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Thread: Joint distribution, find P(X^2 + Y^2 <=1)

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    Joint distribution, find P(X^2 + Y^2 <=1)




    I am trying to teach myself (without a tutor) some degree level Statistics. I think I've answered parts a-c of this question, but would like confirmation, or pointers of where I've gone wrong. Part d I've struggled with, so would like some advice.

    Q:The random variables (X,Y) have joint distribution function FXY(x,y) = xy/4 for 0<x<2, 0<y<2. Obtain a) the joint pdf, b) P(X<1, Y<1), c) P(X<1), d) P(X^2 + Y^2 <=1)

    My answers so far:
    a) fXY(x,y) = (d^2)/dxdy (xy/4)
    = d/dx (x/4)
    = 1/4

    b) P(X<1, Y<1) = (1*1)/4
    = 1/4

    c) P(X<1) = FXY(x, infinity)
    = FXY(1, 2)
    = (1*2)/4
    = 1/2

    d) P(X^2 +Y^2 <=1) = P(Y<= sqrt(1-X^2))
    = ∫ ( ∫ (1/4) dy from 0 to sqrt(1-X^2) ) dx from 0 to 2
    = ∫ (sqrt(1-X^2))/4 dx from 0 to 2

    I'm not sure how to integrate this, and the online integrator I used gave:
    x/2(sqrt(1-X^2)) + 1/2 sin^-1(x)
    which is a problem because you can't put 2 into sin^-1(x).
    Help?

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    for d) you don't need to integrate (plus your limits are not correct)


    you have a uniform distribution in x,y

    what is (the area of a quarter circle of radius 1) * (1/4)
    Dr. Zoidberg: Fry, it's been years since medical school, so remind me. Disemboweling in your species, fatal or non-fatal?

  3. #3
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    Thank you
    Can I check I'm understanding?
    We have (the area of quarter of a circle of radius 1) * (1/4) = pi/16
    because the area under y= sqrt(1-x^2) where x>0 is equivalent to pi/4 and we multiply this by the pdf (which is 1/4) because we want to know the proportion that the area under y= sqrt(1-x^2) is of the whole?

    Therefore P(X^2 + Y^2 <=1) = pi/16

    Out of interest/for future reference, what should my limits for the integration have been? Am I correct in thinking that if I had not had a uniform distribution, I would have had to have integrated?

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    Quote Originally Posted by NicolaR View Post
    Thank you
    Can I check I'm understanding?
    We have (the area of quarter of a circle of radius 1) * (1/4) = pi/16
    because the area under y= sqrt(1-x^2) where x>0 is equivalent to pi/4 and we multiply this by the pdf (which is 1/4) because we want to know the proportion that the area under y= sqrt(1-x^2) is of the whole?

    Therefore P(X^2 + Y^2 <=1) = pi/16

    Out of interest/for future reference, what should my limits for the integration have been? Am I correct in thinking that if I had not had a uniform distribution, I would have had to have integrated?

    that's correct...


    for your last question...you made a minor mistake...


    you had = ∫ ( ∫ (1/4) dy from 0 to sqrt(1-X^2) ) dx from 0 to 2

    you should have = ∫ ( ∫ (1/4) dy from 0 to sqrt(1-X^2) ) dx from 0 to 1
    Dr. Zoidberg: Fry, it's been years since medical school, so remind me. Disemboweling in your species, fatal or non-fatal?

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    I'm a bit confused about why the limits for ∫ dx are 0 to 1. I had 0 to 2 because that is the range of values x can take. Is it 1 because a greater value of x would result in sqrt of a minus number in sqrt(1-x^2)?

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    Quote Originally Posted by NicolaR View Post
    I'm a bit confused about why the limits for ∫ dx are 0 to 1. I had 0 to 2 because that is the range of values x can take. Is it 1 because a greater value of x would result in sqrt of a minus number in sqrt(1-x^2)?
    if x^2+y^2<=1 then

    the largest x can be is 1
    Dr. Zoidberg: Fry, it's been years since medical school, so remind me. Disemboweling in your species, fatal or non-fatal?

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    Oh I see! Thanks a lot, you've really helped me

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