Proving inequality of probability
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I need to prove the simple inequality that $P(A<B) geq P(A< c)times P(B > c)$ where both $A$ and $B$ are random variables and $c$ is a constant.
Can someone help ?
probability probability-theory
edited Dec 6 '18 at 3:40
Andrews
3901317
asked Dec 6 '18 at 2:02
vortex_sparrowvortex_sparrow
11
$endgroup$
add a comment |
$begingroup$
I need to prove the simple inequality that $P(A<B) geq P(A< c)times P(B > c)$ where both $A$ and $B$ are random variables and $c$ is a constant.
Can someone help ?
probability probability-theory
edited Dec 6 '18 at 3:40
Andrews
3901317
asked Dec 6 '18 at 2:02
vortex_sparrowvortex_sparrow
11
$endgroup$
add a comment |
$begingroup$
I need to prove the simple inequality that $P(A<B) geq P(A< c)times P(B > c)$ where both $A$ and $B$ are random variables and $c$ is a constant.
Can someone help ?
probability probability-theory
edited Dec 6 '18 at 3:40
Andrews
3901317
asked Dec 6 '18 at 2:02
vortex_sparrowvortex_sparrow
11
$endgroup$
I need to prove the simple inequality that $P(A<B) geq P(A< c)times P(B > c)$ where both $A$ and $B$ are random variables and $c$ is a constant.
Can someone help ?
probability probability-theory
probability probability-theory
edited Dec 6 '18 at 3:40
Andrews
3901317
asked Dec 6 '18 at 2:02
vortex_sparrowvortex_sparrow
11
edited Dec 6 '18 at 3:40
Andrews
3901317
asked Dec 6 '18 at 2:02
vortex_sparrowvortex_sparrow
11
edited Dec 6 '18 at 3:40
Andrews
3901317
edited Dec 6 '18 at 3:40
Andrews
3901317
edited Dec 6 '18 at 3:40
Andrews
3901317
3901317
asked Dec 6 '18 at 2:02
vortex_sparrowvortex_sparrow
11
asked Dec 6 '18 at 2:02
vortex_sparrowvortex_sparrow
11
asked Dec 6 '18 at 2:02
vortex_sparrowvortex_sparrow
11
11
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Not true. Consider $Omega=[0,1]$, $A(omega)=omega$, $B(omega)=omega-frac12$ and $c=frac14$.
answered Dec 6 '18 at 2:15
Saucy O'PathSaucy O'Path
5,9291627
$endgroup$
$begingroup$
You mean the inequality does not hold ?
$endgroup$
– vortex_sparrow
Dec 6 '18 at 4:40
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Yes, I do.${}{}$
$endgroup$
– Saucy O'Path
Dec 6 '18 at 16:45
add a comment |
$begingroup$
Are you assuming $A, B$ to be independent? If not, if $A=B$ then $P(A<B)=0$ but $P(A<c),P(B>c)>0$ for for $c$. So, I will assume they are independent, discrete (for simplicity).
$P(A<B) = sum_i P(B=i)P(A leq i) $ then $P(A<B) geq sum_{igeq c} P(B=i)P(A leq i) $ for any choice of $c$.
Since $P(A leq i) geq P(A leq c)$ for all $igeq c$,
$P(A<B) geq sum_{igeq c} P(B=i)P(A leq i) geq sum_{igeq c} P(B=i)P(Aleq c)
= P(Bgeq c)P(Aleq c)$.
This is true for all $c$ if $A, B$ are independent. You can produce counterexamples if they are dependent.
answered Dec 6 '18 at 2:16
zerozero
708
$endgroup$
$begingroup$
I don't understand why P(A<= i) >= P(A<= c) , since we have no information about c
$endgroup$
– vortex_sparrow
Dec 6 '18 at 4:40
$begingroup$
Because we are assuming $i geq c$ and ${ A leq c} subset {A leq i }$
$endgroup$
– zero
Dec 7 '18 at 22:26
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Not true. Consider $Omega=[0,1]$, $A(omega)=omega$, $B(omega)=omega-frac12$ and $c=frac14$.
answered Dec 6 '18 at 2:15
Saucy O'PathSaucy O'Path
5,9291627
$endgroup$
$begingroup$
You mean the inequality does not hold ?
