# Proof Regarding Functions of Independent Random Variables

1. Aug 21, 2011

### SpringPhysics

1. The problem statement, all variables and given/known data
Let X and Y be independent random variables. Prove that g(X) and h(Y) are also independent where g and h are functions.

2. Relevant equations
I did some research and somehow stumbled upon how
E(XY) = E(X)E(Y)
is important in the proof.

f(x,y) = f(x)f(y)
F(x,y) = F(x)F(y)

3. The attempt at a solution
I am seriously stuck on this - I do not even know where to begin. I know there was a previous thread (around 7 years old) on this but I did not want to revive an old thread, and there wasn't much response in that thread.

I attempted to determine F(g(x),h(y)) by integration but then I would need f(g(x),h(y)), which (to my knowledge), is unknown unless I use a really messy transformation equation (and the question was stated prior to learning about transformations).

Expectations appear to be the way to go, since they do not require the actual density/probability function of the function of the random variable; however, I do not know of any property of expectations allowing us to deduce that two random variables are independent (since a covariance of 0 does not necessarily imply independence).

If anyone could provide me with somewhere to start, that would be much appreciated.

2. Aug 21, 2011

### LCKurtz

Two random variables X and Y defined on the same sample space are independent if for for subsets A and B of real numbers

P(X ε A, Y ε B) = P(X ε A)P(Y ε B)

You need to prove that for f(X) and g(Y)

Start like this:

P(f(X) ε A, g(Y) ε B) = P(X ε f-1(A), P(Y ε g-1(B))

and see if you can get it from there.

Last edited: Aug 21, 2011
3. Aug 27, 2011

### SpringPhysics

Wouldn't we be assuming that f and g are invertible? Or does that have no bearing on the proof? (I always got stuck because I thought we couldn't apply f or g inverse since we would have to then assume that f and g are invertible functions.)

4. Aug 27, 2011

### Ray Vickson

f and g need not be invertible. f^(-1)(A) is standard notation for the set {w:f(w) \in A}.

Anyway, another approach is to look at the multivariate characteristic function: two random variables Y1 and Y2 are independent iff their joint characteristic function factors: that is, Eexp(i*k1*Y1 + i*k2*Y2) = [E(exp(i*k1*Y1))]*[E(exp(i*k2*Y2))] for all (k1,k2),
where i = sqrt(-1).

RGV