Show that $max{textbf{P}((Acup B)^{c}),textbf{P}(Acap B),textbf{P}(Atriangle B)}geqfrac{4}{9}$












1












$begingroup$


Let $A$ and $B$ be independent events. Show that
begin{align*}
max{textbf{P}((Acup B)^{c}),textbf{P}(Acap B),textbf{P}(Atriangle B)}geqfrac{4}{9}
end{align*}



MY ATTEMPT



Since $textbf{P}(Acap B) = textbf{P}(A)textbf{P}(B)$, we have



begin{align*}
textbf{P}((Acup B)^{c}) & = 1 - textbf{P}(Acup B) = 1 - textbf{P}(A) - textbf{P}(B) + textbf{P}(Acap B)\
& = 1 - textbf{P}(A) - textbf{P}(B) + textbf{P}(A)textbf{P}(B)\
& = (1 - textbf{P}(A)) - textbf{P}(B)(1 - textbf{P}(A))\
& = (1 - textbf{P}(A))(1-textbf{P}(B)) = textbf{P}(A^{c})textbf{P}(B^{c})
end{align*}



Analagously, we get
begin{align*}
textbf{P}(Atriangle B) & = textbf{P}(A) + textbf{P}(B) - 2textbf{P}(Acap B) = textbf{P}(A) + textbf{P}(B) - 2textbf{P}(A)textbf{P}(B)\\
& = textbf{P}(A)(1 - textbf{P}(B)) + textbf{P}(B)(1-textbf{P}(A)) = textbf{P}(A)textbf{P}(B^{c}) + textbf{P}(A^{c})textbf{P}(B)
end{align*}



However, I do not know how to proceed from here. Am I on the right track? Could someone complete the proof? Any help is appreciated. Thanks in advance.






probability probability-theory independence conditional-probability







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Why not write $p=P(A)$ and $q=P(B)$? It would make manipulating your formulae a lot easier.
    $endgroup$
    – Lord Shark the Unknown
    Jan 12 at 4:42










  • $begingroup$
    Try to fix, for example, $P(Acap B) < frac{4}{9}$, and write the sum of the probabilities of the other two events you are considering. This sum has to be greater than some quantity. Can you than deduce, from that, your thesis?
    $endgroup$
    – Matteo
    Jan 12 at 11:54


















1












$begingroup$


Let $A$ and $B$ be independent events. Show that
begin{align*}
max{textbf{P}((Acup B)^{c}),textbf{P}(Acap B),textbf{P}(Atriangle B)}geqfrac{4}{9}
end{align*}



MY ATTEMPT



Since $textbf{P}(Acap B) = textbf{P}(A)textbf{P}(B)$, we have



begin{align*}
textbf{P}((Acup B)^{c}) & = 1 - textbf{P}(Acup B) = 1 - textbf{P}(A) - textbf{P}(B) + textbf{P}(Acap B)\
& = 1 - textbf{P}(A) - textbf{P}(B) + textbf{P}(A)textbf{P}(B)\
& = (1 - textbf{P}(A)) - textbf{P}(B)(1 - textbf{P}(A))\
& = (1 - textbf{P}(A))(1-textbf{P}(B)) = textbf{P}(A^{c})textbf{P}(B^{c})
end{align*}



Analagously, we get
begin{align*}
textbf{P}(Atriangle B) & = textbf{P}(A) + textbf{P}(B) - 2textbf{P}(Acap B) = textbf{P}(A) + textbf{P}(B) - 2textbf{P}(A)textbf{P}(B)\\
& = textbf{P}(A)(1 - textbf{P}(B)) + textbf{P}(B)(1-textbf{P}(A)) = textbf{P}(A)textbf{P}(B^{c}) + textbf{P}(A^{c})textbf{P}(B)
end{align*}



However, I do not know how to proceed from here. Am I on the right track? Could someone complete the proof? Any help is appreciated. Thanks in advance.






probability probability-theory independence conditional-probability







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Why not write $p=P(A)$ and $q=P(B)$? It would make manipulating your formulae a lot easier.
    $endgroup$
    – Lord Shark the Unknown
    Jan 12 at 4:42