$endgroup$
– vortex_sparrow
Dec 6 '18 at 4:40
$begingroup$
Yes, I do.${}{}$
$endgroup$
– Saucy O'Path
Dec 6 '18 at 16:45
add a comment |
$begingroup$
Not true. Consider $Omega=[0,1]$, $A(omega)=omega$, $B(omega)=omega-frac12$ and $c=frac14$.
answered Dec 6 '18 at 2:15
Saucy O'PathSaucy O'Path
5,9291627
$endgroup$
$begingroup$
You mean the inequality does not hold ?
$endgroup$
– vortex_sparrow
Dec 6 '18 at 4:40
$begingroup$
Yes, I do.${}{}$
$endgroup$
– Saucy O'Path
Dec 6 '18 at 16:45
add a comment |
$begingroup$
Not true. Consider $Omega=[0,1]$, $A(omega)=omega$, $B(omega)=omega-frac12$ and $c=frac14$.
answered Dec 6 '18 at 2:15
Saucy O'PathSaucy O'Path
5,9291627
$endgroup$
Not true. Consider $Omega=[0,1]$, $A(omega)=omega$, $B(omega)=omega-frac12$ and $c=frac14$.
answered Dec 6 '18 at 2:15
Saucy O'PathSaucy O'Path
5,9291627
answered Dec 6 '18 at 2:15
Saucy O'PathSaucy O'Path
5,9291627
answered Dec 6 '18 at 2:15
Saucy O'PathSaucy O'Path
5,9291627
answered Dec 6 '18 at 2:15
Saucy O'PathSaucy O'Path
5,9291627
5,9291627
$begingroup$
You mean the inequality does not hold ?
$endgroup$
– vortex_sparrow
Dec 6 '18 at 4:40
$begingroup$
Yes, I do.${}{}$
$endgroup$
– Saucy O'Path
Dec 6 '18 at 16:45
add a comment |
$begingroup$
You mean the inequality does not hold ?
$endgroup$
– vortex_sparrow
Dec 6 '18 at 4:40
$begingroup$
Yes, I do.${}{}$
$endgroup$
– Saucy O'Path
Dec 6 '18 at 16:45
$begingroup$
You mean the inequality does not hold ?
$endgroup$
– vortex_sparrow
Dec 6 '18 at 4:40
$begingroup$
You mean the inequality does not hold ?
$endgroup$
– vortex_sparrow
Dec 6 '18 at 4:40
$begingroup$
Yes, I do.${}{}$
$endgroup$
– Saucy O'Path
Dec 6 '18 at 16:45
$begingroup$
Yes, I do.${}{}$
$endgroup$
– Saucy O'Path
Dec 6 '18 at 16:45
add a comment |
$begingroup$
Are you assuming $A, B$ to be independent? If not, if $A=B$ then $P(A<B)=0$ but $P(A<c),P(B>c)>0$ for for $c$. So, I will assume they are independent, discrete (for simplicity).
$P(A<B) = sum_i P(B=i)P(A leq i) $ then $P(A<B) geq sum_{igeq c} P(B=i)P(A leq i) $ for any choice of $c$.
Since $P(A leq i) geq P(A leq c)$ for all $igeq c$,
$P(A<B) geq sum_{igeq c} P(B=i)P(A leq i) geq sum_{igeq c} P(B=i)P(Aleq c)
= P(Bgeq c)P(Aleq c)$.
This is true for all $c$ if $A, B$ are independent. You can produce counterexamples if they are dependent.
answered Dec 6 '18 at 2:16
zerozero
708
$endgroup$
$begingroup$
I don't understand why P(A<= i) >= P(A<= c) , since we have no information about c
$endgroup$
– vortex_sparrow
Dec 6 '18 at 4:40
$begingroup$
Because we are assuming $i geq c$ and ${ A leq c} subset {A leq i }$
$endgroup$
– zero
Dec 7 '18 at 22:26
add a comment |
$begingroup$
Are you assuming $A, B$ to be independent? If not, if $A=B$ then $P(A<B)=0$ but $P(A<c),P(B>c)>0$ for for $c$. So, I will assume they are independent, discrete (for simplicity).
