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Thread: 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 $\displaystyle xf(x)$. The PDF is just the derivative of the CDF, if continuous.

    $\displaystyle f(x) = 1/x^2$
    $\displaystyle \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)
    $\displaystyle f(x) = -2x\exp(-x^2)$
    $\displaystyle \int_1^\infty xf(x)dx = \int_1^\infty -2x^2 \exp(-x^2)\,dx$

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

    You will probably need to use the standard normal table in your stats book to do $\displaystyle \int_1^\infty \exp(-x^2)\,dx$ 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 $\displaystyle 1/\sqrt{2\pi}$.
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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 $\displaystyle xf(x)$. The PDF is just the derivative of the CDF, if continuous.

    $\displaystyle f(x) = 1/x^2$
    $\displaystyle \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)
    $\displaystyle f(x) = -2x\exp(-x^2)$ 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).

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

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

    You will probably need to use the standard normal table in your stats book to do $\displaystyle \int_1^\infty \exp(-x^2)\,dx$ 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 $\displaystyle 1/\sqrt{2\pi}$.
    ..
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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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