# Independence of random variables after a transformation

We are given four random variables $S:=(S_1,S_2,S_3,S_4)$ defined on $(\Omega, \mathcal A,P)$. The random variables can be viewed as being extracted from a stochastic process. I assume that all combinations of $S_i$'s have strictly positive marginal densities.

I apply transformation $F$ to them which is defined as:

$$F(s_1,s_2,s_3,s_4) := (F_{S_1}(s_1),F_{S_2}(s_2),F_{S_3|S_1}(s_3,s_1),F_{S_4|S_2,S_1}(s_4,s_2,s_1))$$

where

• $F_{S_1}$ is the cdf of $S_1$;
• $F_{S_2}$ is the cdf of $S_2$;
• $F_{S_3|S_1}$ is the cdf of $S_3$ conditional on $S_1$;
• $F_{S_4|S_2,S_1}$ is the cdf of $S_4$ conditional on $S_1$ and $S_2$.

The idea behind the transformation: $F$ is the correct one period ahead density forecast.

My goal is to show that $$Z=(Z_1,Z_2,Z_3,Z_4):=F(S_1,S_2,S_3,S_4)$$ has certain properties. Specifically:

1. all $Z_i$ are uniformly distributed;
2. $Z_1$ and $Z_3$ are independent and $Z_2$ and $Z_4$ are independent.

My attempt

1. I use the following representation of the density of $S$ using the conditional densities.
$$f_{S}(s_1,s_2,s_3,s_4) =f_{S_1}(s_1)f_{S_2|S_1}(s_2,s_1)f_{S_3|S_1,S_2}(s_3,s_2,s_1)f_{S_4|S_1,S_2,S_3}(s_4,s_3,s_2,s_1)$$

2. I use (without a proof) the fact that $F$ is a diffeomorphismus. So using the well known result about a transformation-diffeomorphismus:

$$f_Z(z_1,z_2,z_3,z_4)=\frac {f_{S_1}(s_1)f_{S_2|S_1}(s_2,s_1)f_{S_3|S_1,S_2}(s_3,s_2,s_1)f_{S_4|S_1,S_2,S_3}(s_4,s_3,s_2,s_1)} {f_{S_1}(s_1)f_{S_2}(s_2)f_{S_3|S_1}(s_3,s_1)f_{S_4|S_2,S_1}(s_4,s_2,s_1)}I_{(0,1)^4}(z_1,z_2,z_3,z_4)$$

where all $s_i$ are to be understood as functions of all $z_i$, the relationship between all $s_1$ and all $z_i$ is given by $F^{-1}(z_1,z_2,z_3,z_4) \to (s_1,s_2,s_3,s_4)$, which exists by the assumption above.

1. Now to achieve the goal stated above I need to integrated out from $f_Z$ firstly $z_1,z_3$ to ascertain that $z_2,z_4$ are uniformly distributed and independent, and then to do the same for $z_1,z_3$. But here I am stuck...

For any continuous random variable $X$ with cdf $F$, it is true that $F(X)$ has the uniform distribution on $(0,1)$. "Continuous" means $\Pr(X=x)=0$ for each $x$ (and this is weaker than the existence of a density). That shows that $Z_1$ is uniform under the continuity assumption of (the law of) $S_1$.
Similarly, if one has a regular conditional distribution ${\cal L}(S_3 \mid S_1=s_1)$ that is continuous for all $s_1$ (or almost all, with respect to the law of $S_1$), then $Z_3=F_{S_3\mid S_1}(S_3,S_1)$ has a conditional uniform distribution given $S_1$. Since this distribution does not depend on $S_1$, the marginal ("unconditional") law of $Z_3$ also is the uniform distribution.
To show that $Z_1$ and $Z_3$ are independent, use conditonning with respect to $S_1$: $$\Pr(Z_1 \leq z_1, Z_3 \leq z_3 \mid S_1)={\boldsymbol 1_{Z_1\leq z_1}} \Pr( Z_3 \leq z_3 \mid S_1),$$ because $Z_1$ is $\sigma(S_1)$-measurable (in other words, it is a measurable function of $S_1$). Now, $\Pr( Z_3 \leq z_3 \mid S_1)=z_3$ because $Z_3$ is conditionnally uniform given $S_1$, as we said before. Thus $$\Pr(Z_1 \leq z_1, Z_3 \leq z_3 \mid S_1)={\boldsymbol 1_{Z_1\leq z_1}} z_3$$ and taking the expectation, $$\Pr(Z_1 \leq z_1, Z_3 \leq z_3) = \Pr(Z_1\leq z_1)z_3 = \Pr(Z_1\leq z_1)\Pr(Z_3\leq z_3),$$ by using the fact that $Z_3$ is unconditionnaly uniform too. That shows the independence.
Same reasoning for $Z_2$ and $Z_4$
• Thanks for the answer, I am not getting how one can prove that "the marginal ("unconditional") law of $Z_3$ also is the uniform distribution". – Sergey Zykov Jul 28 '15 at 16:42
• @SergeyZykov The marginal law is an average of the conditional laws. The conditional laws are all the same (uniform for any $s_1$), the average is the same. – Stéphane Laurent Jul 28 '15 at 17:12