Conditional probability integration












1












$begingroup$


Suppose I would like to compute the probability
$hat{p} :=P(max{XY-1,0} > 0, Y in [3,5])$ where $X sim U(-1,1)$ and $Y sim U(-5,5)$ are independent.
To do this first, I considered
$$
hat{p} = P(max{XY-1,0} > 0 | Y in [3, 5])P(Yin [3,5])
$$
Note that $P(Yin [3,5]) = 2/10 = 1/5$.
However, the resulting probability is a function of $Y$, not a number;
$$
P(max{XY-1,0} > 0 | Y in [3, 5]) = P(X > 1/Y | Y in [3, 5]) = (1-1/Y)cdot mathcal{I}_{[3,5]}(Y)
$$
where $mathcal{I}_A(x)$ is an indicator function.
It seems that I can obtain the answer by integrating $Y$, i.e,
$$
hat{p} = P(Yin [3,5]) E_Y[P(max{XY-1,0} > 0 | Y in [3, 5])]
$$
which gives
$$
hat{p} = frac{1}{5}int_{3}^5 (1-1/Y)frac{dy}{10}.
$$



However this seems strange to me as the conditional probability should satisfy
$$
P(Acap B) = P(A|B)P(B)
$$
without integration.
I know that the conditional expectation satisfies
$$
E_Y[E[XY|Y=y]] = E_Y[yE[X|Y=y]] = E[XY].
$$
and it seems that it is highly related to the answer for my question.
However, I am sure how exactly these are related.



Any comments/suggestions/answers will be very appreciated. Thank you.






probability conditional-expectation conditional-probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your expression for $hat p$ is precisely $mathbb P(Acap B) = mathbb P(Amid B)mathbb P(B)$ where the event $A$ is $XY gt 1$ i.e. $max{XY-1,0} > 0$, and the event $B$ is $Yin [3,5]$. You will not be able to avoid integrating the $frac1y$ term when finding $mathbb P(Amid B)$
    $endgroup$
    – Henry
    Dec 14 '18 at 23:05


















1












$begingroup$


Suppose I would like to compute the probability
$hat{p} :=P(max{XY-1,0} > 0, Y in [3,5])$ where $X sim U(-1,1)$ and $Y sim U(-5,5)$ are independent.
To do this first, I considered
$$
hat{p} = P(max{XY-1,0} > 0 | Y in [3, 5])P(Yin [3,5])
$$
Note that $P(Yin [3,5]) = 2/10 = 1/5$.
However, the resulting probability is a function of $Y$, not a number;
$$
P(max{XY-1,0} > 0 | Y in [3, 5]) = P(X > 1/Y | Y in [3, 5]) = (1-1/Y)cdot mathcal{I}_{[3,5]}(Y)
$$
where $mathcal{I}_A(x)$ is an indicator function.
It seems that I can obtain the answer by integrating $Y$, i.e,
$$
hat{p} = P(Yin [3,5]) E_Y[P(max{XY-1,0} > 0 | Y in [3, 5])]
$$
which gives
$$
hat{p} = frac{1}{5}int_{3}^5 (1-1/Y)frac{dy}{10}.
$$



However this seems strange to me as the conditional probability should satisfy
$$
P(Acap B) = P(A|B)P(B)
$$
without integration.
I know that the conditional expectation satisfies
$$
E_Y[E[XY|Y=y]] = E_Y[yE[X|Y=y]] = E[XY].
$$
and it seems that it is highly related to the answer for my question.
However, I am sure how exactly these are related.



Any comments/suggestions/answers will be very appreciated. Thank you.






probability conditional-expectation conditional-probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your expression for $hat p$ is precisely $mathbb P(Acap B) = mathbb P(Amid B)mathbb P(B)$ where the event $A$ is $XY gt 1$ i.e. $max{XY-1,0} > 0$, and the event $B$ is $Yin [3,5]$. You will not be able to avoid integrating the $frac1y$ term when finding $mathbb P(Amid B)$
    $endgroup$
    – Henry
    Dec 14 '18 at 23:05
















