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    Gamma Distribution

    The radii of craters are exponentially distributed with mean 10 feet. Find the mean and variance of the areas.

    I know I'm going to have to use an integral, but how do I know what beta will be? And where will the mean come into play in the integral?

    Thanks!
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  2. #2
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    Quote Originally Posted by fulltwist8 View Post
    The radii of craters are exponentially distributed with mean 10 feet. Find the mean and variance of the areas.

    I know I'm going to have to use an integral, but how do I know what beta will be? And where will the mean come into play in the integral?

    Thanks!
    For starters, the pdf of the radius is given here and it should be clear how the mean comes into play .....

    Now you need to think about the mean and the standard deviation corresponding to the pdf for \pi r^2 = (\sqrt{\pi} r)^2 .....
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    still confused... and now there's a lambda involved?
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    Quote Originally Posted by fulltwist8 View Post
    still confused... and now there's a lambda involved?
    \lambda = \frac{1}{\mu}.

    \mu = 10 \Rightarrow \lambda = \frac{1}{10}.

    Therefore the pdf for the radius is f(r) = \frac{1}{10}\, e^{-r/10}.
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    Quote Originally Posted by mr fantastic View Post
    \lambda = \frac{1}{\mu}.

    \mu = 10 \Rightarrow \lambda = \frac{1}{10}.

    Therefore the pdf for the radius is f(r) = \frac{1}{10}\, e^{-r/10}.
    This is probably just an academic exercise now ....... Anyway, for what it's worth, here's one way of getting the probability distribution function of R^2:

    Note: Once the pdf is known, the mean and variance of R^2 can (in principle) be calculated.


    Let Y = R^2 and let the pdf of Y be g(y).

    Let G(y) = \Pr(Y \leq y). By definition, g(y) = \frac{dG}{dy}.

    Y < y \Rightarrow R^2 < y \Rightarrow -\sqrt{y} < R < \sqrt{y}.

    Therefore:

    G(y) = \Pr(Y \leq y) = \Pr(-\sqrt{y} < R < \sqrt{y}) = \int_{-\sqrt{y}}^{\sqrt{y}} f(r) \, dr.

    But f(r) = \frac{1}{10}\, e^{-r/10} for r \geq 0 and zero elsewhere.

    Therefore:

    G(y) = \Pr(Y \leq y) = \int_{0}^{\sqrt{y}} \frac{1}{10}\, e^{-r/10} \, dr = -\left[ e^{-r/10}\right]_{0}^{\sqrt{y}} = 1 - e^{-\sqrt{y}/10}.

    Therefore:

    g(y) = \frac{dG}{dy} = \frac{1}{20} \, \frac{1}{\sqrt{y}} \, e^{-\sqrt{y}/10}.

    Y = R^2 and r \geq 0 \Rightarrow y \geq 0.

    Therefore:

    {\color{red}g(y) = \frac{1}{20} \, \frac{1}{\sqrt{y}} \, e^{-\sqrt{y}/10}} for {\color{red}y > 0} and zero otherwise

    since \frac{1}{\sqrt{y}} \, e^{-\sqrt{y}/10} is undefined for y = 0.


    E(Y) = E(R^2) = \int_{-\infty}^{\infty} y \, g(y) \, dy = \frac{1}{20} \, \int_{0}^{\infty} \sqrt{y} \, e^{-\sqrt{y}/10} \, dy = 200.


    E(Y^2) = \int_{-\infty}^{\infty} y^2 \, g(y) \, dy = \frac{1}{20} \, \int_{0}^{\infty} y^{3/2} \, e^{-\sqrt{y}/10} \, dy = 240000.


    Var[Y] = Var[R^2] = E[Y^2] - (E[Y])^2 = 240000 - 40000 = 200000.

    sd_Y = 447, which can't be right so there's probably a small mistake somewhere ....


    Then E[\pi R^2] = \pi E[R^2] and Var[\pi R^2] = \pi^2 Var[R^2] using the usual formula.
    Last edited by mr fantastic; March 14th 2008 at 08:02 PM.
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