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

## Main Question or Discussion Point

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:

Related Set Theory, Logic, Probability, Statistics News on Phys.org
mfb
Mentor
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
Mentor
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
Mentor
With the method I described, you don't need any integrals, and you get the right result with two easy derivatives.

mathman