  • $begingroup$
    Try to fix, for example, $P(Acap B) < frac{4}{9}$, and write the sum of the probabilities of the other two events you are considering. This sum has to be greater than some quantity. Can you than deduce, from that, your thesis?
    $endgroup$
    – Matteo
    Jan 12 at 11:54
















1












1








1


0



$begingroup$


Let $A$ and $B$ be independent events. Show that
begin{align*}
max{textbf{P}((Acup B)^{c}),textbf{P}(Acap B),textbf{P}(Atriangle B)}geqfrac{4}{9}
end{align*}



MY ATTEMPT



Since $textbf{P}(Acap B) = textbf{P}(A)textbf{P}(B)$, we have



begin{align*}
textbf{P}((Acup B)^{c}) & = 1 - textbf{P}(Acup B) = 1 - textbf{P}(A) - textbf{P}(B) + textbf{P}(Acap B)\
& = 1 - textbf{P}(A) - textbf{P}(B) + textbf{P}(A)textbf{P}(B)\
& = (1 - textbf{P}(A)) - textbf{P}(B)(1 - textbf{P}(A))\
& = (1 - textbf{P}(A))(1-textbf{P}(B)) = textbf{P}(A^{c})textbf{P}(B^{c})
end{align*}



Analagously, we get
begin{align*}
textbf{P}(Atriangle B) & = textbf{P}(A) + textbf{P}(B) - 2textbf{P}(Acap B) = textbf{P}(A) + textbf{P}(B) - 2textbf{P}(A)textbf{P}(B)\\
& = textbf{P}(A)(1 - textbf{P}(B)) + textbf{P}(B)(1-textbf{P}(A)) = textbf{P}(A)textbf{P}(B^{c}) + textbf{P}(A^{c})textbf{P}(B)
end{align*}



However, I do not know how to proceed from here. Am I on the right track? Could someone complete the proof? Any help is appreciated. Thanks in advance.






probability probability-theory independence conditional-probability







share|cite|improve this question









$endgroup$




Let $A$ and $B$ be independent events. Show that
begin{align*}
max{textbf{P}((Acup B)^{c}),textbf{P}(Acap B),textbf{P}(Atriangle B)}geqfrac{4}{9}
end{align*}



MY ATTEMPT



Since $textbf{P}(Acap B) = textbf{P}(A)textbf{P}(B)$, we have



begin{align*}
textbf{P}((Acup B)^{c}) & = 1 - textbf{P}(Acup B) = 1 - textbf{P}(A) - textbf{P}(B) + textbf{P}(Acap B)\
& = 1 - textbf{P}(A) - textbf{P}(B) + textbf{P}(A)textbf{P}(B)\
& = (1 - textbf{P}(A)) - textbf{P}(B)(1 - textbf{P}(A))\
& = (1 - textbf{P}(A))(1-textbf{P}(B)) = textbf{P}(A^{c})textbf{P}(B^{c})
end{align*}



Analagously, we get
begin{align*}
textbf{P}(Atriangle B) & = textbf{P}(A) + textbf{P}(B) - 2textbf{P}(Acap B) = textbf{P}(A) + textbf{P}(B) - 2textbf{P}(A)textbf{P}(B)\\
& = textbf{P}(A)(1 - textbf{P}(B)) + textbf{P}(B)(1-textbf{P}(A)) = textbf{P}(A)textbf{P}(B^{c}) + textbf{P}(A^{c})textbf{P}(B)
end{align*}



However, I do not know how to proceed from here. Am I on the right track? Could someone complete the proof? Any help is appreciated. Thanks in advance.






probability probability-theory independence conditional-probability




probability probability-theory independence conditional-probability






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 12 at 4:35









user1337user1337

48210




48210








  • 2




    $begingroup$
    Why not write $p=P(A)$ and $q=P(B)$? It would make manipulating your formulae a lot easier.
    $endgroup$
    – Lord Shark the Unknown
    Jan 12 at 4:42










