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Math Help - What is the mean of these two CDFs?

  1. #1
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    What is the mean of these two CDFs?

    1. F(x) = 1 - 1/x (x greater than or equal to 1)

    and

    2. F(x) = 1 - exp(-x^2) (x greater than or equal to 1)

    Many thanks.
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  2. #2
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    1. F(x) = 1 - 1/x (x greater than or equal to 1)

    You can do this in two ways. You can calculate the PDF, f, and then integrate xf(x). The PDF is just the derivative of the CDF, if continuous.

    f(x) = 1/x^2
    \int_1^\infty xf(x)dx = \int_1^\infty 1/x\,dx =\ln(x)|_1^\infty which does not converge.

    2. F(x) = 1 - exp(-x^2) (x greater than or equal to 1)
    f(x) = -2x\exp(-x^2)
    \int_1^\infty xf(x)dx = \int_1^\infty -2x^2 \exp(-x^2)\,dx

    I guess I would try integration by parts with u=x,\,dv=-2x \exp(-x^2)\,dx

    You will probably need to use the standard normal table in your stats book to do \int_1^\infty \exp(-x^2)\,dx which should pop out after the integration by parts step. That integral is almost the area under the curve of the standard normal distribution, but it is missing a 1/\sqrt{2\pi}.
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  3. #3
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    Quote Originally Posted by meymathis View Post
    1. F(x) = 1 - 1/x (x greater than or equal to 1)

    You can do this in two ways. You can calculate the PDF, f, and then integrate xf(x). The PDF is just the derivative of the CDF, if continuous.

    f(x) = 1/x^2
    \int_1^\infty xf(x)dx = \int_1^\infty 1/x\,dx =\ln(x)|_1^\infty which does not converge.

    2. F(x) = 1 - exp(-x^2) (x greater than or equal to 1)
    f(x) = -2x\exp(-x^2) Mr F says: Small typo here. Should be 2x, not -2x .... The bigger problem is that it's not actually a valid cdf .... The support should be [0, +oo), NOT [1, +oo).

    \int_1^\infty xf(x)dx = \int_1^\infty -2x^2 \exp(-x^2)\,dx

    I guess I would try integration by parts with u=x,\,dv=-2x \exp(-x^2)\,dx

    You will probably need to use the standard normal table in your stats book to do \int_1^\infty \exp(-x^2)\,dx which should pop out after the integration by parts step. That integral is almost the area under the curve of the standard normal distribution, but it is missing a 1/\sqrt{2\pi}.
    ..
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  4. #4
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    Darn it. Details, details....

    I'll have you know I checked that the first one was. And this was right after a long back and forth with someone about a CDF not really being a CDF. http://www.mathhelpforum.com/math-he...variables.html

    Thanks Mr. F!
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  5. #5
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    Thanks you two, that was extremely helpful.
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