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Math Help - standard normal density special property?

  1. #1
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    standard normal density special property?

    Hi all,

    Is there a reason why for standard normal distributions,

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

    ?

    Thanks.
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  2. #2
    Junior Member
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    Because \phi(z)=e^{-z^2/2}/\sqrt{2\pi}?
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