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Wikipedia says this about the law of total covariance (http://en.wikipedia.org/wiki/Law_of_total_covariance):

In probability theory, the law of total covariance,[1] covariance decomposition formula, or ECCE states that if $X$, $Y$, and $Z$ are random variables on the same probability space, and the covariance of $X$ and $Y$ is finite, then

$\text{cov}(X,Y)=E(\text{cov}(X,Y∣Z))+\text{cov}(E(X∣Z),E(Y∣Z))$.

It says the following about probability spaces (http://en.wikipedia.org/wiki/Probability_space):

A probability space consists of three parts:[1][2]

  1. A sample space, $Ω$, which is the set of all possible outcomes.
  2. A set of events $\scriptstyle \mathcal{F}$, where each event is a set containing zero or more outcomes.
  3. The assignment of probabilities to the events; that is, a function $P$ from events to probabilities.

I was not aware of the requirement that the random variables $X,Y$ and $Z$ must share the same probability space and, in particular, a sample space.

Is this always true and, if so, why? A proof, or even some examples, showing that the law fails when the variables have different sample spaces would be much appreciated.

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    $\begingroup$ Even if $X, Y, Z$ are defined on three different sample spaces $\Omega_X, \Omega_Y, \Omega_Z$, as soon as you to start talking of their joint behavior (which you are doing as soon as you mention the word covariance), you are, in effect, defining these variables on the product sample space $\Omega_X \times \Omega_Y \times \Omega_Z$. If the product measure is the product of the individual measures, then the covariances are $0$ and the whole calculation of ECCE is trivial. $\endgroup$ – Dilip Sarwate Nov 18 '14 at 14:42
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    $\begingroup$ You cannot even talk about a joint distribution (thus a covariance) if your variables are not defined on the same sample space! And, that is a very technical point which NEVER (maybe except in quantum mecanics ...) creates problems. You dont need to think about it! $\endgroup$ – kjetil b halvorsen Nov 18 '14 at 14:44
  • $\begingroup$ Thanks @DilipSarwate and kjetil. I see my mistake: they are on the same product sample space $\Omega_X \times \Omega_Y \times \Omega_Z$, even when $\Omega_X$,$\Omega_Y$, and $\Omega_Z$ are three different sample spaces. Thanks. $\endgroup$ – Paul Keating Nov 18 '14 at 14:51

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