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Consider the probability space $(\Omega_1,2^{\Omega_1},\mathbb{P}_1)$, where $\Omega_1$ is a sample space, $2^{\Omega_1}$ (powerset of $\Omega_1$) is the $\sigma$-algebra over $\Omega_1$, and $\mathbb{P}_1$ is a probability measure. Likewise, consider the probability space $(\Omega_2,2^{\Omega_2},\mathbb{P}_2)$.

In particular, suppose the first probability space corresponds to the probabilistic experiment of choosing a plaintext at random from the set $\Omega_1 = \{a,b\}$, where $\mathbb{P}_1(a) = 1/4$ and $\mathbb{P}_1(b) = 3/4$. Similarly, suppose that the second probability space corresponds to the probabilistic experiment of choosing a key at random from the set $\Omega_2 = \{K_1,K_2,K_3\}$, where $\mathbb{P}_2(K_1) = 1/2$ and $\mathbb{P}_2(K_2) = \mathbb{P}_2(K_3) = 1/4$.

Now, I'd like to "combine" these two random experiments into a single random experiment whose probability space is such that its sample space is the cross product of the two component sample spaces. That is, the combined probability space is $(\Omega_1 \times \Omega_2, 2^{\Omega_1 \times \Omega_2}, \mathbb{P} )$ that reflects the fact these probabilistic experiments are "independent". That is, $\mathbb{P}(\{(p,k)\}) = \mathbb{P_1}(p)\mathbb{P_2}(k)$. Is there a formal way to do this?

The problem as I see it is that, from what I've learned so far about probability measure theory, the concept of independence applies only to events within the same $\sigma$-algebra. Thus, within this formal framework, it doesn't seem to make sense to talk about the "independence" of the event of picking some plaintext and the event of picking some key, since plaintexts and keys belong to completely different sample spaces (and completely different probability spaces) in this context.

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  • $\begingroup$ I am not sure whether I understand your concern exactly. However, I think you can simply define the product measure. The multiplication identity does not necessarily mean independence and should not be interpreted so. $\endgroup$ – Ran Wang Feb 23 '17 at 19:52
  • $\begingroup$ I suppose I am asking, what does it mean to say that two events belonging to completely different sample spaces are independent? Can this question only be answered by first defining the product measure on these sample spaces? $\endgroup$ – David Smith Feb 23 '17 at 19:54
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    $\begingroup$ In general, I think, you cannot define independence for different sample space, which you have correctly point out. Second, it should be noted even the idea of independence is measure specific, that is you can have two events that are independent in one measure while not so in another measure. $\endgroup$ – Ran Wang Feb 23 '17 at 19:57
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I think you're looking for the product measure

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  • $\begingroup$ In practice, do we use the product measure to combine any number of distinct random experiments? For instance, can the random experiment of flipping a coin 100 times be interpreted as the product of single coin flips? $\endgroup$ – David Smith Feb 23 '17 at 20:00
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    $\begingroup$ For a finite number of distinct experiments, yes. For an infinite number of distinct random experiments, I don't know and would guess there are complications. $\endgroup$ – FalafelPita Feb 23 '17 at 20:09
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    $\begingroup$ This also works for an arbitrary collection of random experiments. My answer here is about a totally different question, but it provides some references to the literature regarding infinite products of probability spaces. (Interestingly, however, you cannot take infinite products of arbitrary measure spaces. Each of the individual measure spaces in the product needs to have measure one.) $\endgroup$ – Josse van Dobben de Bruyn Jul 18 '17 at 20:23

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