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

[TaPaKaH]

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?

 

[mfb]

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?

 

[mathman]

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

 

[mfb]

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?

 

[mathman]

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.

 

[mfb]

With the method I described, you don't need any integrals, and you get the right result with two easy derivatives.

 

[mathman]

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

With a simple change of variables in the integrand: F_σ(x)=F₁(x/σ)
dF₁(x/σ)/dσ = (-x/σ²)f₁(x/σ) = (-x/σ)f_σ(x)

 

[mfb]

That is an interesting method :).