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Math Help - Joint Probability

  1. #1
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    Joint Probability

    Is this correct?

    1). X and Y are independent standard exponential random variables with \lambda = 1
    (a) Find P(X\geq4, Y<3)
    I did P(X\geq4, Y<3) = P(X\geq4)P(Y<3) = e^{-4}e^{-3} = e^{-7}
    (b) Find E(X^2Y)
    I am unsure about this. Any help?
    (2) A random point (X,Y) is distributed uniformly on the square with vertices (1,1), (1,-1), (-1,1), and (-1,-1). The joint pdf is f(x,y)=1/4 on the square.
    (a) Find P(X^2 + Y^2 < 1)
    (b) Find P(2X - Y > 0)

    If someone can just teach me how to do one of the above in (2), I think I can handle the other. Thank you!!
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  2. #2
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    The pdf for an exponential distribution is exp (-x). To find the expectation we multiply by x and integrate to get 1/. To find the expectation of X^2 just square the pdf and multiply that by x and integrate. We get ( ^2)/4. Since X and Y are independent the expectation of the product is the product of the expectation. ie 1/4 (Note the subtlety X^2 IS dependent on X)

    With uniform distribution this becomes a geometry problem. 2a is asking what portion of the square is covered by a circle of radius 1. 2b asks what portion of the square lies under the graph y=2x
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  3. #3
    Moo
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    Hello,
    Quote Originally Posted by VENI View Post
    Is this correct?

    1). X and Y are independent standard exponential random variables with \lambda = 1
    (a) Find P(X\geq4, Y<3)
    I did P(X\geq4, Y<3) = P(X\geq4)P(Y<3) = e^{-4}e^{-3} = e^{-7}
    False.
    P(X<x)=1-e^{-x}
    So P(X>x)=e^{-x}

    hence P(X \geq 4)P(Y<3)=e^{-4} ({\color{red}1-}e^{-3})

    (b) Find E(X^2Y)
    I am unsure about this. Any help?
    Since X and Y are independent, then X and Y are independent.
    Thus E(X^2Y)=E(X^2)E(Y)

    (2) A random point (X,Y) is distributed uniformly on the square with vertices (1,1), (1,-1), (-1,1), and (-1,-1). The joint pdf is f(x,y)=1/4 on the square.
    (a) Find P(X^2 + Y^2 < 1)
    x^2+y^2<1 \Rightarrow \{x \in (-1,1) ~,~ y \in (-\sqrt{1-x^2}~,~\sqrt{1-x^2})\}
    Hence P(X^2+Y^2<1)=P(X \in (-1,1) ~,~ Y \in (-\sqrt{1-x^2}~,~\sqrt{1-x^2}))

    Now remember this definition (I'm sorry, I'm trying to explain so it may be a bit long) :
    The cdf of a rv M is defined as : F_M(m)=P(M\leq m)=\int_{-\infty}^m f_M(t) ~dt, where f_M is the pdf.
    So P(M \in (a,b))=\int_a^b f_M(t) ~dt


    So we have here :
    P(X^2+Y^2<1)=\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} f(u,v) ~ dv ~du, where f is the joint pdf



    As bob000 said, you could also solve it by geometry. And I could explain the above thing more clearly by using indicator functions, but I don't know if you know it.
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