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Math Help - proof uniform distribution

  1. #1
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    proof uniform distribution

    Can some please help me with this proof.
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  2. #2
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    Quote Originally Posted by maths_123 View Post
    Can some please help me with this proof.
    1. Let X = U + 1.

    The cdf of X is given by

    F(x) = \Pr(X < x) = \Pr(U + 1 < x) = \Pr(U < x - 1) = \int_{0}^{x - 1} 1 \, du = \, ....

    Therefore the pdf of X is given by f(x) = \frac{dF}{dx} = \, .... for 1 \leq x \leq 2 and zero elsewhere.

    --------------------------------------------------------------------------------------------------------

    2. Let Y = U^2.

    The cdf of Y is given by

    G(y) = \Pr(Y < y) = \Pr(U^2 < y) = \Pr(-\sqrt{y} < U < \sqrt{y}) = \int_{0}^{\sqrt{y}} 1 \, du = \, ....

    Therefore the pdf of Y is given by g(y) = \frac{dG}{dy} = \, .... for 0 \leq y \leq 1 and zero elsewhere.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    1. Let X = U + 1.

    The cdf of X is given by

    F(x) = \Pr(X < x) = \Pr(U + 1 < x) = \Pr(U < x - 1) = \int_{0}^{x - 1} 1 \, du = \, ....

    Therefore the pdf of X is given by f(x) = \frac{dF}{dx} = \, .... for 1 \leq x \leq 2 and zero elsewhere.

    --------------------------------------------------------------------------------------------------------

    2. Let Y = U^2.

    The cdf of Y is given by

    G(y) = \Pr(Y < y) = \Pr(U^2 < y) = \Pr(-\sqrt{y} < U < \sqrt{y}) = \int_{0}^{\sqrt{y}} 1 \, du = \, ....

    Therefore the pdf of Y is given by g(y) = \frac{dG}{dy} = \, .... for 0 \leq y \leq 1 and zero elsewhere.

    Thanks for the help, but i still don't understand how this shows the U+1 is a uniform distribution but u^2 does not.
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  4. #4
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    Quote Originally Posted by maths_123 View Post
    Thanks for the help, but i still don't understand how this shows the U+1 is a uniform distribution but u^2 does not.
    Did you do the integrations that I left you to get the cdf in each case? Did you differentiate the cdf in each case to get the pdf? If you've done these things then it should be obvious.

    Please post all that you've done in response to my first post.
    Last edited by mr fantastic; December 7th 2008 at 03:34 PM. Reason: Corrected a spellng error .... Mr Spelling Bee, that's me.
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