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Math Help - Two random variables...

  1. #1
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    Two random variables...

    P1. Lets say we have 2 random variables X~Uniformly(0,1) and Y~Uniformly(0,2)
    Determine the probability P(X<Y).
    I suspect the answer is 1/4, but I do now know how to formally get to this result.

    P2. the same as above but this time X~Exp(1) and Y~Exp(2)
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  2. #2
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    Quote Originally Posted by madflame991 View Post
    P1. Lets say we have 2 random variables X~Uniformly(0,1) and Y~Uniformly(0,2)
    Determine the probability P(X<Y).
    I suspect the answer is 1/4, but I do now know how to formally get to this result.

    P2. the same as above but this time X~Exp(1) and Y~Exp(2)

    \displaystyle P(X<Y)=\int_{x=0}^1\int_{y=x}^2 f(x,y)\ \;dydx

    CB
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  3. #3
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    That makes sense. Thanks a lot!
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  4. #4
    MHF Contributor matheagle's Avatar
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    and for the second one I'd use

    \int_0^{\infty}\int_0^y f(x,y)dxdy
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    So... who should f(x,y) be? In the first case I saw there was no need for f(x,y)
    From what I understood the second should be:

    \int_0^{\infty}\int_{e^y}^{\infty} 2*e^{-2*x}dxdy

    Is this right?
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  6. #6
    MHF Contributor matheagle's Avatar
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    you need to review densities
    and you need f(x,y) in both cases
    in the first problem

    f(y)=1/2 and f(x)=1 since these are uniforms
    By indep the joint density is the product.

    in the second question f_X(x)=e^{-x}
    and the density of Y can eithe be f_Y(y)=2e^{-2y} or (1/2)e^{-y/2}

    depending how you define the exponential distribution.
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  7. #7
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    Ok, I know who f_X(x)=e^{-x} and f_Y(x)=2e^{-2x} are, but how do I get from them to the f(x,y) mentioned above?

    Sorry, I probably look like a big noob, but I really need to know these to pass an exam. Thx for your time
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by madflame991 View Post
    Ok, I know who f_X(x)=e^{-x} and f_Y(x)=2e^{-2x} are, but how do I get from them to the f(x,y) mentioned above?

    Sorry, I probably look like a big noob, but I really need to know these to pass an exam. Thx for your time
    Since they are (by assumption) independent the joint distribution is the product of the marginal distributions.

    f_{X,Y}(x,y)=f_X(x)f_Y(y)

    CB
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