Mixing
Compare two random variables and their probability with respect to same number
$begingroup$
Problem
Given random variables $X$, $Y$, if I know $Xgeq Y$, what can I say about
$Pr[Xgeq a]$ and $Pr[Ygeq a]$ where $ageq 0$.
What I Have Done
$Pr[Xgeq a]$ and $Pr[Ygeq a]$ could be evaluated with integral but since I do not know the pdf of $X$ and $Y$,this approach could not give me
any result.
probability random-variables
asked Jan 26 at 1:07
Mr.RobotMr.Robot
384110
$endgroup$
add a comment |
$begingroup$
Problem
Given random variables $X$, $Y$, if I know $Xgeq Y$, what can I say about
$Pr[Xgeq a]$ and $Pr[Ygeq a]$ where $ageq 0$.
What I Have Done
$Pr[Xgeq a]$ and $Pr[Ygeq a]$ could be evaluated with integral but since I do not know the pdf of $X$ and $Y$,this approach could not give me
any result.
probability random-variables
asked Jan 26 at 1:07
Mr.RobotMr.Robot
384110
$endgroup$
$begingroup$
I think you can establish an inequality between the two probabilities. Note that the event $Ygeq a$ implies $X geq a$.
$endgroup$
– user440191
Jan 26 at 1:13
add a comment |
$begingroup$
Problem
Given random variables $X$, $Y$, if I know $Xgeq Y$, what can I say about
$Pr[Xgeq a]$ and $Pr[Ygeq a]$ where $ageq 0$.
What I Have Done
$Pr[Xgeq a]$ and $Pr[Ygeq a]$ could be evaluated with integral but since I do not know the pdf of $X$ and $Y$,this approach could not give me
any result.
probability random-variables
asked Jan 26 at 1:07
Mr.RobotMr.Robot
384110
$endgroup$
Problem
Given random variables $X$, $Y$, if I know $Xgeq Y$, what can I say about
$Pr[Xgeq a]$ and $Pr[Ygeq a]$ where $ageq 0$.
What I Have Done
$Pr[Xgeq a]$ and $Pr[Ygeq a]$ could be evaluated with integral but since I do not know the pdf of $X$ and $Y$,this approach could not give me
any result.
probability random-variables
probability random-variables
asked Jan 26 at 1:07
Mr.RobotMr.Robot
384110
asked Jan 26 at 1:07
Mr.RobotMr.Robot
384110
asked Jan 26 at 1:07
Mr.RobotMr.Robot
384110
asked Jan 26 at 1:07
Mr.RobotMr.Robot
384110
asked Jan 26 at 1:07
Mr.RobotMr.Robot
384110
384110
$begingroup$
I think you can establish an inequality between the two probabilities. Note that the event $Ygeq a$ implies $X geq a$.
$endgroup$
– user440191
Jan 26 at 1:13
add a comment |
$begingroup$
I think you can establish an inequality between the two probabilities. Note that the event $Ygeq a$ implies $X geq a$.
$endgroup$
– user440191
Jan 26 at 1:13
$begingroup$
I think you can establish an inequality between the two probabilities. Note that the event $Ygeq a$ implies $X geq a$.
$endgroup$
– user440191
Jan 26 at 1:13
$begingroup$
I think you can establish an inequality between the two probabilities. Note that the event $Ygeq a$ implies $X geq a$.
$endgroup$
– user440191
Jan 26 at 1:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that
$$
Ygeq aimplies Xgeq Ygeq a
$$
whence
$$
(Yge a)subseteq(Xgeq a)
$$
and so by monotonicicity of measure (namely, $Asubseteq Bimplies P(A)leq P(B)$) we have that
$$
P(Ygeq a)leq P(Xge a)
$$
answered Jan 26 at 1:16
Foobaz JohnFoobaz John
22.7k41452
$endgroup$
$begingroup$
Thank you for your help! But does it generally hold that if $ARightarrow B$, then $Asubseteq B$ and therefore $Pr[A]leq Pr[B]$.
$endgroup$
– Mr.Robot
Jan 26 at 20:06
$begingroup$
Yes. If $Asubseteq B$, then since $B=Acup (Bsetminus A)$ (which is a disjoint union), it follows that $P(B)=P(A)+P(Bsetminus A)geq P(A)$.
$endgroup$
– Foobaz John
Jan 26 at 21:46
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that
$$
Ygeq aimplies Xgeq Ygeq a
$$
whence
$$
(Yge a)subseteq(Xgeq a)
$$
and so by monotonicicity of measure (namely, $Asubseteq Bimplies P(A)leq P(B)$) we have that
$$
P(Ygeq a)leq P(Xge a)
$$
answered Jan 26 at 1:16
Foobaz JohnFoobaz John
22.7k41452
$endgroup$
$begingroup$
Thank you for your help! But does it generally hold that if $ARightarrow B$, then $Asubseteq B$ and therefore $Pr[A]leq Pr[B]$.
