Cumulative distribution function properties in terms of complements
What is the expression $$int_{frac{1}{3}}^1int_{frac{1}{3}}^1f(x,y)dxdy$$in terms of CDF $F$ when $f$ is its pdf and bivariate distribution as support $[0,1]times[0,1]$?
I thought it was $1-F(1/3,1/3)$ but someone said it is wrong.
probability probability-theory probability-distributions
asked Nov 26 at 22:32
Jeremy Kwak
92
add a comment |
What is the expression $$int_{frac{1}{3}}^1int_{frac{1}{3}}^1f(x,y)dxdy$$in terms of CDF $F$ when $f$ is its pdf and bivariate distribution as support $[0,1]times[0,1]$?
I thought it was $1-F(1/3,1/3)$ but someone said it is wrong.
probability probability-theory probability-distributions
asked Nov 26 at 22:32
Jeremy Kwak
92
add a comment |
What is the expression $$int_{frac{1}{3}}^1int_{frac{1}{3}}^1f(x,y)dxdy$$in terms of CDF $F$ when $f$ is its pdf and bivariate distribution as support $[0,1]times[0,1]$?
I thought it was $1-F(1/3,1/3)$ but someone said it is wrong.
probability probability-theory probability-distributions
asked Nov 26 at 22:32
Jeremy Kwak
92
What is the expression $$int_{frac{1}{3}}^1int_{frac{1}{3}}^1f(x,y)dxdy$$in terms of CDF $F$ when $f$ is its pdf and bivariate distribution as support $[0,1]times[0,1]$?
I thought it was $1-F(1/3,1/3)$ but someone said it is wrong.
probability probability-theory probability-distributions
probability probability-theory probability-distributions
asked Nov 26 at 22:32
Jeremy Kwak
92
asked Nov 26 at 22:32
Jeremy Kwak
92
asked Nov 26 at 22:32
Jeremy Kwak
92
asked Nov 26 at 22:32
Jeremy Kwak
92
asked Nov 26 at 22:32
Jeremy Kwak
92
92
add a comment |
add a comment |
1 Answer
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$1−F(1,1/3)-F(1/3,1)+F(1/3,1/3)$
Just look at it geometrically: the square $[0,1]times[0,1]$ is split into 4 rectangles by a horizontal line $y=1/3$ and a vertical line $x=1/3$.
$F(x,y)$ is the integral over the rectangle with vertices $(0,0)$ and $(x,y)$.
To get the integral you need, you have to take the whole square ($F(1,1)=1$) and remove two thin rectangles ($F(1,1/3)$ and $F(1/3,1)$), but now you removed the small square ($F(1/3,1/3)$) twice, so you need to re-add it.
answered Nov 26 at 22:53
sds
3,5281129
add a comment |
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1 Answer
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$1−F(1,1/3)-F(1/3,1)+F(1/3,1/3)$
Just look at it geometrically: the square $[0,1]times[0,1]$ is split into 4 rectangles by a horizontal line $y=1/3$ and a vertical line $x=1/3$.
$F(x,y)$ is the integral over the rectangle with vertices $(0,0)$ and $(x,y)$.
To get the integral you need, you have to take the whole square ($F(1,1)=1$) and remove two thin rectangles ($F(1,1/3)$ and $F(1/3,1)$), but now you removed the small square ($F(1/3,1/3)$) twice, so you need to re-add it.
answered Nov 26 at 22:53
sds
3,5281129
add a comment |
$1−F(1,1/3)-F(1/3,1)+F(1/3,1/3)$
Just look at it geometrically: the square $[0,1]times[0,1]$ is split into 4 rectangles by a horizontal line $y=1/3$ and a vertical line $x=1/3$.
$F(x,y)$ is the integral over the rectangle with vertices $(0,0)$ and $(x,y)$.
To get the integral you need, you have to take the whole square ($F(1,1)=1$) and remove two thin rectangles ($F(1,1/3)$ and $F(1/3,1)$), but now you removed the small square ($F(1/3,1/3)$) twice, so you need to re-add it.
answered Nov 26 at 22:53
sds
3,5281129
add a comment |
$1−F(1,1/3)-F(1/3,1)+F(1/3,1/3)$
Just look at it geometrically: the square $[0,1]times[0,1]$ is split into 4 rectangles by a horizontal line $y=1/3$ and a vertical line $x=1/3$.
$F(x,y)$ is the integral over the rectangle with vertices $(0,0)$ and $(x,y)$.
To get the integral you need, you have to take the whole square ($F(1,1)=1$) and remove two thin rectangles ($F(1,1/3)$ and $F(1/3,1)$), but now you removed the small square ($F(1/3,1/3)$) twice, so you need to re-add it.
answered Nov 26 at 22:53
sds
3,5281129
$1−F(1,1/3)-F(1/3,1)+F(1/3,1/3)$
Just look at it geometrically: the square $[0,1]times[0,1]$ is split into 4 rectangles by a horizontal line $y=1/3$ and a vertical line $x=1/3$.
$F(x,y)$ is the integral over the rectangle with vertices $(0,0)$ and $(x,y)$.
To get the integral you need, you have to take the whole square ($F(1,1)=1$) and remove two thin rectangles ($F(1,1/3)$ and $F(1/3,1)$), but now you removed the small square ($F(1/3,1/3)$) twice, so you need to re-add it.
answered Nov 26 at 22:53
sds
3,5281129
answered Nov 26 at 22:53
sds
3,5281129
answered Nov 26 at 22:53
sds
3,5281129
answered Nov 26 at 22:53
sds
3,5281129
3,5281129
add a comment |
add a comment |
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