Hi all,

Is there a reason why for standard normal distributions,

$\displaystyle -\phi'(z)=z\phi(z)$

?

Thanks.

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- Dec 3rd 2008, 07:45 PMsoloakewlstickstandard normal density special property?
Hi all,

Is there a reason why for standard normal distributions,

$\displaystyle -\phi'(z)=z\phi(z)$

?

Thanks. - Dec 3rd 2008, 08:06 PMcl85
Because $\displaystyle \phi(z)=e^{-z^2/2}/\sqrt{2\pi}$?