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Math Help - Stats problem

  1. #1
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    Stats problem

    A random variable X has cdf of the form F(x) = \frac{1}{1+e^{-\frac{x-\mu}{\sigma}}}, x is real, \mu is real and \sigma > 0.

    Find the pdf function of X.

    Find the form of the moment generating function Y = \frac{X-\mu}{\sigma} and hence calculate the variance of X

    [Note : the gamma function, \Gamma{(t)} is such that \Gamma''{(1)} - \Gamma'{(1)} = \frac{\pi^2}{6}.

    Letting z = {e^{-\frac{x-\mu}{\sigma}}}.

    The pdf turns out to be f(x) = \frac{d}{dx}F(x) = \frac{z}{\sigma (1+z)^2}.

    Using the fact that if Y = aX+b then  M_y{(t)} = e^{bt}M_x{(at)},
    It turns out that  M_y{(t)} = e^{\frac{-t\mu}{\sigma}} \int^{\infty}_{-\infty} e^{\frac{tx}{\sigma}} f(x) dx.
    Using dx = -\frac{\sigma dz}{z} and  (z e^{\frac{-\mu}{\sigma}})^{-t} = e^{\frac{tx}{\sigma}}
    We then have M_y{(t)} = \int^{\infty}_{-\infty}\frac{z^{-t}z}{\sigma (1+z)^2}dx = \int^{\infty}_{0} \frac{z^{-t}}{(1+z)^2}dz which, according to Maple, is equal to \pi csc(\pi t) t.
    But then I end up with dividing by zero errors when trying to calculate  M'_y{(1)}.
    Plus I don't see how nor where to use the note given in the exercice.

    Any help would be much appreciated. Thank you.
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  2. #2
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    Quote Originally Posted by Simply82358 View Post
    A random variable X has cdf of the form F(x) = \frac{1}{1+e^{-\frac{x-\mu}{\sigma}}}, x is real, \mu is real and \sigma > 0.

    Find the pdf function of X.

    Find the form of the moment generating function Y = \frac{X-\mu}{\sigma} and hence calculate the variance of X

    [Note : the gamma function, \Gamma{(t)} is such that \Gamma''{(1)} - \Gamma'{(1)} = \frac{\pi^2}{6}.

    Letting z = {e^{-\frac{x-\mu}{\sigma}}}.

    The pdf turns out to be f(x) = \frac{d}{dx}F(x) = \frac{z}{\sigma (1+z)^2}.

    Using the fact that if Y = aX+b then  M_y{(t)} = e^{bt}M_x{(at)},
    It turns out that  M_y{(t)} = e^{\frac{-t\mu}{\sigma}} \int^{\infty}_{-\infty} e^{\frac{tx}{\sigma}} f(x) dx.
    Using dx = -\frac{\sigma dz}{z} and  (z e^{\frac{-\mu}{\sigma}})^{-t} = e^{\frac{tx}{\sigma}}
    We then have M_y{(t)} = \int^{\infty}_{-\infty}\frac{z^{-t}z}{\sigma (1+z)^2}dx = \int^{\infty}_{0} \frac{z^{-t}}{(1+z)^2}dz which, according to Maple, is equal to \pi csc(\pi t) t.
    But then I end up with dividing by zero errors when trying to calculate  M'_y{(1)}.
    Plus I don't see how nor where to use the note given in the exercice.

    Any help would be much appreciated. Thank you.
    Your calculations look OK and you got m_Y(t) = \frac{\pi \, t}{\sin (\pi t)}.

    Note that \lim_{t \rightarrow 0} m_Y(t) = 1 which is a good sign.

    But I don't know why you're trying to calculate the derivative of m_Y(t) at t = 1 ......

    To get E(Y) and E(Y^2) (and hence Var(Y) and hence Var(X)) you need to calculate the derivatives of m_Y(t) at t = 0 .....


    And the fact that above I said \lim_{t \rightarrow 0} m_Y(t) = 1 and not m_Y(0) = 1 should give you the clue to what has to be done .....


    You need to calculate the limiting values of the derivatives of m_Y(t) as t \rightarrow 0.
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  3. #3
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    Sorry I meant 0, this was a typo.
    Thank you for the help.
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