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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...
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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 $Z_4$

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  • $\begingroup$ 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". $\endgroup$ – Sergey Zykov Jul 28 '15 at 16:42
  • $\begingroup$ @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. $\endgroup$ – Stéphane Laurent Jul 28 '15 at 17:12
  • $\begingroup$ I tried to put the intuition you gave to me in the previous comment in more rigorous terms here. Do you mind to take a look and let me know whether I got you right? $\endgroup$ – Sergey Zykov Aug 7 '15 at 17:33
  • $\begingroup$ after some time spent with probability books I see the mathematical background of the proposed solution, now I guess I can accept the answer. $\endgroup$ – Sergey Zykov Aug 17 '15 at 8:28

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