  • $begingroup$
    Try to fix, for example, $P(Acap B) < frac{4}{9}$, and write the sum of the probabilities of the other two events you are considering. This sum has to be greater than some quantity. Can you than deduce, from that, your thesis?
    $endgroup$
    – Matteo
    Jan 12 at 11:54
















  • 2




    $begingroup$
    Why not write $p=P(A)$ and $q=P(B)$? It would make manipulating your formulae a lot easier.
    $endgroup$
    – Lord Shark the Unknown
    Jan 12 at 4:42










  • $begingroup$
    Try to fix, for example, $P(Acap B) < frac{4}{9}$, and write the sum of the probabilities of the other two events you are considering. This sum has to be greater than some quantity. Can you than deduce, from that, your thesis?
    $endgroup$
    – Matteo
    Jan 12 at 11:54










2




2




$begingroup$
Why not write $p=P(A)$ and $q=P(B)$? It would make manipulating your formulae a lot easier.
$endgroup$
– Lord Shark the Unknown
Jan 12 at 4:42




$begingroup$
Why not write $p=P(A)$ and $q=P(B)$? It would make manipulating your formulae a lot easier.
$endgroup$
– Lord Shark the Unknown
Jan 12 at 4:42












$begingroup$
Try to fix, for example, $P(Acap B) < frac{4}{9}$, and write the sum of the probabilities of the other two events you are considering. This sum has to be greater than some quantity. Can you than deduce, from that, your thesis?
$endgroup$
– Matteo
Jan 12 at 11:54






$begingroup$
Try to fix, for example, $P(Acap B) < frac{4}{9}$, and write the sum of the probabilities of the other two events you are considering. This sum has to be greater than some quantity. Can you than deduce, from that, your thesis?
$endgroup$
– Matteo
Jan 12 at 11:54












1 Answer
1






active

oldest

votes


















1












$begingroup$

As in the comment by Lord Shark The Unknown you can first set
$$ P(A) = p$$
and
$$P(B) = q.$$
Then the thesis becomes the following.




Show that the inequalities
begin{eqnarray}
pq &<& frac{4}{9}\
(1-p)(1-q) &<& frac{4}{9}\
p(1-q) + q(1-p) &<& frac{4}{9}
end{eqnarray}
cannot hold simultaneously.




Performing the following change of variables
begin{equation}
begin{cases}
p = frac{1}{2}(Q+P+1)\
q = frac{1}{2}(Q-P+1)
end{cases}
end{equation}
leads to the corresponding inequalities in terms of $P$ and $Q$, i.e.
begin{eqnarray}
(Q+1)^2 -P^2 &<& frac{16}{9}tag{1}label{uno}\
(Q-1)^2 -P^2 &<& frac{16}{9}tag{2}label{due}\
Q^2-P^2>frac{1}{9}tag{3}label{tre}
end{eqnarray}
which are areas delimited by hyperbolae (in red, green, and black respectively in the Figure). The red dashed area corresponds to the set of points $(P,Q)$ satisfying conditions eqref{uno} and eqref{due}. The black dashed areas mark te set of points that satisfy condition eqref{tre}. Therefore the three inequalities cannot be all true. And the thesis follows.



enter image description here






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In the first place, thanks for the response. Is there any intuition about the given substitutions, though?
    $endgroup$
    – user1337
    Jan 13 at 18:59












  • $begingroup$
    @user1337, I noticed that expressing the conditions with respect to $(p,q)$ lead to curves that had a nicer representation once rotated by $frac{pi}{4}$ and with origin shifted by $left(frac{1}{2}, frac{1}{2}right)$, resulting in the substitution I proposed you. Maybe there are nicer way to approach the problem, of course.
    $endgroup$
    – Matteo
    Jan 13 at 19:18










  • $begingroup$
    @user1337, as a matter of fact, a rotation of $frac{pi}{4}$ should be enough to "see" the result. That is, ignoring the scaling factor, $P=p-q$ and $Q=p+q$.
    $endgroup$
    – Matteo
    Jan 13 at 19:45










  • $begingroup$
    Matteo, thank you very much for the explanation.
    $endgroup$
    – user1337
    Jan 13 at 19:46










  • $begingroup$
    @user1337, how about a vote up, then ;)
    $endgroup$
    – Matteo
    Jan 14 at 8:55












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

As in the comment by Lord Shark The Unknown you can first set
$$ P(A) = p$$
and
$$P(B) = q.$$
Then the thesis becomes the following.




