# How to parametrize two dependent random variables

I read the book 'Numerical Methods for Stochastic Computations: A Spectral Method Approach' by Dongbin Xiu, Chapter 4. The main goal of the problem is, given random variables $Y_1, ...,Y_n$ (on a probability space, say, $\Omega$), find independent random variables $Z_1,...,Z_d, 1\le d\le n$, so that $Y=T(Z)$ for some transformation function $T$.

Note that two random variables $Y_1, Y_2$ are independent if $P(Y_1\in B_1 \cap Y_2\in B_2)=P(Y_1\in B_1)P(Y_2\in B_2)$ for any measurable sets $B_1, B_2$, where $P$ is a probability measure (Don't be confused with linear independence).

Anyway, my problem is when $n=2$. If $Y_1, Y_2$ are independent, we can take $Z_i=Y_i$.
When $Y_1$ and $Y_2$ are not independent, the book says that there exists a (nonzero) function $f$ such that $$f(Y_1,Y_2)=0$$ so that we can find a random variable $Z(\omega)$ to parametrize $Y_1=a_1\circ Z, Y_2=a_2\circ Z$ and $f(a_1,a_2)=0$. Or equivalently, we may find a function $g$ so that $Y_2=g(Y_1)$, so that $Z=Y_1, Y_2=g\circ Z$. (So we found one variable $Z$, which is cleary satisfies the independent statement, and a transfrom $a_i$ or $g$)

First, I don't know how to find such functions $f, g, a, b$ and second, I don't know why those two results are equivalent. Any references will be appreciated!

• Something is very wrong here. Consider for example any non independent couple $(Y_1,Y_2)$ whose PDF has full support in $\mathbb R^2$, then $$P(f(Y_1,Y_2)=0)=1$$ only if $f=0$ almost everywhere for the Lebesgue measure on $\mathbb R^2$. Thus the assertion that "there exists a nonzero function $f$ such that..." is wrong. Later on, the non independence of $(Y_1,Y_2)$ does not imply that "we may find a function $g$ so that $Y_2=g(Y_1)$", by the same counterexample. Either you miscopied the book, or the book is wrong. – Did Nov 25 '17 at 11:07
• I fully agree with you. I checked whether there is additional statements, but no. I just hope there is any related articles are something, but I think it is a wrong question. Thank you very much! – CSH Nov 25 '17 at 11:20