I have two questions about transformation functions in probability theory. I wrote my answers to them but I'm not sure if I am right.
Let's look at the theorem below (I took it from here (the last)).
Theorem. Let X be an absolutely continuous random variable with probability density function fX(x) and support RX={x∈R:f(x)>0}. Let g:R→R be one-to-one and differentiable on the support of X. If
dydg−1(y)=0,∀y∈g(RX)
then the probability density of Y is
fY(y)=fX(g−1(y))∣∣∣∣∣dydg−1(y)∣∣∣∣∣,∀y∈g(RX).
Questions.
1) Can I choose any open subset A, RX⊆A⊆R which contains support RX as the domain of g in the thorem?
As I understood R is a "natural domain" (i.e. the largest possible domain) for g but we can always limit it for a specific function g on practice. I think set A must be open to ensure the existence of the derivative of g in all points of RX (set RX can be closed).
For example it will be convenient to choose A=(0,∞) as the domain for a function g:A→R, g(x)=ln(x) if X can have only positive values.
Or I have to choose R as the domain and somehow define function g on (−∞,0] ?
2) Why the theorem requires function g to be one-to-one instead of to be invertible?
I think that the domain of a function g−1 is exactly the set RY=g(RX). Hence if g is one-to-one then g will be invertible function. Even if g is formally not an onto function (not all points in codomaing=R have pre-images).
Please correct me if I'm wrong. Thanks.
Let's look at the theorem below (I took it from here (the last)).
Theorem. Let X be an absolutely continuous random variable with probability density function fX(x) and support RX={x∈R:f(x)>0}. Let g:R→R be one-to-one and differentiable on the support of X. If
dydg−1(y)=0,∀y∈g(RX)
then the probability density of Y is
fY(y)=fX(g−1(y))∣∣∣∣∣dydg−1(y)∣∣∣∣∣,∀y∈g(RX).
Questions.
1) Can I choose any open subset A, RX⊆A⊆R which contains support RX as the domain of g in the thorem?
As I understood R is a "natural domain" (i.e. the largest possible domain) for g but we can always limit it for a specific function g on practice. I think set A must be open to ensure the existence of the derivative of g in all points of RX (set RX can be closed).
For example it will be convenient to choose A=(0,∞) as the domain for a function g:A→R, g(x)=ln(x) if X can have only positive values.
Or I have to choose R as the domain and somehow define function g on (−∞,0] ?
2) Why the theorem requires function g to be one-to-one instead of to be invertible?
I think that the domain of a function g−1 is exactly the set RY=g(RX). Hence if g is one-to-one then g will be invertible function. Even if g is formally not an onto function (not all points in codomaing=R have pre-images).
Please correct me if I'm wrong. Thanks.