# Derivation of a well-known property of the standard normal distribution

I have found the following property of the standard normal distribution:

$$\int_r^\infty xf(x) dx = f(r)$$

where $$f(.)$$ is the pmf of the standard normal distribution.

$$f(r)$$ and $$F(r)$$ are defined as follows:

$$f(r) = \cfrac{1}{\sqrt {2π}} e^{-r^2/2}$$ ; $$\int_r^\infty f(x) dx$$

$$F(.)$$ is the cdf of the standard normal distribution

Does anybody knows how this property is derived or where I can find this derivation?

Steven

It is obtained by just noting that $$-xe^{-x^{2}/2}$$ is the derivative of $$e^{-x^{2}/2}$$, so $$\int_r^{\infty} xf(x)dx =f(r)$$.
Since $$f(r)=\frac{1}{\sqrt{2\pi}}\exp-\frac{r^2}{2}$$, $$f^\prime(r)=-rf(r)$$. The desired result follows by integration, together with $$f(-\infty)=0=\Bbb E X=\int_{-\infty}^\infty xf(x)dx$$.