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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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  • $\begingroup$ Also asked on stats.SE where it has already received an answer. $\endgroup$ – Dilip Sarwate Jul 20 '15 at 12:49
  • $\begingroup$ The answer on stats.SE was not accepted because I saw it as not detailed/ rigorous enough. So all further answers are welcome. $\endgroup$ – Sergey Zykov Jul 29 '15 at 8:27

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