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Math Help - [SOLVED] joint density function /joint probability

  1. #1
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    Question [SOLVED] joint density function /joint probability

    3.55 walpole

    Given the joint density function

    f (x,y) = 6-x-y/ 8 , 0<x<2, 2<y<4
    0, elsewhere,

    find P (1<Y<3|X=1).

    i have got g(x)=(3-x)/4 and also f(y|x) = (6-x-y) / (2(3-x))

    i substitute x=1 into (6-x-y) / (2(3-x)) and i got (5-y)

    so i integrate (5-y) to dy from 1 to 3.
    but it is not right.

    did i miss something ?
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  2. #2
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    Quote Originally Posted by hazel View Post
    3.55 walpole

    Given the joint density function

    f (x,y) = 6-x-y/ 8 , 0<x<2, 2<y<4
    0, elsewhere,

    find P (1<Y<3|X=1).

    i have got g(x)=(3-x)/4 and also f(y|x) = (6-x-y) / (2(3-x))

    i substitute x=1 into (6-x-y) / (2(3-x)) and i got (5-y) / 2

    so i integrate (5-y) to dy from 1 to 3. Should be 2 to 3
    but it is not right.

    did i miss something ?
    .
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  3. #3
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    The probability must be < 1.

    Try \int_{0}^{2}\int_{2}^{3}\frac{6-x-y}{8}dydx
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  4. #4
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    Quote Originally Posted by hazel View Post
    3.55 walpole

    Given the joint density function

    f (x,y) = 6-x-y/ 8 , 0<x<2, 2<y<4
    0, elsewhere,
    Did you intend this to be (6- x- y)/8 or 6- x- (y/8)?
    Either way it is impossible. for f(x,y) to be a joint density function, you must have \int \int f(x,y)dx dy= 1
    but \int_{x=0}^2 \int_{y=0}^4 \frac{6- x- y}{8}dy dx= 3.
    \int_{x=0}^2\int_{y=0}^4 (6- x- \frac{y}{9})dy dx is even worse!

    find P (1<Y<3|X=1).

    i have got g(x)=(3-x)/4 and also f(y|x) = (6-x-y) / (2(3-x))

    i substitute x=1 into (6-x-y) / (2(3-x)) and i got (5-y)

    so i integrate (5-y) to dy from 1 to 3.
    but it is not right.

    did i miss something ?
    If f(x,y)= (6- x- y)/24 for 0< x< 2, 0< y< 4, and 0 elsewhere, then it is a joint probability density.

    With x= 1, we have P(0< y< 4|X= 1)= C\int_{y=0}^4 \frac{5- y}{24}dy= C(1/2)= 1 so C= 2.

    And, in that case, P(1< Y< 3|X= 1) is given by [tex]2\int_{y=1}^3 \frac{5- y}{24}dy
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  5. #5
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    sorry , guess i integrated wrongly.
    it should be integrate (6-x-y) / 2(3-x) dy from 2 to 3, where x=1
    and i should get :
    =1/4 (5-y) dy from 2 to 3
    = [5y - y^2] from 2 to 3
    =5/8
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  6. #6
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    just some clarification: for this question , is there an error in the question ?

    Original question :
    (3.55 walpole)

    Given the joint density function

    f (x,y) = (6-x-y)/ 8 , 0<x<2, 2<y<4 , 0, elsewhere,

    find P (1<Y<3|X=1).


    Should it be : Find P(2<Y<3 | X=1) ? since Y<2 in the given condition. it can't be "1" in P(1<Y<3.....)


    so when solving ,
    g(x) = int 2-4 (6-x-y)/8 dy
    =( 3-x )/4 for 0< x <2

    and for P (2 <Y<3 |X=1)
    = int 2-3 (6-x-y) /8 div (3-x) /4 , X=1
    = 5/8 (answer)

    am i right ?
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