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Distribution of two continuous problems



The 2019 Stack Overflow Developer Survey Results Are Inthe distribution of distance between two random points from $U(0,1)^3$Probability of Normal DistributionMean and Variance both equal to $lambda$ for a Poisson DistributionJoint distribution of $(X,min(X,Y))$ for $X$ and $Y$ i.i.d. uniform on $(0,1)$Question about the bounds of two independent random variables.Discrete uniform distribution over two disjoint intervalsComposition of Continuous Random Variables, finding distributionUniform distribution dependent on another uniform distributionHow good can you approximate a continuous distribution by replacing trailing digits with uniformly random digits?exponential rv's, gamma distribution










0












$begingroup$


question: (x,y) are distributed via $f(x,y)=6xy^2$ in $[0,1] times [0,1]$. What is the $P(x y^3 le 1/2)$?



My solution:



fx = integral(f(x,y))dy = 2x So Fx=x^2



fy = integral(f(x,y))dx = 3y^2 So Fx=y^3



P(xy^3 <= 1/2) =



p(x <= 1/2*y^-3) =



Fx(1/2*y^-3) =



(1/2*y^-3)^2 =



1/4y^-6 (y from 0 to 1) so P = 1/4
but it is wrong.



Could you please take a look and say what's my fault here?






probability statistics functions probability-distributions random-variables







share|cite|improve this question









New contributor




Fatemehhh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$











  • $begingroup$
    Again, please edit and use MathJax to properly format math expressions.
    $endgroup$
    – Lee David Chung Lin
    Apr 8 at 1:16















0












$begingroup$


question: (x,y) are distributed via $f(x,y)=6xy^2$ in $[0,1] times [0,1]$. What is the $P(x y^3 le 1/2)$?



My solution:



fx = integral(f(x,y))dy = 2x So Fx=x^2



fy = integral(f(x,y))dx = 3y^2 So Fx=y^3



P(xy^3 <= 1/2) =



p(x <= 1/2*y^-3) =



Fx(1/2*y^-3) =



(1/2*y^-3)^2 =



1/4y^-6 (y from 0 to 1) so P = 1/4
but it is wrong.



Could you please take a look and say what's my fault here?






probability statistics functions probability-distributions random-variables







share|cite|improve this question









New contributor




Fatemehhh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    Again, please edit and use MathJax to properly format math expressions.
    $endgroup$
    – Lee David Chung Lin
    Apr 8 at 1:16













0












0








0





$begingroup$


question: (x,y) are distributed via $f(x,y)=6xy^2$ in $[0,1] times [0,1]$. What is the $P(x y^3 le 1/2)$?



My solution:



fx = integral(f(x,y))dy = 2x So Fx=x^2



fy = integral(f(x,y))dx = 3y^2 So Fx=y^3



P(xy^3 <= 1/2) =



p(x <= 1/2*y^-3) =



Fx(1/2*y^-3) =



(1/2*y^-3)^2 =



1/4y^-6 (y from 0 to 1) so P = 1/4
but it is wrong.



Could you please take a look and say what's my fault here?






probability statistics functions probability-distributions random-variables







share|cite|improve this question









New contributor




Fatemehhh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




question: (x,y) are distributed via $f(x,y)=6xy^2$ in $[0,1] times [0,1]$. What is the $P(x y^3 le 1/2)$?



My solution:



fx = integral(f(x,y))dy = 2x So Fx=x^2



fy = integral(f(x,y))dx = 3y^2 So Fx=y^3



P(xy^3 <= 1/2) =



p(x <= 1/2*y^-3) =



Fx(1/2*y^-3) =



(1/2*y^-3)^2 =



1/4y^-6 (y from 0 to 1) so P = 1/4
but it is wrong.



Could you please take a look and say what's my fault here?






probability statistics functions probability-distributions random-variables




probability statistics functions probability-distributions random-variables






share|cite|improve this question









New contributor




Fatemehhh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Fatemehhh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited Apr 8 at 1:46









David G. Stork

12.1k41836




12.1k41836






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asked Apr 8 at 0:24









FatemehhhFatemehhh

102




102




New contributor




Fatemehhh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Fatemehhh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Fatemehhh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











  • $begingroup$
    Again, please edit and use MathJax to properly format math expressions.
    $endgroup$
    – Lee David Chung Lin
    Apr 8 at 1:16
















  • $begingroup$
    Again, please edit and use MathJax to properly format math expressions.
    $endgroup$
    – Lee David Chung Lin
    Apr 8 at 1:16















$begingroup$
Again, please edit and use MathJax to properly format math expressions.
$endgroup$
– Lee David Chung Lin
Apr 8 at 1:16




$begingroup$
Again, please edit and use MathJax to properly format math expressions.
$endgroup$
– Lee David Chung Lin
Apr 8 at 1:16










1 Answer
1






active

oldest

votes


















0












$begingroup$

Why would you bother calculating the marginal distributions? The point of the exercise is to consider the region in $(x,y) in [0,1] times [0,1]$ such that $xy^3 le 1/2$, and to integrate the joint density over this region.



