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A derivative equality for density+distribution funs of N(0,1)

  1. Oct 15, 2013 #1
    An equality for density+distribution funs of N(0,1)

    Reading through a book, I met the following equality (##F## is cumulative distribuion function, ##f## is density function)
    $$\frac{d}{d\sigma}F_{N(0,\sigma^2)}(x)=\frac{-x}{\sigma}f_{N(0,\sigma^2)}(x)$$
    which was given without any futher explanations (assumed obvious, I guess) but I have a hard time figuring out why it holds.

    Could anyone, please, provide a hint on how to prove it?
     
    Last edited: Oct 15, 2013
  2. jcsd
  3. Oct 15, 2013 #2

    mfb

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    What happens if you derive both sides with respect to x? You can exchange the order of the derivatives.
    What happens for x -> minus infinity?
     
  4. Oct 15, 2013 #3

    mathman

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    It looks like you are dealing with a normal distribution. dF/dσ = (dF/dx)dx/dσ, where dx/dσ=-x/σ
     
  5. Oct 15, 2013 #4

    mfb

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    Why don't we have to treat σ and x as two independent variables here?
     
  6. Oct 16, 2013 #5

    mathman

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    I may have mislead you. However I looked at it using brute force and the result comes out. It requires integration by parts in the middle of the derivation, since you will get a terms looking similar to F but with x2 in the integrand.
     
  7. Oct 16, 2013 #6

    mfb

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    With the method I described, you don't need any integrals, and you get the right result with two easy derivatives.
     
  8. Oct 17, 2013 #7

    mathman

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    Notation simplification: Fσ(x) and fσ(x) normal distribution and density functions with mean 0 and std. dev. σ.

    With a simple change of variables in the integrand: Fσ(x)=F1(x/σ)
    dF1(x/σ)/dσ = (-x/σ2)f1(x/σ) = (-x/σ)fσ(x)
     
  9. Oct 17, 2013 #8

    mfb

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    That is an interesting method :).
     
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