$P(A<B) = sum_i P(B=i)P(A leq i) $ then $P(A<B) geq sum_{igeq c} P(B=i)P(A leq i) $ for any choice of $c$.
Since $P(A leq i) geq P(A leq c)$ for all $igeq c$,
$P(A<B) geq sum_{igeq c} P(B=i)P(A leq i) geq sum_{igeq c} P(B=i)P(Aleq c)
= P(Bgeq c)P(Aleq c)$.
This is true for all $c$ if $A, B$ are independent. You can produce counterexamples if they are dependent.
answered Dec 6 '18 at 2:16
zerozero
708
$endgroup$
$begingroup$
I don't understand why P(A<= i) >= P(A<= c) , since we have no information about c
$endgroup$
– vortex_sparrow
Dec 6 '18 at 4:40
$begingroup$
Because we are assuming $i geq c$ and ${ A leq c} subset {A leq i }$
$endgroup$
– zero
Dec 7 '18 at 22:26
add a comment |
$begingroup$
Are you assuming $A, B$ to be independent? If not, if $A=B$ then $P(A<B)=0$ but $P(A<c),P(B>c)>0$ for for $c$. So, I will assume they are independent, discrete (for simplicity).
$P(A<B) = sum_i P(B=i)P(A leq i) $ then $P(A<B) geq sum_{igeq c} P(B=i)P(A leq i) $ for any choice of $c$.
Since $P(A leq i) geq P(A leq c)$ for all $igeq c$,
$P(A<B) geq sum_{igeq c} P(B=i)P(A leq i) geq sum_{igeq c} P(B=i)P(Aleq c)
= P(Bgeq c)P(Aleq c)$.
This is true for all $c$ if $A, B$ are independent. You can produce counterexamples if they are dependent.
answered Dec 6 '18 at 2:16
zerozero
708
$endgroup$
Are you assuming $A, B$ to be independent? If not, if $A=B$ then $P(A<B)=0$ but $P(A<c),P(B>c)>0$ for for $c$. So, I will assume they are independent, discrete (for simplicity).
$P(A<B) = sum_i P(B=i)P(A leq i) $ then $P(A<B) geq sum_{igeq c} P(B=i)P(A leq i) $ for any choice of $c$.
Since $P(A leq i) geq P(A leq c)$ for all $igeq c$,
$P(A<B) geq sum_{igeq c} P(B=i)P(A leq i) geq sum_{igeq c} P(B=i)P(Aleq c)
= P(Bgeq c)P(Aleq c)$.
This is true for all $c$ if $A, B$ are independent. You can produce counterexamples if they are dependent.
answered Dec 6 '18 at 2:16
zerozero
708
answered Dec 6 '18 at 2:16
zerozero
708
answered Dec 6 '18 at 2:16
zerozero
708
answered Dec 6 '18 at 2:16
zerozero
708
708
$begingroup$
I don't understand why P(A<= i) >= P(A<= c) , since we have no information about c
$endgroup$
– vortex_sparrow
Dec 6 '18 at 4:40
$begingroup$
Because we are assuming $i geq c$ and ${ A leq c} subset {A leq i }$
$endgroup$
– zero
Dec 7 '18 at 22:26
add a comment |
$begingroup$
I don't understand why P(A<= i) >= P(A<= c) , since we have no information about c
$endgroup$
– vortex_sparrow
Dec 6 '18 at 4:40
$begingroup$
Because we are assuming $i geq c$ and ${ A leq c} subset {A leq i }$
$endgroup$
– zero
Dec 7 '18 at 22:26
$begingroup$
I don't understand why P(A<= i) >= P(A<= c) , since we have no information about c
$endgroup$
– vortex_sparrow
Dec 6 '18 at 4:40
$begingroup$
I don't understand why P(A<= i) >= P(A<= c) , since we have no information about c
$endgroup$
– vortex_sparrow
Dec 6 '18 at 4:40
$begingroup$
Because we are assuming $i geq c$ and ${ A leq c} subset {A leq i }$
$endgroup$
– zero
Dec 7 '18 at 22:26
$begingroup$
Because we are assuming $i geq c$ and ${ A leq c} subset {A leq i }$
$endgroup$
– zero
Dec 7 '18 at 22:26
add a comment |
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