1












1








1





$begingroup$


Suppose I would like to compute the probability
$hat{p} :=P(max{XY-1,0} > 0, Y in [3,5])$ where $X sim U(-1,1)$ and $Y sim U(-5,5)$ are independent.
To do this first, I considered
$$
hat{p} = P(max{XY-1,0} > 0 | Y in [3, 5])P(Yin [3,5])
$$
Note that $P(Yin [3,5]) = 2/10 = 1/5$.
However, the resulting probability is a function of $Y$, not a number;
$$
P(max{XY-1,0} > 0 | Y in [3, 5]) = P(X > 1/Y | Y in [3, 5]) = (1-1/Y)cdot mathcal{I}_{[3,5]}(Y)
$$
where $mathcal{I}_A(x)$ is an indicator function.
It seems that I can obtain the answer by integrating $Y$, i.e,
$$
hat{p} = P(Yin [3,5]) E_Y[P(max{XY-1,0} > 0 | Y in [3, 5])]
$$
which gives
$$
hat{p} = frac{1}{5}int_{3}^5 (1-1/Y)frac{dy}{10}.
$$



However this seems strange to me as the conditional probability should satisfy
$$
P(Acap B) = P(A|B)P(B)
$$
without integration.
I know that the conditional expectation satisfies
$$
E_Y[E[XY|Y=y]] = E_Y[yE[X|Y=y]] = E[XY].
$$
and it seems that it is highly related to the answer for my question.
However, I am sure how exactly these are related.



Any comments/suggestions/answers will be very appreciated. Thank you.






probability conditional-expectation conditional-probability







share|cite|improve this question











$endgroup$




Suppose I would like to compute the probability
$hat{p} :=P(max{XY-1,0} > 0, Y in [3,5])$ where $X sim U(-1,1)$ and $Y sim U(-5,5)$ are independent.
To do this first, I considered
$$
hat{p} = P(max{XY-1,0} > 0 | Y in [3, 5])P(Yin [3,5])
$$
Note that $P(Yin [3,5]) = 2/10 = 1/5$.
However, the resulting probability is a function of $Y$, not a number;
$$
P(max{XY-1,0} > 0 | Y in [3, 5]) = P(X > 1/Y | Y in [3, 5]) = (1-1/Y)cdot mathcal{I}_{[3,5]}(Y)
$$
where $mathcal{I}_A(x)$ is an indicator function.
It seems that I can obtain the answer by integrating $Y$, i.e,
$$
hat{p} = P(Yin [3,5]) E_Y[P(max{XY-1,0} > 0 | Y in [3, 5])]
$$
which gives
$$
hat{p} = frac{1}{5}int_{3}^5 (1-1/Y)frac{dy}{10}.
$$



However this seems strange to me as the conditional probability should satisfy
$$
P(Acap B) = P(A|B)P(B)
$$
without integration.
I know that the conditional expectation satisfies
$$
E_Y[E[XY|Y=y]] = E_Y[yE[X|Y=y]] = E[XY].
$$
and it seems that it is highly related to the answer for my question.
However, I am sure how exactly these are related.



Any comments/suggestions/answers will be very appreciated. Thank you.






probability conditional-expectation conditional-probability




probability conditional-expectation conditional-probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 23:32







induction601

















asked Dec 14 '18 at 21:48









induction601induction601

1,295314




1,295314












  • $begingroup$
    Your expression for $hat p$ is precisely $mathbb P(Acap B) = mathbb P(Amid B)mathbb P(B)$ where the event $A$ is $XY gt 1$ i.e. $max{XY-1,0} > 0$, and the event $B$ is $Yin [3,5]$. You will not be able to avoid integrating the $frac1y$ term when finding $mathbb P(Amid B)$
    $endgroup$
    – Henry
    Dec 14 '18 at 23:05




















  • $begingroup$
    Your expression for $hat p$ is precisely $mathbb P(Acap B) = mathbb P(Amid B)mathbb P(B)$ where the event $A$ is $XY gt 1$ i.e. $max{XY-1,0} > 0$, and the event $B$ is $Yin [3,5]$. You will not be able to avoid integrating the $frac1y$ term when finding $mathbb P(Amid B)$
    $endgroup$
    – Henry
    Dec 14 '18 at 23:05


















$begingroup$
Your expression for $hat p$ is precisely $mathbb P(Acap B) = mathbb P(Amid B)mathbb P(B)$ where the event $A$ is $XY gt 1$ i.e. $max{XY-1,0} > 0$, and the event $B$ is $Yin [3,5]$. You will not be able to avoid integrating the $frac1y$ term when finding $mathbb P(Amid B)$
$endgroup$
– Henry
Dec 14 '18 at 23:05






$begingroup$
Your expression for $hat p$ is precisely $mathbb P(Acap B) = mathbb P(Amid B)mathbb P(B)$ where the event $A$ is $XY gt 1$ i.e. $max{XY-1,0} > 0$, and the event $B$ is $Yin [3,5]$. You will not be able to avoid integrating the $frac1y$ term when finding $mathbb P(Amid B)$
$endgroup$
– Henry
Dec 14 '18 at 23:05












1 Answer
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oldest

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$begingroup$

It seems all these conditioning/deconditioning/reconditioning steps only serve to muddy the waters for you.