Show that the inequalities
begin{eqnarray}
pq &<& frac{4}{9}\
(1-p)(1-q) &<& frac{4}{9}\
p(1-q) + q(1-p) &<& frac{4}{9}
end{eqnarray}
cannot hold simultaneously.




Performing the following change of variables
begin{equation}
begin{cases}
p = frac{1}{2}(Q+P+1)\
q = frac{1}{2}(Q-P+1)
end{cases}
end{equation}
leads to the corresponding inequalities in terms of $P$ and $Q$, i.e.
begin{eqnarray}
(Q+1)^2 -P^2 &<& frac{16}{9}tag{1}label{uno}\
(Q-1)^2 -P^2 &<& frac{16}{9}tag{2}label{due}\
Q^2-P^2>frac{1}{9}tag{3}label{tre}
end{eqnarray}
which are areas delimited by hyperbolae (in red, green, and black respectively in the Figure). The red dashed area corresponds to the set of points $(P,Q)$ satisfying conditions eqref{uno} and eqref{due}. The black dashed areas mark te set of points that satisfy condition eqref{tre}. Therefore the three inequalities cannot be all true. And the thesis follows.



enter image description here






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In the first place, thanks for the response. Is there any intuition about the given substitutions, though?
    $endgroup$
    – user1337
    Jan 13 at 18:59












  • $begingroup$
    @user1337, I noticed that expressing the conditions with respect to $(p,q)$ lead to curves that had a nicer representation once rotated by $frac{pi}{4}$ and with origin shifted by $left(frac{1}{2}, frac{1}{2}right)$, resulting in the substitution I proposed you. Maybe there are nicer way to approach the problem, of course.
    $endgroup$
    – Matteo
    Jan 13 at 19:18










  • $begingroup$
    @user1337, as a matter of fact, a rotation of $frac{pi}{4}$ should be enough to "see" the result. That is, ignoring the scaling factor, $P=p-q$ and $Q=p+q$.
    $endgroup$
    – Matteo
    Jan 13 at 19:45










  • $begingroup$
    Matteo, thank you very much for the explanation.
    $endgroup$
    – user1337
    Jan 13 at 19:46










  • $begingroup$
    @user1337, how about a vote up, then ;)
    $endgroup$
    – Matteo
    Jan 14 at 8:55
















1












$begingroup$

As in the comment by Lord Shark The Unknown you can first set
$$ P(A) = p$$
and
$$P(B) = q.$$
Then the thesis becomes the following.




Show that the inequalities
begin{eqnarray}
pq &<& frac{4}{9}\
(1-p)(1-q) &<& frac{4}{9}\
p(1-q) + q(1-p) &<& frac{4}{9}
end{eqnarray}
cannot hold simultaneously.




Performing the following change of variables
begin{equation}
begin{cases}
p = frac{1}{2}(Q+P+1)\
q = frac{1}{2}(Q-P+1)
end{cases}
end{equation}
leads to the corresponding inequalities in terms of $P$ and $Q$, i.e.
begin{eqnarray}
(Q+1)^2 -P^2 &<& frac{16}{9}tag{1}label{uno}\
(Q-1)^2 -P^2 &<& frac{16}{9}tag{2}label{due}\
Q^2-P^2>frac{1}{9}tag{3}label{tre}
end{eqnarray}
which are areas delimited by hyperbolae (in red, green, and black respectively in the Figure). The red dashed area corresponds to the set of points $(P,Q)$ satisfying conditions eqref{uno} and eqref{due}. The black dashed areas mark te set of points that satisfy condition eqref{tre}. Therefore the three inequalities cannot be all true. And the thesis follows.



enter image description here






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In the first place, thanks for the response. Is there any intuition about the given substitutions, though?
    $endgroup$
    – user1337
    Jan 13 at 18:59












