_l].$$
But note that $mathbb{P}[A_l] ge mathbb{P}[A'_l]$ for each $l$, because $A'_l subseteq A_l$. Can you finish from here.
answered Jan 11 at 22:40
MikeMike
3,951412
$endgroup$
Hint: Let $A_1,A_2 ldots, A_r$ be a collection of events. Then
$$A_1 lor A_2 lor A_3 lor ldots lor A_r = A'_1 lor A'_2 ldots lor A'_r,$$,
where
$$A_1 doteq A_1; A'_l doteq A_l setminus (A_1 lor A_2 ldots lor A_{l-1}) text{for each } l=1,2,ldots,r $$
So
$$mathbb{P}[A_1 lor A_2 lor A_3 lor ldots lor A_r] = mathbb{P}[A'_1 lor A'_2 ldots lor A'_r].$$
And as the $A'_l$s are mutually disjoint,
$$ mathbb{P}[A'_1 lor A'_2 ldots lor A'_r] = sum_{l=1}^r mathbb{P}[A'_l]$$
Yielding
$$mathbb{P}[A_1 lor A_2 lor A_3 lor ldots lor A_r] = mathbb{P}[A'_1 lor A'_2 ldots lor A'_r] = sum_{l=1}^r mathbb{P}[A'_l].$$
But note that $mathbb{P}[A_l] ge mathbb{P}[A'_l]$ for each $l$, because $A'_l subseteq A_l$. Can you finish from here.
answered Jan 11 at 22:40
MikeMike
3,951412
answered Jan 11 at 22:40
MikeMike
3,951412
answered Jan 11 at 22:40
MikeMike
3,951412
answered Jan 11 at 22:40
MikeMike
3,951412
3,951412
add a comment |
add a comment |
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$begingroup$
Does anyone have comments on the downvote? I'm not really trying to break rules here, would appreciate if someone could let me know what I'm doing wrong
$endgroup$
– Hinton
Jan 11 at 22:40
$begingroup$
Is the notation $vee$ common for unions $cup$ in certain regions of the world? This is a rarity, to me.
$endgroup$
– LoveTooNap29
Jan 12 at 1:27
$begingroup$
I have way more education in logic than probability so I used the notation that seemed most familiar to me
$endgroup$
– Hinton
Jan 12 at 22:08