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Let $(X,\mathcal{F}_X,\mathbb{P})$ and $(Y,\mathcal{F}_Y,\mathbb{Q})$ be two probability spaces. I know that the expectation of random variable $Z:X\rightarrow \mathbb{R}$ is affected by the random variable $W:Y\rightarrow \mathbb{R}$. So I want to work with a conditional expectation - something like $\mathbb{E}(Z\mid W)$.

So my questions are :

  1. Is it OK to define conditional expectation if the random variables are from different probability spaces?

  2. If so, would I have all the usual properties of a conditional expectations?

  3. If not, how should I handle it? Should I consider the product of two probability spaces?

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  • $\begingroup$ The question of what is the underlying probability space is a modeling problem and isnt really important. $\endgroup$
    – random0620
    Commented Jun 20 at 6:03
  • $\begingroup$ When you see conditional expectation, just think its some function of $W$. $\endgroup$
    – random0620
    Commented Jun 20 at 6:04
  • $\begingroup$ In general on an abstract probability space there's no straightforward way to calculate it. If you know the joint density then there are simple formulas. $\endgroup$
    – random0620
    Commented Jun 20 at 6:05
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    $\begingroup$ A random variable on one space is measured with respect to events on that space, it does not make sense to say it is affected by something outside the space. You need both $W$ and $Z$ to be on the same space to take expectation of one with respect to another. Product space concepts are used to model a problem. In this case, they would make $Z$ and $W$ independent. If you have some other relationship between $Z$ and $W$ then you are already assuming they are on the same space. $\endgroup$
    – Michael
    Commented Jun 27 at 0:30

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The answer to the question 1 is no. So we need to somehow merge the two probability spaces $X$ and $Y$ into a single one, where both $Z$ and $W$ can coexist.

This can be easily achieved if only the distributions of $Z$ and $W$ are involved, instead of the whole construction of $Z$ and $W$. However, in general, we need more specifications about $Z$ and $W$. If such details are provided, I am willing to edit post to provide a further answer.

$W$ is continuous, nonnegative random variable, $Z$ is discrete with $\mathbb{P}(Z=a)=p$, and $\mathbb{P}(Z=b)=1-p$. $\mathbb{E}(W|Z)=\alpha Z\,\text{a.s.}$, where $\alpha>0$.

Under the situation above (and under $0<a<b$), we can construct $Z$ and $W$ as follows:

  • Prepare a probability space as $([0,1],\mathcal{B}([0,1]),\lambda)$ and denote $\Omega:=[0,1]$.
  • Construct $Z$ as $$ Z(\omega):=a1_{[0,p]}(\omega)+b1_{[p,1]}(\omega),\qquad\omega\in\Omega. $$
  • Construct $W$ as the piecewise linear function connecting the points $(0,0),(p,2\alpha a),(1,2\alpha b-2\alpha a)$.

This construction leads to $Z$ being discrete, $W$ continuous and nonnegative, and $\mathbb{E}(W|Z)=\alpha Z\;\text{a.s.}$

Note that the conditional expectation $\mathbb{E}(W|Z)$ (conditioned upon a discrete variable $Z$) is defined to be $$ \mathbb{E}(W|Z)(\omega)=\mathbb{E}(1_{\{Z=a\}}W)1_{\{Z=a\}}(\omega)+\mathbb{E}(1_{\{Z=b\}}W)1_{\{Z=b\}}(\omega).$$

My favorite references include Jacod and Protter (2004) Chapter 23, and Dudley (2002) Chapter 10.

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  • $\begingroup$ Thank you for your answer. Here are details on $X$ and $Z$. If you could help me merge the spaces, I would greatly appreciate it. $X$ is continuous, nonnegative random variable, $Z$ is discrete with $\mathbb{P}(Z=a)=p$, and $\mathbb{P}(Z=b)=1-p$. $\mathbb{E}(X)=\alpha Z$, where $\alpha>0$. $\endgroup$
    – curiosity
    Commented Jun 20 at 10:54
  • $\begingroup$ Thank you for your clarification. The terms 'continuous', 'nonnegative', and formulae involving $\mathbb{P}$ and $\mathbb{E}$ are all conditions on the distributions of $Z$ and $W$, not the specific construction as a mapping from $X$ or $Y$ to $\mathbb{R}$. In the situations like this, there exists a probability space that accommodate both $Z$ and $W$. You can think of this as a finite dimensional version of Kolmogorov extension theorem. $\endgroup$
    – Hirofumi Shiba
    Commented Jun 21 at 3:42
  • $\begingroup$ Just pretend we are dealing with such constructed $Z$ and $W$ from the outset, and forget about which probability space they actually live in. $\endgroup$
    – Hirofumi Shiba
    Commented Jun 21 at 3:43
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    $\begingroup$ Anyway, assuming you mean $\mathbb{E}(X|Z)=\alpha Z$, it would be easy to construct an example if we take $[0,1]$ as the accommodating probability space for $Z,W$. We can just take $Z:=a1_{[0,p]}+b1_{[p,1]}$ and $X:=\alpha Z$. $\endgroup$
    – Hirofumi Shiba
    Commented Jun 21 at 3:53
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    $\begingroup$ Of course! Its argument is any point in the sample space $\Omega=[0,1]$. We consider this $\Omega$ as a probability space $(\Omega,\mathcal{B}([0,1]),\lambda)$ with the usual Bore $\sigma$-algebra $\mathcal{B}([0,1])$ and the Lebesgue measure $\lambda$ on $[0,1]$. $\endgroup$
    – Hirofumi Shiba
    Commented Jun 21 at 4:10

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