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Math Help - Logistic distribution

  1. #1
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    Logistic distribution

    Let X have a logistic distribution with p.d.f.
    f(x) = e^(-x)/(1 + e^(-x))^2 -infinty <x<infinity

    Show that
    Y = 1/(1 + e^(-x))

    has a U (0, 1) distribution. (Hint: Find P (Y ≤ y) = P(1/(1 + e^(-x)<=y)
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  2. #2
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    Quote Originally Posted by affelix View Post
    Let X have a logistic distribution with p.d.f.
    f(x) = e^(-x)/(1 + e^(-x))^2 -infinty <x<infinity

    Show that
    Y = 1/(1 + e^(-x))

    has a U (0, 1) distribution. (Hint: Find P (Y ≤ y) = P(1/(1 + e^(-x)<=y)
    First note that 0 < y < 1 since 0 < \frac{1}{1 + e^{-x}} < 1 for -\infty < x < + \infty.

    Then the cdf of Y is given by


    F(y) = \Pr(Y \leq y) = \Pr \left( \frac{1}{1 + e^{-X}} \leq y \right) = \Pr\left(1 + e^{-X} \geq \frac{1}{y} \right)


     = \Pr\left( X \leq \ln \left( \frac{y}{1 - y}\right)\right)


    = \int_{-\infty}^{\ln \left( \frac{y}{1 - y}\right)} f(x) \, dx.


    Your job is to calculate this integral (for 0 < y < 1 the answer is y, by the way) and then calculate f(y) = \frac{dF}{dy}. Note also that details are important ....
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