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Math Help - Continuous Random Variable Question

  1. #1
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    Continuous Random Variable Question

    Let U and V be independent and identically distributed uniformly on the interval [0,1]. Show that for 0<x<1: P(x<V<U^2)=\frac{1}{3}-x+\frac{2}{3} x^{\frac{3}{2}}.

    So far I have worked out the cdf, and hence pdf of U^2. I thought about P(x<V<U^2)=P(V<U^2)-P(V<x) but a) this doesn't give me the correct answer and b) I don't actually think P(V<U^2) can be calculated.

    Could someone give me a hint as to where to go?
    Thanks.
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  2. #2
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    Re: Continuous Random Variable Question

    why dont you think P(U < V^2) can be calculated?

    P(U < V^2) = \int_{V=0}^{V=1} \int_{U=0}^{U=V^2} f(u,v) dU dV

    And similarly

    P(x < U < V^2) = \int_{V=x}^{V=1} \int_{U=x}^{U=V^2} f(u,v) dU dV

    (i think)

    Assuming that integral is correct (check!), can you evaluate it?
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  3. #3
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    Re: Continuous Random Variable Question

    Hello !
    Quote Originally Posted by SpringFan25 View Post
    why dont you think P(U < V^2) can be calculated?

    P(U < V^2) = \int_{V=0}^{V=1} \int_{U=0}^{U=V^2} f(u,v) dU dV

    And similarly

    P(x < U < V^2) = \int_{V=x}^{V=1} \int_{U=x}^{U=V^2} f(u,v) dU dV

    (i think)

    Assuming that integral is correct (check!), can you evaluate it?
    There's something wrong with your capital letters. It should be small letters within the integrals !

    The integral would then rather be :

    P(x < U < V^2) = \int_{v=x}^{v=\infty} \int_{u=x}^{u=v^2} f(u,v) \ du \ dv

    (since we have put f without any information, the indicator functions are still "in" it, so the boundaries go up to the infinity for v)

    And f(u,v) is the joint pdf of U and V, which is the product of their pdf since they're independent.
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    Re: Continuous Random Variable Question

    your right, but i rebel against lower case letters on thursdays

    i think its reasonable to put the limits given in the question, its seems like matter of choice whether you define the pdf with an indicator function or over a limited range. (using an indicator function is an interesting idea actually, ive only ever sseen it done this way, but your way is neat).
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  5. #5
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    Re: Continuous Random Variable Question

    Shouldn't the limits for v be \sqrt x to 1?

    I also prefer having the integration limits reflect the support of the pdf so that people don't make stupid mistakes when I tell them how to do a problem.
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  6. #6
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    Re: Continuous Random Variable Question

    yes i think you're right about the \sqrt{x}
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  7. #7
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    Re: Continuous Random Variable Question

    Thank you so much, came out as the right answer in the end!!
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