How good is this bound?
$begingroup$
Suppose we have two events $A, B subset Omega$ in a probability space $(Omega, mathcal{F}, mathbf{P})$ and we want to know the probability of $A cap B$.
If $A$ and $B$ are independent, of course $mathbf{P}(A cap B) = mathbf{P}(A)mathbf{P}(B)$.
However, what can we say without independence? It seems that with Cauchy-Schwarz we have that
$$
mathbf{P}(A cap B) = mathbf{E} 1_A 1_B leq sqrt{mathbf{P}(A) mathbf{P}(B)}.
$$
Is this a good bound? Is it useful? That's a soft question, but I guess I am wondering if it is really useful at all.
probability-theory soft-question
asked Jan 3 at 4:45
Drew BradyDrew Brady
712315
$endgroup$
add a comment |
$begingroup$
Suppose we have two events $A, B subset Omega$ in a probability space $(Omega, mathcal{F}, mathbf{P})$ and we want to know the probability of $A cap B$.
If $A$ and $B$ are independent, of course $mathbf{P}(A cap B) = mathbf{P}(A)mathbf{P}(B)$.
However, what can we say without independence? It seems that with Cauchy-Schwarz we have that
$$
mathbf{P}(A cap B) = mathbf{E} 1_A 1_B leq sqrt{mathbf{P}(A) mathbf{P}(B)}.
$$
Is this a good bound? Is it useful? That's a soft question, but I guess I am wondering if it is really useful at all.
probability-theory soft-question
asked Jan 3 at 4:45
Drew BradyDrew Brady
712315
$endgroup$
add a comment |
$begingroup$
Suppose we have two events $A, B subset Omega$ in a probability space $(Omega, mathcal{F}, mathbf{P})$ and we want to know the probability of $A cap B$.
If $A$ and $B$ are independent, of course $mathbf{P}(A cap B) = mathbf{P}(A)mathbf{P}(B)$.
However, what can we say without independence? It seems that with Cauchy-Schwarz we have that
$$
mathbf{P}(A cap B) = mathbf{E} 1_A 1_B leq sqrt{mathbf{P}(A) mathbf{P}(B)}.
$$
Is this a good bound? Is it useful? That's a soft question, but I guess I am wondering if it is really useful at all.
probability-theory soft-question
asked Jan 3 at 4:45
Drew BradyDrew Brady
712315
$endgroup$
Suppose we have two events $A, B subset Omega$ in a probability space $(Omega, mathcal{F}, mathbf{P})$ and we want to know the probability of $A cap B$.
If $A$ and $B$ are independent, of course $mathbf{P}(A cap B) = mathbf{P}(A)mathbf{P}(B)$.
However, what can we say without independence? It seems that with Cauchy-Schwarz we have that
$$
mathbf{P}(A cap B) = mathbf{E} 1_A 1_B leq sqrt{mathbf{P}(A) mathbf{P}(B)}.
$$
Is this a good bound? Is it useful? That's a soft question, but I guess I am wondering if it is really useful at all.
probability-theory soft-question
probability-theory soft-question
asked Jan 3 at 4:45
Drew BradyDrew Brady
712315
asked Jan 3 at 4:45
Drew BradyDrew Brady
712315
edited Jan 3 at 4:52
Drew Brady
asked Jan 3 at 4:45
Drew BradyDrew Brady
712315
asked Jan 3 at 4:45
Drew BradyDrew Brady
712315
asked Jan 3 at 4:45
Drew BradyDrew Brady
712315
712315
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
This bound is not really useful because $min({bf P}(A), {bf P}(B))$ is better. Note that $$sqrt{{bf P}(A) {bf P}(B)} ge min({bf P}(A),{bf P}(B)) ge {bf P}(A cap B)$$
answered Jan 3 at 4:57
Robert IsraelRobert Israel
321k23210462
$endgroup$
$begingroup$
Fair. Recording another thought here, in case others find it useful: you might ask, okay what if $1 < p < infty$, $q$ solves $1/p + 1/q = 1$? Holder's inequality also gives you the bound $mathbf{P}(A cap B) leq mathbf{P}(A)^{1/p} mathbf{P}(B)^{1/q}$, but a similar argument above holds; this is also not useful.
