Hi I'm working with a model of gaussian demand the following is supposed to be a readily derived property however I was unable to reproduce it myself. Why is $f(x)>xF(x)$ for $F(x)=\int_{x}^{\infty}\dfrac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}t^2}dt$ where $f(x)=\dfrac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}dx$ is the first derivative of F(x) with respect to x.
Thanks for your help
Let $x>0$. Note that $\int_x^{\infty} e^{-t^{2}/2}dt < \frac 1 x \int_x^{\infty} t e^{-t^{2}/2}dt$ and the integral here can be evaluated easily since the derivative of $e^{-t^{2}/2}$ is $-te^{-t^{2}/2}$. The inequality is trivially true of $x \leq 0$. .