To compute the probability of an event depending on $(X,Y)$, all that is needed is the joint distribution of $(X,Y)$, and we have it, hence we can simply write $$p=P(XY>1,3leqslant Yleqslant 5)=iint_{mathbb R^2}mathbf 1_{xy>1,3leqslant yleqslant5},f_{X,Y}(x,y),dxdy$$ Using $$f_{X,Y}(x,y)=frac1{20},mathbf 1_{-1<x<1,-5<y<5}$$ this yields $$20p=int_3^5int_{1/y}^1dxdy=int_3^5left(1-frac1yright)dy=left.(y-log y)right|_{y=3}^{y=5}$$ Thus, finally, $$p=frac1{20}left(2-logfrac53right)$$






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    $begingroup$

    It seems all these conditioning/deconditioning/reconditioning steps only serve to muddy the waters for you.



    To compute the probability of an event depending on $(X,Y)$, all that is needed is the joint distribution of $(X,Y)$, and we have it, hence we can simply write $$p=P(XY>1,3leqslant Yleqslant 5)=iint_{mathbb R^2}mathbf 1_{xy>1,3leqslant yleqslant5},f_{X,Y}(x,y),dxdy$$ Using $$f_{X,Y}(x,y)=frac1{20},mathbf 1_{-1<x<1,-5<y<5}$$ this yields $$20p=int_3^5int_{1/y}^1dxdy=int_3^5left(1-frac1yright)dy=left.(y-log y)right|_{y=3}^{y=5}$$ Thus, finally, $$p=frac1{20}left(2-logfrac53right)$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      It seems all these conditioning/deconditioning/reconditioning steps only serve to muddy the waters for you.



      To compute the probability of an event depending on $(X,Y)$, all that is needed is the joint distribution of $(X,Y)$, and we have it, hence we can simply write $$p=P(XY>1,3leqslant Yleqslant 5)=iint_{mathbb R^2}mathbf 1_{xy>1,3leqslant yleqslant5},f_{X,Y}(x,y),dxdy$$ Using $$f_{X,Y}(x,y)=frac1{20},mathbf 1_{-1<x<1,-5<y<5}$$ this yields $$20p=int_3^5int_{1/y}^1dxdy=int_3^5left(1-frac1yright)dy=left.(y-log y)right|_{y=3}^{y=5}$$ Thus, finally, $$p=frac1{20}left(2-logfrac53right)$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        It seems all these conditioning/deconditioning/reconditioning steps only serve to muddy the waters for you.



        To compute the probability of an event depending on $(X,Y)$, all that is needed is the joint distribution of $(X,Y)$, and we have it, hence we can simply write $$p=P(XY>1,3leqslant Yleqslant 5)=iint_{mathbb R^2}mathbf 1_{xy>1,3leqslant yleqslant5},f_{X,Y}(x,y),dxdy$$ Using $$f_{X,Y}(x,y)=frac1{20},mathbf 1_{-1<x<1,-5<y<5}$$ this yields $$20p=int_3^5int_{1/y}^1dxdy=int_3^5left(1-frac1yright)dy=left.(y-log y)right|_{y=3}^{y=5}$$ Thus, finally, $$p=frac1{20}left(2-logfrac53right)$$






        share|cite|improve this answer









        $endgroup$



        It seems all these conditioning/deconditioning/reconditioning steps only serve to muddy the waters for you.



        To compute the probability of an event depending on $(X,Y)$, all that is needed is the joint distribution of $(X,Y)$, and we have it, hence we can simply write $$p=P(XY>1,3leqslant Yleqslant 5)=iint_{mathbb R^2}mathbf 1_{xy>1,3leqslant yleqslant5},f_{X,Y}(x,y),dxdy$$ Using $$f_{X,Y}(x,y)=frac1{20},mathbf 1_{-1<x<1,-5<y<5}$$ this yields $$20p=int_3^5int_{1/y}^1dxdy=int_3^5left(1-frac1yright)dy=left.(y-log y)right|_{y=3}^{y=5}$$ Thus, finally, $$p=frac1{20}left(2-logfrac53right)$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 16 '18 at 23:59









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