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The logistic distribution is associated with the CDF $F_X(x)=\dfrac{1}{1+e^{-x}}$, $-\infty$<$x$<$\infty$. Find the PDF of the logistic distribution and show it is symmetric about $x$=$0$.

Taking the derivative of $F_X(x) = \dfrac{e^{-x}}{(1+e^{-x})^2}$ I am not sure how to show it is symmetric about $x=0$. Is it setting $x$ to $0$?

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The function is symmetric about $x=0$ if for any $x$, $f(-x)=f(x)$. You have the pdf $$ f_X(x)=\frac{e^{-x}}{(1+e^{-x})^2}. $$ Take $f_X(-x)$: $$ f_X(-x)=\frac{e^{x}}{(1+e^{x})^2}. $$ Multiply both numerator and denominator by $e^{-2x}$: $$ f_X(-x)=\frac{e^{x}e^{-2x}}{(1+e^{x})^2e^{-2x}}=\frac{e^{-x}}{\bigl((1+e^{x})e^{-x}\bigr)^2}=\frac{e^{-x}}{(1+e^{-x})^2}=f_X(x). $$

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No, showing that $f(~~)$ is symmetric about $0$ is done by demonstrating that $f(0+x)=f(0-x)$ for all $x$.

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  • $\begingroup$ ...and in general $f(~)$ is symmetric about $c$ when $f(c{+}x)=f(c{-}x)$ for all $x$. $\endgroup$ Feb 12, 2020 at 2:33

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