Find the PDF of the logistic distribution with CDF $F_X(x)=\frac{1}{1+e^{-x}}$

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 $$0$$?

2 Answers

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 $$e^{-2x}$$: $$f_X(-x)=\frac{e^{x}e^{-2x}}{(1+e^{x})^2e^{-2x}}=\frac{e^{-x}}{\bigl((1+e^{x})e^{-x}\bigr)^2}=\frac{e^{-x}}{(1+e^{-x})^2}=f_X(x).$$

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

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