  • $begingroup$
    @user1337, I noticed that expressing the conditions with respect to $(p,q)$ lead to curves that had a nicer representation once rotated by $frac{pi}{4}$ and with origin shifted by $left(frac{1}{2}, frac{1}{2}right)$, resulting in the substitution I proposed you. Maybe there are nicer way to approach the problem, of course.
    $endgroup$
    – Matteo
    Jan 13 at 19:18










  • $begingroup$
    @user1337, as a matter of fact, a rotation of $frac{pi}{4}$ should be enough to "see" the result. That is, ignoring the scaling factor, $P=p-q$ and $Q=p+q$.
    $endgroup$
    – Matteo
    Jan 13 at 19:45










  • $begingroup$
    Matteo, thank you very much for the explanation.
    $endgroup$
    – user1337
    Jan 13 at 19:46










  • $begingroup$
    @user1337, how about a vote up, then ;)
    $endgroup$
    – Matteo
    Jan 14 at 8:55














1












1








1





$begingroup$

As in the comment by Lord Shark The Unknown you can first set
$$ P(A) = p$$
and
$$P(B) = q.$$
Then the thesis becomes the following.




Show that the inequalities
begin{eqnarray}
pq &<& frac{4}{9}\
(1-p)(1-q) &<& frac{4}{9}\
p(1-q) + q(1-p) &<& frac{4}{9}
end{eqnarray}
cannot hold simultaneously.




Performing the following change of variables
begin{equation}
begin{cases}
p = frac{1}{2}(Q+P+1)\
q = frac{1}{2}(Q-P+1)
end{cases}
end{equation}
leads to the corresponding inequalities in terms of $P$ and $Q$, i.e.
begin{eqnarray}
(Q+1)^2 -P^2 &<& frac{16}{9}tag{1}label{uno}\
(Q-1)^2 -P^2 &<& frac{16}{9}tag{2}label{due}\
Q^2-P^2>frac{1}{9}tag{3}label{tre}
end{eqnarray}
which are areas delimited by hyperbolae (in red, green, and black respectively in the Figure). The red dashed area corresponds to the set of points $(P,Q)$ satisfying conditions eqref{uno} and eqref{due}. The black dashed areas mark te set of points that satisfy condition eqref{tre}. Therefore the three inequalities cannot be all true. And the thesis follows.



enter image description here






share|cite|improve this answer









$endgroup$



As in the comment by Lord Shark The Unknown you can first set
$$ P(A) = p$$
and
$$P(B) = q.$$
Then the thesis becomes the following.




Show that the inequalities
begin{eqnarray}
pq &<& frac{4}{9}\
(1-p)(1-q) &<& frac{4}{9}\
p(1-q) + q(1-p) &<& frac{4}{9}
end{eqnarray}
cannot hold simultaneously.




Performing the following change of variables
begin{equation}
begin{cases}
p = frac{1}{2}(Q+P+1)\
q = frac{1}{2}(Q-P+1)
end{cases}
end{equation}
leads to the corresponding inequalities in terms of $P$ and $Q$, i.e.
begin{eqnarray}
(Q+1)^2 -P^2 &<& frac{16}{9}tag{1}label{uno}\
(Q-1)^2 -P^2 &<& frac{16}{9}tag{2}label{due}\
Q^2-P^2>frac{1}{9}tag{3}label{tre}
end{eqnarray}
which are areas delimited by hyperbolae (in red, green, and black respectively in the Figure). The red dashed area corresponds to the set of points $(P,Q)$ satisfying conditions eqref{uno} and eqref{due}. The black dashed areas mark te set of points that satisfy condition eqref{tre}. Therefore the three inequalities cannot be all true. And the thesis follows.



enter image description here







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 13 at 16:50









MatteoMatteo

1,3121313




1,3121313












  • $begingroup$
    In the first place, thanks for the response. Is there any intuition about the given substitutions, though?
    $endgroup$
    – user1337
    Jan 13 at 18:59