To this end, consider the curve $xy^3 = 1/2$. Clearly, when $y = 1$, $x = 1/2$, and when $x = 1$, $y = (1/2)^1/3$. There are no other points on the boundary of the unit square that intersect this curve. Since $(0,0)$ satisfies the inequality, we know the region of interest is bounded by $0 le y le 1$, $0 le x le 1$, $xy^3 le 1/2$, and the vertices $(0,0)$, $(1,0)$, $(1, (1/2)^1/3)$, $(1/2, 1)$, $(0,1)$. Consequently it is easier to compute the complementary probability:



$$Pr[xy^3 > 1/2] = int_x=1/2^1 int_y = (2x)^-1/3^1 6xy^2 , dy , dx.$$






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Thank you so much.
    $endgroup$
    – Fatemehhh
    Apr 8 at 1:46











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









0












$begingroup$

Why would you bother calculating the marginal distributions? The point of the exercise is to consider the region in $(x,y) in [0,1] times [0,1]$ such that $xy^3 le 1/2$, and to integrate the joint density over this region.



To this end, consider the curve $xy^3 = 1/2$. Clearly, when $y = 1$, $x = 1/2$, and when $x = 1$, $y = (1/2)^1/3$. There are no other points on the boundary of the unit square that intersect this curve. Since $(0,0)$ satisfies the inequality, we know the region of interest is bounded by $0 le y le 1$, $0 le x le 1$, $xy^3 le 1/2$, and the vertices $(0,0)$, $(1,0)$, $(1, (1/2)^1/3)$, $(1/2, 1)$, $(0,1)$. Consequently it is easier to compute the complementary probability:



$$Pr[xy^3 > 1/2] = int_x=1/2^1 int_y = (2x)^-1/3^1 6xy^2 , dy , dx.$$






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Thank you so much.
    $endgroup$
    – Fatemehhh
    Apr 8 at 1:46















0












$begingroup$

Why would you bother calculating the marginal distributions? The point of the exercise is to consider the region in $(x,y) in [0,1] times [0,1]$ such that $xy^3 le 1/2$, and to integrate the joint density over this region.



To this end, consider the curve $xy^3 = 1/2$. Clearly, when $y = 1$, $x = 1/2$, and when $x = 1$, $y = (1/2)^1/3$. There are no other points on the boundary of the unit square that intersect this curve. Since $(0,0)$ satisfies the inequality, we know the region of interest is bounded by $0 le y le 1$, $0 le x le 1$, $xy^3 le 1/2$, and the vertices $(0,0)$, $(1,0)$, $(1, (1/2)^1/3)$, $(1/2, 1)$, $(0,1)$. Consequently it is easier to compute the complementary probability:



$$Pr[xy^3 > 1/2] = int_x=1/2^1 int_y = (2x)^-1/3^1 6xy^2 , dy , dx.$$






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Thank you so much.
    $endgroup$
    – Fatemehhh
    Apr 8 at 1:46













0












0








0





$begingroup$

Why would you bother calculating the marginal distributions? The point of the exercise is to consider the region in $(x,y) in [0,1] times [0,1]$ such that $xy^3 le 1/2$, and to integrate the joint density over this region.



To this end, consider the curve $xy^3 = 1/2$. Clearly, when $y = 1$, $x = 1/2$, and when $x = 1$, $y = (1/2)^1/3$. There are no other points on the boundary of the unit square that intersect this curve. Since $(0,0)$ satisfies the inequality, we know the region of interest is bounded by $0 le y le 1$, $0 le x le 1$, $xy^3 le 1/2$, and the vertices $(0,0)$, $(1,0)$, $(1, (1/2)^1/3)$, $(1/2, 1)$, $(0,1)$. Consequently it is easier to compute the complementary probability:



$$Pr[xy^3 > 1/2] = int_x=1/2^1 int_y = (2x)^-1/3^1 6xy^2 , dy , dx.$$






share|cite|improve this answer









$endgroup$



Why would you bother calculating the marginal distributions? The point of the exercise is to consider the region in $(x,y) in [0,1] times [0,1]$ such that $xy^3 le 1/2$, and to integrate the joint density over this region.



To this end, consider the curve $xy^3 = 1/2$. Clearly, when $y = 1$, $x = 1/2$, and when $x = 1$, $y = (1/2)^1/3$. There are no other points on the boundary of the unit square that intersect this curve. Since $(0,0)$ satisfies the inequality, we know the region of interest is bounded by $0 le y le 1$, $0 le x le 1$, $xy^3 le 1/2$, and the vertices $(0,0)$, $(1,0)$, $(1, (1/2)^1/3)$, $(1/2, 1)$, $(0,1)$. Consequently it is easier to compute the complementary probability:



$$Pr[xy^3 > 1/2] = int_x=1/2^1 int_y = (2x)^-1/3^1 6xy^2 , dy , dx.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Apr 8 at 1:19









heropupheropup

65.5k865104




65.5k865104











  • $begingroup$
    Thank you so much.
    $endgroup$
    – Fatemehhh
    Apr 8 at 1:46
















  • $begingroup$
    Thank you so much.
    $endgroup$
    – Fatemehhh
    Apr 8 at 1:46















$begingroup$
Thank you so much.
$endgroup$
– Fatemehhh
Apr 8 at 1:46




$begingroup$
Thank you so much.
$endgroup$
– Fatemehhh
Apr 8 at 1:46










Fatemehhh is a new contributor. Be nice, and check out our Code of Conduct.









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