How to find bounds of integral of a C.D.F?
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Suppose f(x,y) = {1/2 for 0<=x<=2 and 0<= y <= 1}
and 0 otherwise.
Find Pr(X>=Y). So everything to the right of x = y is where to integrate on
I believe you can find the bounds two ways:
1. Integrate on x first then on y. $$int_{0}^{1} int_{y}^{2}1/2 ,dx,dy$$ and consequently the answer is 3/4.
- Integrate on y first then on x. To me the bounds are intuitively supposed to be: $$int_{0}^{2} int_{o}^{x}1/2 ,dy,dx$$
But that is not correct because the answer is 3/4.
Can someone explain how the to derive the bounds for the second method?
probability-distributions bounds-of-integration
asked Nov 16 at 23:45
Sir lethian
163
add a comment |
up vote
0
down vote
favorite
Suppose f(x,y) = {1/2 for 0<=x<=2 and 0<= y <= 1}
and 0 otherwise.
Find Pr(X>=Y). So everything to the right of x = y is where to integrate on
I believe you can find the bounds two ways:
1. Integrate on x first then on y. $$int_{0}^{1} int_{y}^{2}1/2 ,dx,dy$$ and consequently the answer is 3/4.
- Integrate on y first then on x. To me the bounds are intuitively supposed to be: $$int_{0}^{2} int_{o}^{x}1/2 ,dy,dx$$
But that is not correct because the answer is 3/4.
Can someone explain how the to derive the bounds for the second method?
probability-distributions bounds-of-integration
asked Nov 16 at 23:45
Sir lethian
163
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose f(x,y) = {1/2 for 0<=x<=2 and 0<= y <= 1}
and 0 otherwise.
Find Pr(X>=Y). So everything to the right of x = y is where to integrate on
I believe you can find the bounds two ways:
1. Integrate on x first then on y. $$int_{0}^{1} int_{y}^{2}1/2 ,dx,dy$$ and consequently the answer is 3/4.
- Integrate on y first then on x. To me the bounds are intuitively supposed to be: $$int_{0}^{2} int_{o}^{x}1/2 ,dy,dx$$
But that is not correct because the answer is 3/4.
Can someone explain how the to derive the bounds for the second method?
probability-distributions bounds-of-integration
asked Nov 16 at 23:45
Sir lethian
163
Suppose f(x,y) = {1/2 for 0<=x<=2 and 0<= y <= 1}
and 0 otherwise.
Find Pr(X>=Y). So everything to the right of x = y is where to integrate on
I believe you can find the bounds two ways:
1. Integrate on x first then on y. $$int_{0}^{1} int_{y}^{2}1/2 ,dx,dy$$ and consequently the answer is 3/4.
- Integrate on y first then on x. To me the bounds are intuitively supposed to be: $$int_{0}^{2} int_{o}^{x}1/2 ,dy,dx$$
But that is not correct because the answer is 3/4.
Can someone explain how the to derive the bounds for the second method?
probability-distributions bounds-of-integration
probability-distributions bounds-of-integration
asked Nov 16 at 23:45
Sir lethian
163
asked Nov 16 at 23:45
Sir lethian
163
edited Nov 17 at 0:23
asked Nov 16 at 23:45
Sir lethian
163
asked Nov 16 at 23:45
Sir lethian
163
asked Nov 16 at 23:45
Sir lethian
163
163
add a comment |
add a comment |
1 Answer
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You have to split the integral, since from $0 le x le 1$, $y$ goes between $0$ and $x$, but for $1 le x le 2$, $y$ goes between $0$ and $1$.
So the correct integral should be
$$int_0^1{int_0^x{1/2 dy dx}} + int_1^2{int_0^1{1/2 dy dx}}$$
$$= 1/4+1/2 = 3/4$$
answered Nov 17 at 0:54
helper
524212
A much simpler way of integrating is to use the fact that $f(x,y)$ is constant, and just calculate the area where $X ge Y$ and multiply by the value of $f$ (i.e., $1/2$)
– helper
Nov 17 at 0:56
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You have to split the integral, since from $0 le x le 1$, $y$ goes between $0$ and $x$, but for $1 le x le 2$, $y$ goes between $0$ and $1$.
So the correct integral should be
$$int_0^1{int_0^x{1/2 dy dx}} + int_1^2{int_0^1{1/2 dy dx}}$$
$$= 1/4+1/2 = 3/4$$
answered Nov 17 at 0:54
helper
524212
A much simpler way of integrating is to use the fact that $f(x,y)$ is constant, and just calculate the area where $X ge Y$ and multiply by the value of $f$ (i.e., $1/2$)
– helper
Nov 17 at 0:56
add a comment |
up vote
0
down vote
You have to split the integral, since from $0 le x le 1$, $y$ goes between $0$ and $x$, but for $1 le x le 2$, $y$ goes between $0$ and $1$.
So the correct integral should be
$$int_0^1{int_0^x{1/2 dy dx}} + int_1^2{int_0^1{1/2 dy dx}}$$
$$= 1/4+1/2 = 3/4$$
answered Nov 17 at 0:54
helper
524212
A much simpler way of integrating is to use the fact that $f(x,y)$ is constant, and just calculate the area where $X ge Y$ and multiply by the value of $f$ (i.e., $1/2$)
– helper
Nov 17 at 0:56
add a comment |
up vote
0
down vote
up vote
0
down vote
You have to split the integral, since from $0 le x le 1$, $y$ goes between $0$ and $x$, but for $1 le x le 2$, $y$ goes between $0$ and $1$.
So the correct integral should be
$$int_0^1{int_0^x{1/2 dy dx}} + int_1^2{int_0^1{1/2 dy dx}}$$
$$= 1/4+1/2 = 3/4$$
answered Nov 17 at 0:54
helper
524212
You have to split the integral, since from $0 le x le 1$, $y$ goes between $0$ and $x$, but for $1 le x le 2$, $y$ goes between $0$ and $1$.
So the correct integral should be
$$int_0^1{int_0^x{1/2 dy dx}} + int_1^2{int_0^1{1/2 dy dx}}$$
$$= 1/4+1/2 = 3/4$$
answered Nov 17 at 0:54
helper
524212
answered Nov 17 at 0:54
helper
524212
answered Nov 17 at 0:54
helper
524212
answered Nov 17 at 0:54
helper
524212
524212
A much simpler way of integrating is to use the fact that $f(x,y)$ is constant, and just calculate the area where $X ge Y$ and multiply by the value of $f$ (i.e., $1/2$)
– helper
Nov 17 at 0:56
add a comment |
A much simpler way of integrating is to use the fact that $f(x,y)$ is constant, and just calculate the area where $X ge Y$ and multiply by the value of $f$ (i.e., $1/2$)
– helper
Nov 17 at 0:56
A much simpler way of integrating is to use the fact that $f(x,y)$ is constant, and just calculate the area where $X ge Y$ and multiply by the value of $f$ (i.e., $1/2$)
– helper
Nov 17 at 0:56
A much simpler way of integrating is to use the fact that $f(x,y)$ is constant, and just calculate the area where $X ge Y$ and multiply by the value of $f$ (i.e., $1/2$)
– helper
Nov 17 at 0:56
add a comment |
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