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**Simply82358** A random variable $\displaystyle X$ has cdf of the form $\displaystyle F(x) = \frac{1}{1+e^{-\frac{x-\mu}{\sigma}}}$, $\displaystyle x$ is real, $\displaystyle \mu$ is real and $\displaystyle \sigma > 0$.

Find the pdf function of $\displaystyle X$.

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

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

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

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

Using the fact that if $\displaystyle Y = aX+b$ then $\displaystyle M_y{(t)} = e^{bt}M_x{(at)}$,

It turns out that $\displaystyle M_y{(t)} = e^{\frac{-t\mu}{\sigma}} \int^{\infty}_{-\infty} e^{\frac{tx}{\sigma}} f(x) dx.$

Using $\displaystyle dx = -\frac{\sigma dz}{z}$ and $\displaystyle (z e^{\frac{-\mu}{\sigma}})^{-t} = e^{\frac{tx}{\sigma}} $

We then have $\displaystyle M_y{(t)} = \int^{\infty}_{-\infty}\frac{z^{-t}z}{\sigma (1+z)^2}dx = \int^{\infty}_{0} \frac{z^{-t}}{(1+z)^2}dz$ which, according to Maple, is equal to $\displaystyle \pi csc(\pi t) t$.

But then I end up with dividing by zero errors when trying to calculate $\displaystyle M'_y{(1)}$.

Plus I don't see how nor where to use the note given in the exercice.

Any help would be much appreciated. Thank you.