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Thread: Conditional Distribution over the unit Disk

  1. #1
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    Conditional Distribution over the unit Disk




    Suppose the Random Variables U and V are uniform (-1,1). How can I show that their conditional distribution, given U^2+V^2 <1, is


    f_{U,V|U^2+V^2<1} (u,v|w<1) =1/{\pi} \quad u^2+v^2<1

    Can this be derived using the conditional distribution quotient? I would appreciate it if someone could explain all necessary steps.

    Thank you.

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    Re: Conditional Distribution over the unit Disk

    I suppose U, V are given as independent.

    First of all you will need to derive the conditional joint CDF, then differentiate it to obtain the conditional joint pdf.

    i.e. to calculate the following first:

    [math] F_{U,V|U^2+V^2<1}(u,v) [/math

    = \Pr\{U \leq u, V \leq v|U^2 + V^2 < 1\}

    = \frac {\Pr\{U \leq u, V \leq v, U^2 + V^2 < 1\}} {\Pr\{U^2 + V^2 < 1\}}

    = \frac {4} {\pi} \iint\limits_{\substack{x < u, y < v \\ x^2+y^2 < 1}} f_{U,V}(x,y) dxdy

    = \frac {1} {\pi} \iint\limits_{\substack{x < u, y < v \\ x^2+y^2 < 1}} dxdy

    Now the remaining part is just a calculus problem. Sketch the region in your mind

    When -1 < u < 0, the above double integral can be expressed as

    \int_{-\sqrt{1-u^2}}^v u + \sqrt{1 - y^2}dy

    and by Fundamental Theorem of Calculus,

    \frac {\partial^2} {\partial u \partial v} \int_{-\sqrt{1-u^2}}^v u + \sqrt{1 - y^2}dy

    = \frac {\partial} {\partial u} (u + \sqrt{1 - v^2})

    = 1

    When 0 < u < 1, we have an additional extra term

    2\int_{-1}^{-\sqrt{1-u^2}} \sqrt{1 - y^2} dy

    but since it is independent of v, the derivative vanish.

    Therefore the result follows.

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    Re: Conditional Distribution over the unit Disk


    Thank you but can you please explain the integrals a bit? I'm having some trouble comprehending the expressions and their limits.

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