$endgroup$
– Drew Brady
Jan 3 at 6:47
add a comment |
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1 Answer
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oldest
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1 Answer
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active
oldest
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$begingroup$
This bound is not really useful because $min({bf P}(A), {bf P}(B))$ is better. Note that $$sqrt{{bf P}(A) {bf P}(B)} ge min({bf P}(A),{bf P}(B)) ge {bf P}(A cap B)$$
answered Jan 3 at 4:57
Robert IsraelRobert Israel
321k23210462
$endgroup$
$begingroup$
Fair. Recording another thought here, in case others find it useful: you might ask, okay what if $1 < p < infty$, $q$ solves $1/p + 1/q = 1$? Holder's inequality also gives you the bound $mathbf{P}(A cap B) leq mathbf{P}(A)^{1/p} mathbf{P}(B)^{1/q}$, but a similar argument above holds; this is also not useful.
$endgroup$
– Drew Brady
Jan 3 at 6:47
add a comment |
$begingroup$
This bound is not really useful because $min({bf P}(A), {bf P}(B))$ is better. Note that $$sqrt{{bf P}(A) {bf P}(B)} ge min({bf P}(A),{bf P}(B)) ge {bf P}(A cap B)$$
answered Jan 3 at 4:57
Robert IsraelRobert Israel
321k23210462
$endgroup$
$begingroup$
Fair. Recording another thought here, in case others find it useful: you might ask, okay what if $1 < p < infty$, $q$ solves $1/p + 1/q = 1$? Holder's inequality also gives you the bound $mathbf{P}(A cap B) leq mathbf{P}(A)^{1/p} mathbf{P}(B)^{1/q}$, but a similar argument above holds; this is also not useful.
$endgroup$
– Drew Brady
Jan 3 at 6:47
add a comment |
$begingroup$
This bound is not really useful because $min({bf P}(A), {bf P}(B))$ is better. Note that $$sqrt{{bf P}(A) {bf P}(B)} ge min({bf P}(A),{bf P}(B)) ge {bf P}(A cap B)$$
answered Jan 3 at 4:57
Robert IsraelRobert Israel
321k23210462
$endgroup$
This bound is not really useful because $min({bf P}(A), {bf P}(B))$ is better. Note that $$sqrt{{bf P}(A) {bf P}(B)} ge min({bf P}(A),{bf P}(B)) ge {bf P}(A cap B)$$
answered Jan 3 at 4:57
Robert IsraelRobert Israel
321k23210462
answered Jan 3 at 4:57
Robert IsraelRobert Israel
321k23210462
answered Jan 3 at 4:57
Robert IsraelRobert Israel
321k23210462
answered Jan 3 at 4:57
Robert IsraelRobert Israel
321k23210462
321k23210462
$begingroup$
Fair. Recording another thought here, in case others find it useful: you might ask, okay what if $1 < p < infty$, $q$ solves $1/p + 1/q = 1$? Holder's inequality also gives you the bound $mathbf{P}(A cap B) leq mathbf{P}(A)^{1/p} mathbf{P}(B)^{1/q}$, but a similar argument above holds; this is also not useful.
$endgroup$
– Drew Brady
Jan 3 at 6:47
add a comment |
$begingroup$
Fair. Recording another thought here, in case others find it useful: you might ask, okay what if $1 < p < infty$, $q$ solves $1/p + 1/q = 1$? Holder's inequality also gives you the bound $mathbf{P}(A cap B) leq mathbf{P}(A)^{1/p} mathbf{P}(B)^{1/q}$, but a similar argument above holds; this is also not useful.
$endgroup$
– Drew Brady
Jan 3 at 6:47
$begingroup$
Fair. Recording another thought here, in case others find it useful: you might ask, okay what if $1 < p < infty$, $q$ solves $1/p + 1/q = 1$? Holder's inequality also gives you the bound $mathbf{P}(A cap B) leq mathbf{P}(A)^{1/p} mathbf{P}(B)^{1/q}$, but a similar argument above holds; this is also not useful.
$endgroup$
– Drew Brady
Jan 3 at 6:47
$begingroup$
Fair. Recording another thought here, in case others find it useful: you might ask, okay what if $1 < p < infty$, $q$ solves $1/p + 1/q = 1$? Holder's inequality also gives you the bound $mathbf{P}(A cap B) leq mathbf{P}(A)^{1/p} mathbf{P}(B)^{1/q}$, but a similar argument above holds; this is also not useful.
$endgroup$
– Drew Brady
Jan 3 at 6:47
add a comment |
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