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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

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It is obtained by just noting that $-xe^{-x^{2}/2}$ is the derivative of $e^{-x^{2}/2}$, so $\int_r^{\infty} xf(x)dx =f(r)$.

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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 $f(-\infty)=0=\Bbb E X=\int_{-\infty}^\infty xf(x)dx$.

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  • $\begingroup$ integration by parts? $\endgroup$ – Steven31415 Apr 26 '19 at 9:17
  • $\begingroup$ @FromTheoryToPractice No, I just meant the fundamental theorem of calculus. $\endgroup$ – J.G. Apr 26 '19 at 9:17

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