Find f(u,v) = joint pdf of U and V.

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A circular dartboard has a radius of 1 foot. When a dart is thrown at the board, where it sticks is uniformly distributed on the face of the board, meaning that probability is proportional to area. (And the probability for the whole board is 1, since throws which miss the board are repeated until the dart hits the board.) Hitting points for different darts are independent.

(a). Throw 2 darts. Let X be the distance to the center for the first dart, in feet. Let Y be the distance to the center for the second dart. Let U = min(X,Y) and V = max(X,Y).

Find f(u,v) = joint pdf of U and V. Remember to say where f(u,v) is positive.

Attached is the solution for this question.

However, I do not understand how we got f(u,v)= 8uv. Would you guys mind explaining? Or is there any simpler way to do this? Something that is more clear.

Thanks

probability

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asked Jul 28 at 6:21

ibuntu

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A circular dartboard has a radius of 1 foot. When a dart is thrown at the board, where it sticks is uniformly distributed on the face of the board, meaning that probability is proportional to area. (And the probability for the whole board is 1, since throws which miss the board are repeated until the dart hits the board.) Hitting points for different darts are independent.

(a). Throw 2 darts. Let X be the distance to the center for the first dart, in feet. Let Y be the distance to the center for the second dart. Let U = min(X,Y) and V = max(X,Y).

Find f(u,v) = joint pdf of U and V. Remember to say where f(u,v) is positive.

Attached is the solution for this question.

However, I do not understand how we got f(u,v)= 8uv. Would you guys mind explaining? Or is there any simpler way to do this? Something that is more clear.

Thanks

probability

share|cite|improve this question

asked Jul 28 at 6:21

ibuntu

566

add a comment | 

up vote
0
down vote

favorite

2

up vote
0
down vote

favorite

2

2

A circular dartboard has a radius of 1 foot. When a dart is thrown at the board, where it sticks is uniformly distributed on the face of the board, meaning that probability is proportional to area. (And the probability for the whole board is 1, since throws which miss the board are repeated until the dart hits the board.) Hitting points for different darts are independent.

(a). Throw 2 darts. Let X be the distance to the center for the first dart, in feet. Let Y be the distance to the center for the second dart. Let U = min(X,Y) and V = max(X,Y).

Find f(u,v) = joint pdf of U and V. Remember to say where f(u,v) is positive.

Attached is the solution for this question.

However, I do not understand how we got f(u,v)= 8uv. Would you guys mind explaining? Or is there any simpler way to do this? Something that is more clear.

Thanks

probability

share|cite|improve this question

asked Jul 28 at 6:21

ibuntu

566

A circular dartboard has a radius of 1 foot. When a dart is thrown at the board, where it sticks is uniformly distributed on the face of the board, meaning that probability is proportional to area. (And the probability for the whole board is 1, since throws which miss the board are repeated until the dart hits the board.) Hitting points for different darts are independent.

(a). Throw 2 darts. Let X be the distance to the center for the first dart, in feet. Let Y be the distance to the center for the second dart. Let U = min(X,Y) and V = max(X,Y).

Find f(u,v) = joint pdf of U and V. Remember to say where f(u,v) is positive.

Attached is the solution for this question.

However, I do not understand how we got f(u,v)= 8uv. Would you guys mind explaining? Or is there any simpler way to do this? Something that is more clear.

Thanks

probability

share|cite|improve this question

asked Jul 28 at 6:21

ibuntu

566

share|cite|improve this question

share|cite|improve this question

asked Jul 28 at 6:21

ibuntu

566

asked Jul 28 at 6:21

ibuntu

566

asked Jul 28 at 6:21

ibuntu

566

566

add a comment | 

add a comment | 

1 Answer
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Firstly let's calculate the CDF of $X$: $$F_X(x) = P(X le x) = fracpi x^2pi r^2 = x^2, 0 le x le 1$$ hence the PDF of $X$ is $f_X(x) = 2x, 0 le x le 1$ then the joint PDF of $X, Y$ is $$f(x, y) = f_X(x) f_Y(y) = 4xy, 0 le x le 1, 0 le y le 1$$

Now, compute the joint CDF of $U, V$:

$$P(min(X, Y) le u, max(X, Y) le v) = \ P(min(X, Y) le u, max(X, Y) le v|X>Y) times P(X>Y) + P(min(X, Y) le u, max(X, Y) le v|X le Y) times P(X le Y) = \ = P(Y le u, X le v | X <Y) timesfrac12 + P(X le u, Yle v|X le Y) times frac12 = \ = fracu^2v^2frac12 times frac12 + fracv^2u^2frac12 times frac12 = 2u^2v^2, 0 le u le v le 1$$

Finally the joint PDF of $U, V$ is $$fracpartial^2partial upartial v 2u^2v^2 = 8uv, 0 le u le v le 1$$

