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Math Help - Uniform Distribution

  1. #1
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    Uniform Distribution

    A circle of radius r has area A=(pi)r^2. If random circle has a radius that is uniformly distributed over the interval (0,1). What are the mean and variance of the area of the circle?

    Sol.

    First I integrated (pi)r^2 from 0 to 1 and got pi/3. How do we find mean and variance now? thanks
    Last edited by somestudent2; April 10th 2008 at 09:02 AM.
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  2. #2
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    Quote Originally Posted by somestudent2 View Post
    A circle of radius r has area A=(pi)r^2. If random circle has a radius that is uniformly distributed over the interval (0,1). What are the mean and variance of the area of the circle?

    Sol.

    First I integrated (pi)r^2 from 0 to 1 and got pi/3. How do we find mean and variance now? thanks
    Find the mean and variance of the radius first.
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  3. #3
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    the mean is 1/2 and variance 1/12 for radius. Do we just substitute that in A=(pi)r^2?
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  4. #4
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    Quote Originally Posted by somestudent2 View Post
    the mean is 1/2 and variance 1/12 for radius. Do we just substitute that in A=(pi)r^2?
    I believe so, but don't hold me to that answer just yet...
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  5. #5
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    You REALLY should go through it once.

    If Y = \pi r^{2}

    And if r is Uniform on [0,1]

    Then f(r) = 1 on [0,1]

    Y is defined on [0,\pi]

    G(Y) = Pr(Y \leq y) = Pr(\pi r^{2} \leq y) = Pr\left(-\sqrt{\frac{y}{\pi}} \leq r \leq \sqrt{\frac{y}{\pi}}\right) = \int_{-\sqrt{\frac{y}{\pi}}}^{\sqrt{\frac{y}{\pi}}}f(x) dx

    G(y) = \int_{0}^{\sqrt{\frac{y}{\pi}}}f(x) dx = \int_{0}^{\sqrt{\frac{y}{\pi}}} 1 dx = \sqrt{\frac{y}{\pi}}

    Thus g(y) = G'(y) = \frac{1}{2\sqrt{\pi}\sqrt{y}} on [0,\pi]

    Checking:

    \int_{0}^{\pi}\frac{1}{2\sqrt{\pi}\sqrt{y}} dy = 1 -- Good

    E[Y] = \int_{0}^{\pi}\frac{y}{2\sqrt{\pi}\sqrt{y}} dy = \frac{1}{3}\pi

    E[Y^{2}] = \int_{0}^{\pi}\frac{y^{2}}{2\sqrt{\pi}\sqrt{y}} dy = \frac{1}{5}\pi^{2}

    Var(y) = \frac{\pi^{2}}{5} - \left(\frac{\pi}{3}\right)^{2} = \frac{4}{45}\pi^{2}


    Of course, if you were paying attention to all that, it may suggest to you the following:

    E[\pi r^{2}] = \int_{0}^{1}\pi r^{2} dr = \frac{1}{3} \pi

    E\left[\left(\pi r^{2}\right)^{2}\right] = \int_{0}^{1}\left(\pi r^{2}\right)^{2} dr = \frac{1}{5} \pi^{2}

    Those should look familiar.
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