How Can I Prove the Independence of Functions of Independent Random Variables?

[franz32]

Hello, I need some help on the independence of random variables...
"How do I prove that if X and Y are two independent random variables, then U=g(X) and V = h(Y) are also independent?"

- Isn`t that if random variables X and Y are independent, it implies
that f(x,y) = g(x)h(y) and vice versa? Also, note that g(x) and h(y) are
two marginals. But what I don`t understand is that what does it mean to
have U = g(X) to be a capital "X"?

- {then U=g(X) and V = h(Y) are also independent} what am I supposed to
show in this proof? And lastly, what is my first step/strategy in proving
this? Hope you can give me hints.. =)

 

[matt grime]

U is g of the random variable X. You do this all the time, such as when working out the variance: it is
E(X^2)-E(X)^2, So there's a function of a random variable there (X^2).

Are these continuous of discrete R.V.'s? Not that it matters two much. If it 's continuous look at the pdf of the joind dist. since g is a function of X alone, and h a function of Y alone the double integrals INT dxdy split as int dx int dy. If discrete replace integrals with sums.

 

[franz32]

Sir matt grime/anyone... =]
I hope someone can guide me.
I want to prove first the continuous. So, the joint pdf can be described as

f[g(X),h(Y)] = INTaINTb g(x)h(y) dx dy -> am I right here?
where a and b are arbitrary intervals.
= INTa h(y) [INTb g(x)dx] dy -> h(y) is treated as a constant.
= [INTb g(x)dx] [INTa h(y)dy] -> [INTb g(x)dx] is now a constant
= g(X) h(Y)

I believe I got screwed up in my notations... is this the proof? I hope it is.. but can someone help me edit this... will I use u's and v's here?... I think not.

For the discrete case...

f(g(X), h(Y)) = P(U = g(X), V = h(Y)) = P(U = g(X)) P(V = h(Y)) = g(X)h(Y)?

Is this the right proof? I hope someone can help me.. =]

 

[relinquished™]

f[g(X),h(Y)] = INTaINTb g(x)h(y) dx dy -> am I right here?
where a and b are arbitrary intervals.
= INTa h(y) [INTb g(x)dx] dy -> h(y) is treated as a constant.
= [INTb g(x)dx] [INTa h(y)dy] -> [INTb g(x)dx] is now a constant
= g(X) h(Y)

I believe you made a mistake here, it's not

$$f[g(X),h(Y)] = g(x) \cdot h(y)$$

but it's

$$<br /> f(X,Y) = g(x) \cdot h(y)<br />$$

You were on the right track, but it should be

$$f(U,V) = g(U)h(V) = g(g(x)) \cdot h(h(y))$$

I believe that U and V (g(x) and h(y), respectively) should be independent since Y cannot influence g(x) and X cannot influence h(y) since X and Y are independent. I just don't know how to prove it in mathematical notation, but it's worth a try =)