  • $begingroup$
    @user1337, I noticed that expressing the conditions with respect to $(p,q)$ lead to curves that had a nicer representation once rotated by $frac{pi}{4}$ and with origin shifted by $left(frac{1}{2}, frac{1}{2}right)$, resulting in the substitution I proposed you. Maybe there are nicer way to approach the problem, of course.
    $endgroup$
    – Matteo
    Jan 13 at 19:18










  • $begingroup$
    @user1337, as a matter of fact, a rotation of $frac{pi}{4}$ should be enough to "see" the result. That is, ignoring the scaling factor, $P=p-q$ and $Q=p+q$.
    $endgroup$
    – Matteo
    Jan 13 at 19:45










  • $begingroup$
    Matteo, thank you very much for the explanation.
    $endgroup$
    – user1337
    Jan 13 at 19:46










  • $begingroup$
    @user1337, how about a vote up, then ;)
    $endgroup$
    – Matteo
    Jan 14 at 8:55


















  • $begingroup$
    In the first place, thanks for the response. Is there any intuition about the given substitutions, though?
    $endgroup$
    – user1337
    Jan 13 at 18:59












  • $begingroup$
    @user1337, I noticed that expressing the conditions with respect to $(p,q)$ lead to curves that had a nicer representation once rotated by $frac{pi}{4}$ and with origin shifted by $left(frac{1}{2}, frac{1}{2}right)$, resulting in the substitution I proposed you. Maybe there are nicer way to approach the problem, of course.
    $endgroup$
    – Matteo
    Jan 13 at 19:18










  • $begingroup$
    @user1337, as a matter of fact, a rotation of $frac{pi}{4}$ should be enough to "see" the result. That is, ignoring the scaling factor, $P=p-q$ and $Q=p+q$.
    $endgroup$
    – Matteo
    Jan 13 at 19:45










  • $begingroup$
    Matteo, thank you very much for the explanation.
    $endgroup$
    – user1337
    Jan 13 at 19:46










  • $begingroup$
    @user1337, how about a vote up, then ;)
    $endgroup$
    – Matteo
    Jan 14 at 8:55
















$begingroup$
In the first place, thanks for the response. Is there any intuition about the given substitutions, though?
$endgroup$
– user1337
Jan 13 at 18:59






$begingroup$
In the first place, thanks for the response. Is there any intuition about the given substitutions, though?
$endgroup$
– user1337
Jan 13 at 18:59














$begingroup$
@user1337, I noticed that expressing the conditions with respect to $(p,q)$ lead to curves that had a nicer representation once rotated by $frac{pi}{4}$ and with origin shifted by $left(frac{1}{2}, frac{1}{2}right)$, resulting in the substitution I proposed you. Maybe there are nicer way to approach the problem, of course.
$endgroup$
– Matteo
Jan 13 at 19:18




$begingroup$
@user1337, I noticed that expressing the conditions with respect to $(p,q)$ lead to curves that had a nicer representation once rotated by $frac{pi}{4}$ and with origin shifted by $left(frac{1}{2}, frac{1}{2}right)$, resulting in the substitution I proposed you. Maybe there are nicer way to approach the problem, of course.
$endgroup$
– Matteo
Jan 13 at 19:18












$begingroup$
@user1337, as a matter of fact, a rotation of $frac{pi}{4}$ should be enough to "see" the result. That is, ignoring the scaling factor, $P=p-q$ and $Q=p+q$.
$endgroup$
– Matteo
Jan 13 at 19:45




$begingroup$
@user1337, as a matter of fact, a rotation of $frac{pi}{4}$ should be enough to "see" the result. That is, ignoring the scaling factor, $P=p-q$ and $Q=p+q$.
$endgroup$
– Matteo
Jan 13 at 19:45












$begingroup$
Matteo, thank you very much for the explanation.
$endgroup$
– user1337
Jan 13 at 19:46




$begingroup$
Matteo, thank you very much for the explanation.
$endgroup$
– user1337
Jan 13 at 19:46












$begingroup$
@user1337, how about a vote up, then ;)
$endgroup$
– Matteo
Jan 14 at 8:55




$begingroup$
@user1337, how about a vote up, then ;)
$endgroup$
– Matteo
Jan 14 at 8:55


















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