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answered Jul 28 at 19:17

D F

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Firstly let's calculate the CDF of $X$: $$F_X(x) = P(X le x) = fracpi x^2pi r^2 = x^2, 0 le x le 1$$ hence the PDF of $X$ is $f_X(x) = 2x, 0 le x le 1$ then the joint PDF of $X, Y$ is $$f(x, y) = f_X(x) f_Y(y) = 4xy, 0 le x le 1, 0 le y le 1$$

Now, compute the joint CDF of $U, V$:

$$P(min(X, Y) le u, max(X, Y) le v) = \ P(min(X, Y) le u, max(X, Y) le v|X>Y) times P(X>Y) + P(min(X, Y) le u, max(X, Y) le v|X le Y) times P(X le Y) = \ = P(Y le u, X le v | X <Y) timesfrac12 + P(X le u, Yle v|X le Y) times frac12 = \ = fracu^2v^2frac12 times frac12 + fracv^2u^2frac12 times frac12 = 2u^2v^2, 0 le u le v le 1$$

Finally the joint PDF of $U, V$ is $$fracpartial^2partial upartial v 2u^2v^2 = 8uv, 0 le u le v le 1$$

share|cite|improve this answer

answered Jul 28 at 19:17

D F

1,0551218

add a comment | 

up vote
1
down vote



accepted

Firstly let's calculate the CDF of $X$: $$F_X(x) = P(X le x) = fracpi x^2pi r^2 = x^2, 0 le x le 1$$ hence the PDF of $X$ is $f_X(x) = 2x, 0 le x le 1$ then the joint PDF of $X, Y$ is $$f(x, y) = f_X(x) f_Y(y) = 4xy, 0 le x le 1, 0 le y le 1$$

Now, compute the joint CDF of $U, V$:

$$P(min(X, Y) le u, max(X, Y) le v) = \ P(min(X, Y) le u, max(X, Y) le v|X>Y) times P(X>Y) + P(min(X, Y) le u, max(X, Y) le v|X le Y) times P(X le Y) = \ = P(Y le u, X le v | X <Y) timesfrac12 + P(X le u, Yle v|X le Y) times frac12 = \ = fracu^2v^2frac12 times frac12 + fracv^2u^2frac12 times frac12 = 2u^2v^2, 0 le u le v le 1$$

Finally the joint PDF of $U, V$ is $$fracpartial^2partial upartial v 2u^2v^2 = 8uv, 0 le u le v le 1$$

share|cite|improve this answer

answered Jul 28 at 19:17

D F

1,0551218

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Firstly let's calculate the CDF of $X$: $$F_X(x) = P(X le x) = fracpi x^2pi r^2 = x^2, 0 le x le 1$$ hence the PDF of $X$ is $f_X(x) = 2x, 0 le x le 1$ then the joint PDF of $X, Y$ is $$f(x, y) = f_X(x) f_Y(y) = 4xy, 0 le x le 1, 0 le y le 1$$

Now, compute the joint CDF of $U, V$:

$$P(min(X, Y) le u, max(X, Y) le v) = \ P(min(X, Y) le u, max(X, Y) le v|X>Y) times P(X>Y) + P(min(X, Y) le u, max(X, Y) le v|X le Y) times P(X le Y) = \ = P(Y le u, X le v | X <Y) timesfrac12 + P(X le u, Yle v|X le Y) times frac12 = \ = fracu^2v^2frac12 times frac12 + fracv^2u^2frac12 times frac12 = 2u^2v^2, 0 le u le v le 1$$

Finally the joint PDF of $U, V$ is $$fracpartial^2partial upartial v 2u^2v^2 = 8uv, 0 le u le v le 1$$

share|cite|improve this answer

answered Jul 28 at 19:17

D F

1,0551218

Firstly let's calculate the CDF of $X$: $$F_X(x) = P(X le x) = fracpi x^2pi r^2 = x^2, 0 le x le 1$$ hence the PDF of $X$ is $f_X(x) = 2x, 0 le x le 1$ then the joint PDF of $X, Y$ is $$f(x, y) = f_X(x) f_Y(y) = 4xy, 0 le x le 1, 0 le y le 1$$

Now, compute the joint CDF of $U, V$:

$$P(min(X, Y) le u, max(X, Y) le v) = \ P(min(X, Y) le u, max(X, Y) le v|X>Y) times P(X>Y) + P(min(X, Y) le u, max(X, Y) le v|X le Y) times P(X le Y) = \ = P(Y le u, X le v | X <Y) timesfrac12 + P(X le u, Yle v|X le Y) times frac12 = \ = fracu^2v^2frac12 times frac12 + fracv^2u^2frac12 times frac12 = 2u^2v^2, 0 le u le v le 1$$

Finally the joint PDF of $U, V$ is $$fracpartial^2partial upartial v 2u^2v^2 = 8uv, 0 le u le v le 1$$

share|cite|improve this answer

answered Jul 28 at 19:17

D F

1,0551218

share|cite|improve this answer

share|cite|improve this answer

answered Jul 28 at 19:17

D F

1,0551218

answered Jul 28 at 19:17

D F

1,0551218

answered Jul 28 at 19:17

D F

1,0551218

1,0551218

add a comment | 

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