$endgroup$
– Mr.Robot
Jan 26 at 20:06
$begingroup$
Yes. If $Asubseteq B$, then since $B=Acup (Bsetminus A)$ (which is a disjoint union), it follows that $P(B)=P(A)+P(Bsetminus A)geq P(A)$.
$endgroup$
– Foobaz John
Jan 26 at 21:46
add a comment |
$begingroup$
Note that
$$
Ygeq aimplies Xgeq Ygeq a
$$
whence
$$
(Yge a)subseteq(Xgeq a)
$$
and so by monotonicicity of measure (namely, $Asubseteq Bimplies P(A)leq P(B)$) we have that
$$
P(Ygeq a)leq P(Xge a)
$$
answered Jan 26 at 1:16
Foobaz JohnFoobaz John
22.7k41452
$endgroup$
$begingroup$
Thank you for your help! But does it generally hold that if $ARightarrow B$, then $Asubseteq B$ and therefore $Pr[A]leq Pr[B]$.
$endgroup$
– Mr.Robot
Jan 26 at 20:06
$begingroup$
Yes. If $Asubseteq B$, then since $B=Acup (Bsetminus A)$ (which is a disjoint union), it follows that $P(B)=P(A)+P(Bsetminus A)geq P(A)$.
$endgroup$
– Foobaz John
Jan 26 at 21:46
add a comment |
$begingroup$
Note that
$$
Ygeq aimplies Xgeq Ygeq a
$$
whence
$$
(Yge a)subseteq(Xgeq a)
$$
and so by monotonicicity of measure (namely, $Asubseteq Bimplies P(A)leq P(B)$) we have that
$$
P(Ygeq a)leq P(Xge a)
$$
answered Jan 26 at 1:16
Foobaz JohnFoobaz John
22.7k41452
$endgroup$
Note that
$$
Ygeq aimplies Xgeq Ygeq a
$$
whence
$$
(Yge a)subseteq(Xgeq a)
$$
and so by monotonicicity of measure (namely, $Asubseteq Bimplies P(A)leq P(B)$) we have that
$$
P(Ygeq a)leq P(Xge a)
$$
answered Jan 26 at 1:16
Foobaz JohnFoobaz John
22.7k41452
answered Jan 26 at 1:16
Foobaz JohnFoobaz John
22.7k41452
answered Jan 26 at 1:16
Foobaz JohnFoobaz John
22.7k41452
answered Jan 26 at 1:16
Foobaz JohnFoobaz John
22.7k41452
22.7k41452
$begingroup$
Thank you for your help! But does it generally hold that if $ARightarrow B$, then $Asubseteq B$ and therefore $Pr[A]leq Pr[B]$.
$endgroup$
– Mr.Robot
Jan 26 at 20:06
$begingroup$
Yes. If $Asubseteq B$, then since $B=Acup (Bsetminus A)$ (which is a disjoint union), it follows that $P(B)=P(A)+P(Bsetminus A)geq P(A)$.
$endgroup$
– Foobaz John
Jan 26 at 21:46
add a comment |
$begingroup$
Thank you for your help! But does it generally hold that if $ARightarrow B$, then $Asubseteq B$ and therefore $Pr[A]leq Pr[B]$.
$endgroup$
– Mr.Robot
Jan 26 at 20:06
$begingroup$
Yes. If $Asubseteq B$, then since $B=Acup (Bsetminus A)$ (which is a disjoint union), it follows that $P(B)=P(A)+P(Bsetminus A)geq P(A)$.
$endgroup$
– Foobaz John
Jan 26 at 21:46
$begingroup$
Thank you for your help! But does it generally hold that if $ARightarrow B$, then $Asubseteq B$ and therefore $Pr[A]leq Pr[B]$.
$endgroup$
– Mr.Robot
Jan 26 at 20:06
$begingroup$
Thank you for your help! But does it generally hold that if $ARightarrow B$, then $Asubseteq B$ and therefore $Pr[A]leq Pr[B]$.
$endgroup$
– Mr.Robot
Jan 26 at 20:06
$begingroup$
Yes. If $Asubseteq B$, then since $B=Acup (Bsetminus A)$ (which is a disjoint union), it follows that $P(B)=P(A)+P(Bsetminus A)geq P(A)$.
$endgroup$
– Foobaz John
Jan 26 at 21:46
$begingroup$
Yes. If $Asubseteq B$, then since $B=Acup (Bsetminus A)$ (which is a disjoint union), it follows that $P(B)=P(A)+P(Bsetminus A)geq P(A)$.
$endgroup$
– Foobaz John
Jan 26 at 21:46
add a comment |
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$begingroup$
I think you can establish an inequality between the two probabilities. Note that the event $Ygeq a$ implies $X geq a$.
$endgroup$
– user440191
Jan 26 at 1:13