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Math Help - Verifying that Uniform and Pareto distributions are probability densities

  1. #1
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    Verifying that Uniform and Pareto distributions are probability densities

    Show that the following functions are probability densities, i.e.

    f(x)\geq 0 and \int_{-\infty}^{\infty}f(x) dx =1

    Uniform distribution:

    f(x)=\left\{\begin{array}{cc}\frac{1}{b-a} & \mbox{ for } a\leq x \leq b\\0 & \mbox{ otherwise }\end{array}\right.

    Pareto distribution: For k>0

    f(x)=\left\{\begin{array}{cc}\frac{k}{x^{k+1}} & \mbox{ for } x \geq 1\\0 & \mbox{ otherwise }\end{array}\right.
    Last edited by acevipa; June 8th 2010 at 07:07 PM.
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  2. #2
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    Quote Originally Posted by acevipa View Post
    Show that the following functions are probability densities, i.e.

    f(x)\geq 0 and \int_{-\infty}^{\infty}f(x) dx =1

    Uniform distribution:

    f(x)=\left\{\begin{array}{cc}\frac{1}{b-a} & \mbox{ for } a\leq x \leq b\\0 & \mbox{ otherwise }\end{array}\right.

    Pareto distribution: For k>0

    f(x)=\left\{\begin{array}{cc}\frac{k}{x^{k+1}} & \mbox{ for } x \geq 1\\0 & \mbox{ otherwise }\end{array}\right.
    Where exactly are you stuck here?
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    Could someone please verify my working?

    Uniform Distirbution:

    \int_{-\infty}^{\infty}\frac{1}{b-a}dx = \int_{a}^{b}\frac{1}{b-a}dx

    =\left[\frac{x}{b-a}\right]_a^b

    =\frac{b}{b-a}-\frac{a}{b-a}

    =\frac{b-a}{b-a}

    =1

    \int_{-\infty}^{\infty}\frac{1}{b-a}dx=1

    Why is f(x) \geq 0

    Now for a Pareto distribution:

    \int_{-\infty}^{\infty}\frac{k}{x^{k+1}}dx = \int_{1}^{\infty}\frac{k}{x^{k+1}}dx

    =k\int_{1}^{\infty}\frac{1}{x^{k+1}}dx

    =k \left[\frac{1}{-kx^k}\right]_1^{\infty}

    =-\left[\frac{1}{x^k}\right]_1^{\infty}

    =-\left[\frac{1}{\infty^k}-\frac{1}{1^k}\right] Not too sure if you sub \infty in since it's a definite integral. What's the accepted way of evaluating this indefinite integral

    =-[0-1]

    \int_{-\infty}^{\infty} \frac{k}{x^{k+1}}dx=1

    Since k>0 and x\geq 1  \Longrightarrow x^{k+1}\geq 0

    \Longrightarrow \frac{k}{x^{k+1}} \geq 0
    Last edited by acevipa; June 9th 2010 at 12:50 AM.
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  4. #4
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    Quote Originally Posted by acevipa View Post
    Could someone please verify my working?

    Uniform Distirbution:

    \int_{-\infty}^{\infty}\frac{1}{b-a}dx = \int_{a}^{b}\frac{1}{b-a}dx

    =\left[\frac{x}{b-a}\right]_a^b

    =\frac{b}{b-a}-\frac{a}{b-a}

    =\frac{b-a}{b-a}

    =1

    \int_{-\infty}^{\infty}\frac{1}{b-a}dx=1

    Why is f(x) \geq 0 Mr F says: Although Russell and Whitehead would disagree (if they were still alive), it should be obvious that f(x) = 1/(b - a) > 0 if b > a ....


    Now for a Pareto distribution:

    \int_{-\infty}^{\infty}\frac{k}{x^{k+1}}dx = \int_{1}^{\infty}\frac{k}{x^{k+1}}dx

    =k\int_{1}^{\infty}\frac{1}{x^{k+1}}dx

    =k \left[\frac{1}{-kx^k}\right]_1^{\infty}

    =-\left[\frac{1}{x^k}\right]_1^{\infty}

    =-\left[\frac{1}{\infty^k}-\frac{1}{1^k}\right] Mr F says: This is poor notation. The integral is an improper integral and needs to be carefully calculated using limits. Do you know how to do that?

    =-[0-1]

    \int_{-\infty}^{infty}\frac{k}{x^{k+1}}dx=1
    It looks OK (except for my second red note).
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    It looks OK (except for my second red note).
    Ok, yeah your first red note seems kind of obvious now.

    I thought my notation wasn't that good. How would you write it as?
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  6. #6
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    Quote Originally Posted by acevipa View Post
    Ok, yeah your first red note seems kind of obvious now.

    I thought my notation wasn't that good. How would you write it as?
    = k \lim_{\alpha \to +\infty} \int_{1}^{\alpha}\frac{1}{x^{k+1}}dx

    =k \lim_{\alpha \to +\infty} \left[\frac{1}{-kx^k}\right]_1^{\alpha}

    =- \lim_{\alpha \to +\infty} \left[\frac{1}{x^k}\right]_1^{\alpha}

    =- \lim_{\alpha \to + \infty}\left[\frac{1}{\alpha^k}-\frac{1}{1^k}\right]

    =-[0-1] = 1.
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    = k \lim_{\alpha \to +\infty} \int_{1}^{\alpha}\frac{1}{x^{k+1}}dx

    =k \lim_{\alpha \to +\infty} \left[\frac{1}{-kx^k}\right]_1^{\alpha}

    =- \lim_{\alpha \to +\infty} \left[\frac{1}{x^k}\right]_1^{\alpha}

    =- \lim_{\alpha \to + \infty}\left[\frac{1}{\alpha^k}-\frac{1}{1^k}\right]

    =-[0-1] = 1.
    Thanks a lot. That makes more sense
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