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Math Help - Uniform

  1. #1
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    Uniform

    Let X have the uniform distribution (0,2) and let the conditional distribution of Y given X=x be uniform on (0,x^2).

    a) Find the condition expectation and variance of Y given X=x. USe these to find the marginal expectation and variance of Y...

    Many thanks
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  2. #2
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    Quote Originally Posted by brd_7 View Post
    Let X have the uniform distribution (0,2) and let the conditional distribution of Y given X=x be uniform on (0,x^2).

    a) Find the condition expectation and variance of Y given X=x. USe these to find the marginal expectation and variance of Y...

    Many thanks
    The pdf of X is f_X(x) = \frac{1}{2} for 0 \leq x \leq 2 and zero elsewhere.

    The conditional distribution of Y given X = x is f_Y(y | x) = \frac{1}{x^2} for 0 \leq y \leq x^2 and zero elsewhere.

    Now apply the standard definitions and do the necessary calculations:

    E(Y | X = x) = \int_0^{x^2} y \, f_Y(y | x) \, dy.

    E(Y^2 | X = x) = \int_0^{x^2} y^2 \, f_Y(y | x) \, dy.

    Var(Y | X = x) = E(Y^2 | X = x) - [E(Y | X = x)]^2.


    E(Y) = E[E(Y | X = x)] = \int_{-\infty}^{+\infty} E(Y | X = x) \, f_X(x) \, dx = \int_0^2 E(Y | X = x) \, \frac{1}{2} \, dx.

    To get Var(Y), read this: Law of total variance - Wikipedia, the free encyclopedia and do the necessary computation.
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  3. #3
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    Ive got it all upto the Variance of Y.. Ive got an example thats with numbers, but i cant seem to figure out what to do, and its especially confusing because of it being a uniform distribution.. Any ideas?

    And i just need to do the Var(Y) now..
    Last edited by brd_7; October 11th 2008 at 03:54 AM.
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  4. #4
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    Quote Originally Posted by brd_7 View Post
    Ive got it all upto the Variance of Y.. Ive got an example thats with numbers, but i cant seem to figure out what to do, and its especially confusing because of it being a uniform distribution.. Any ideas?

    And i just need to do the Var(Y) now..
    To save the wheel getting re-invented, if you give all your answers (with working would be ideal) it will be easier to show how to get the variance.
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  5. #5
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    Sorry, i did write down the answers, but realised they were wrong, made a careless mistake. Anyway.. I got the following..

    x^2/2

    x^4/3

    x^4/12

    1/3

    With just Var(Y) to work out..

    Many Thanks
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  6. #6
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    Quote Originally Posted by brd_7 View Post
    Sorry, i did write down the answers, but realised they were wrong, made a careless mistake. Anyway.. I got the following..

    x^2/2

    x^4/3

    x^4/12

    1/3

    With just Var(Y) to work out..

    Many Thanks
    So you need to calculate E[Var(Y | X)] = E\left[\frac{X^4}{12}\right] and Var[E(Y | X)] = Var\left[ \frac{X^2}{2}\right].

    E\left[\frac{X^4}{12}\right] = \int_0^2 \left( \frac{x^4}{12}\right) \left(\frac{1}{2} \right) \, dx.



    Var\left[ \frac{X^2}{2}\right]:

    You need the pdf of U = \frac{X^2}{2}. This can be got by calculating G(u), the cdf of U. Then g(u) = \frac{dG}{du}, where g(u) is the pdf of U.

    Then Var \left[ \frac{X^2}{2}\right] = Var (U) = E(U^2) - [E(U)]^2.


    G(u) = \Pr(U < u) = \Pr\left( \frac{X^2}{2} < u\right) = \Pr( -\sqrt{2u} < X < \sqrt{2u}) = \Pr(0 < X < \sqrt{2u}) since 0 \leq x \leq 2

    = \int_0^{\sqrt{2u}} \frac{1}{2} \, dx = \, .... for 0 \leq u \leq 2.

    Therefore g(u) = \frac{dG}{du} = \, ..... for 0 \leq u \leq 2 and zero elsewhere.
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