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Thread: Joint density function

  1. #1
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    Joint density function




    This is from Exercise 5.26 from Mathematical Statistics with Applications v6.

    f (y1, y2) = {2, 0 <= y1 <= 1, 0 <= y2 <= 1, 0 <= y1 + y2 <= 1
    {0, elsewhere

    Find :
    P ( Y1 ≥ | Y2 = ) =
    P ( Y1 ≥ | Y2 ≤ ) =

    So I know I have to do something with the f(y1, y2) = f(y1, y2) / f (y2) but I really am not sure how to get there. Any help would be ace. Cheers.

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    Quote Originally Posted by statsdud View Post
    This is from Exercise 5.26 from Mathematical Statistics with Applications v6.

    f (y1, y2) = {2, 0 <= y1 <= 1, 0 <= y2 <= 1, 0 <= y1 + y2 <= 1
    {0, elsewhere

    Find :
    P ( Y1 ≥ &#189; | Y2 = &#188; ) =
    P ( Y1 ≥ &#189; | Y2 ≤ &#188; ) =

    So I know I have to do something with the f(y1, y2) = f(y1, y2) / f (y2) but I really am not sure how to get there. Any help would be ace. Cheers.
    you mean : f(y1|y2) = f(y1, y2) / f (y2)

    you can use that or just draw a picture and calculate the ration of the areas (since the density is constant on its support) at least for the first one.

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    Quote Originally Posted by statsdud View Post
    This is from Exercise 5.26 from Mathematical Statistics with Applications v6.

    f (y1, y2) = {2, 0 <= y1 <= 1, 0 <= y2 <= 1, 0 <= y1 + y2 <= 1
    {0, elsewhere

    Find :
    P ( Y1 ≥ | Y2 = ) =
    P ( Y1 ≥ | Y2 ≤ ) =

    So I know I have to do something with the f(y1, y2) = f(y1, y2) / f (y2) but I really am not sure how to get there. Any help would be ace. Cheers.
    Do you know how to calculate f_Y2(y2)?

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    f_Y2(y2) - Wouldn't it just stay at 2?

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    Quote Originally Posted by statsdud View Post
    f_Y2(y2) - Wouldn't it just stay at 2?
    No








    ...........

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    Do I have to sub y2 into it? So something like 2y2? Or do I have to integrate with respect to y2 to get there?

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    Quote Originally Posted by statsdud View Post
    Do I have to sub y2 into it? So something like 2y2? Or do I have to integrate with respect to y2 to get there?
    you have to integrate w.r.t. y1 (you are finding the marginal distribution)

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    Ok so f(y1|y2) = f(y1, y2) / f (y2) would be 2 / 2y1 = 1/y1 ?

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    Ok so then it integrates between 0 and y2 so it would be 1/y2 instead yeah?

    So I'd sub in y2 = 1/4 and then do the integration between 0 and 1/2 to get to the probability?

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    Quote Originally Posted by statsdud View Post
    Ok so then it integrates between 0 and y2 so it would be 1/y2 instead yeah?

    So I'd sub in y2 = 1/4 and then do the integration between 0 and 1/2 to get to the probability?
    no...

    since

    0 <= y1 + y2 <= 1


    and y1>=0 and y2>=0
    we have that

    0<=y1<=1-y2

    so integrate from 0 to 1-y2

  11. #11
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    So:

    f( Y1 | Y2 ) = f( Y1 , Y2 ) / f2( Y2 )
    = 2 / 2 ( 1 y2 )
    = 1 / 1 y2

    P ( Y1 ≥ | Y2 = ) = ∫ 1 / ( 1 ) dy1
    = 1/21 ∫ 1 / ( 1 ) dy1
    = 1/21 ∫ 4/3 dy1
    = [ 4/3 y1 ] 1/21
    = 4/3 - 2/3
    = 2/3 ?

    but is it different for the ≤ ?

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    Quote Originally Posted by statsdud View Post
    So:

    f( Y1 | Y2 ) = f( Y1 , Y2 ) / f2( Y2 )
    = 2 / 2 ( 1 y2 )
    = 1 / 1 y2

    P ( Y1 ≥ | Y2 = ) = ∫ 1 / ( 1 ) dy1
    = 1/21 ∫ 1 / ( 1 ) dy1
    = 1/21 ∫ 4/3 dy1
    = [ 4/3 y1 ] 1/21
    = 4/3 - 2/3
    = 2/3 ?

    but is it different for the ≤ ?
    what is 1/21 in 1/21 ∫ 1 / ( 1 ) dy1?


    also the answer isn't 2/3

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    The intergratal between 1 and 1/2 of 1/(1-1/4)

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    Quote Originally Posted by statsdud View Post
    The intergratal between 1 and 1/2 of 1/(1-1/4)
    you cant go all the way to 1

    remember we have 0<=y1+y2<=1

    and y2=1/4 so the largest y1 can be is 3/4

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    Ahhh ok, so if I change that in the integral, everything else is ok?

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