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

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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?
 
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  • #2
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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?
 
  • #3
mathman
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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?
It looks like you are dealing with a normal distribution. dF/dσ = (dF/dx)dx/dσ, where dx/dσ=-x/σ
 
  • #4
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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
Why don't we have to treat σ and x as two independent variables here?
 
  • #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.
 
  • #6
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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.
 
  • #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)
 
  • #8
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That is an interesting method :).
 

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