Assume that $$X=(X_1, ..., X_n),: (\Omega, A,P)\to (\{0,1\}^n, 2^{{\{0,1\}}^n})$$ and $$Y=(Y_1, ..., Y_n):(\Omega, A,P)\to (\{0,1\}^n, 2^{{\{0,1\}}^n})$$are two random Variables that have binary RVs as their components (Therefore $X_i(\omega)\in\{0,1\}, Y_i(\omega) \in \{0,1\}$) and both ($X$ and $Y$) are exchangeable, i.e. $$P((X_1, ..., X_n)=(x_1, ..., x_n))= P((X_{\sigma(1)}, ..., X_{\sigma(n)})=(x_1, ..., x_n))$$


$$P((Y_1, ..., Y_n)=(y_1, ..., y_n))= P((Y_{\sigma(1)}, ..., Y_{\sigma(n)})=(y_1, ..., y_n))$$ for all permutations $\sigma$.

My question is whether it holds that $Z=(X_1Y_1, ..., X_nY_n)$ is exchangeable?

Or framed differently which assumptions are neccessary for $Z$ to be exchangeable?

  • $\begingroup$ It looks there is at least one typographical error in your question: do you really mean that the last component of $Z$ is "$Y_nY_n$?" The notation is opaque: are you claiming $X$ is random variable with binary components and $Y$ is a random variable whose components are binary functions of binary $n$-vectors? When you state a problem this abstractly, (1) it's crucial you get everything exactly right and (2) you should consider posting it on the math site instead. $\endgroup$
    – whuber
    Oct 16 '18 at 16:18
  • $\begingroup$ Thanks for pointing this out. I will clarify the notation $\endgroup$
    – Sebastian
    Oct 16 '18 at 16:25

The product does not have to be exchangeable. The following counterexample will show what can go wrong and why.

We will specify the joint distributions $P_1$ of $(X_1,Y_1)$ and $P_2$ of $(X_2,Y_2)$ and assume each of these bivariate random variables is independent. Thus, the $X_i$ will be exchangeable provided they are identically distributed, and likewise for the $Y_i.$ All variables will be Bernoulli variables: by definition, their probabilities will be concentrated on the set $\{0,1\}.$

Let $P_1(0,0) = P_1(1,1) = 1/2$ and $P_2(x,y)=1/4$ for $x,y\in\{0,1\}.$

Since all marginal distributions are Bernoulli$(1/2),$ the marginal exchangeability assumption holds. But now compute that $\Pr(X_1Y_1=0) = 1/2$ and $\Pr(X_2Y_2=0)=3/4,$ showing the products have different distributions (and therefore cannot be exchangeable).

This shows that the joint distribution matters.

However, the joint distributions could differ, yet the products could be exchangeable, so exchangeability of the bivariate random variables $(X_i,Y_i)$, albeit a sufficient condition for exchangeability of the products $X_iY_i,$ is not a necessary condition.

An example of this is given by ternary variables with values in $\{-1,0,1\}.$ For instance, consider the following probabilities:

$$P_1((-1,y)) = 1/6\quad(y\in\{-1,0,1\});\quad P_1((1,-1)) = P_1((1,1))=1/4$$


$$P_2((x,y)) = P_1((-x,y)).$$

It is straightforward to check that the marginal distributions of the $X_i$ assign equal probabilities of $1/2$ to $\pm 1,$ the marginal distributions of the $Y_i$ have probability vectors $(5/12, 1/6, 5/12),$ and that the distribution of the $X_iY_i$ is the same as that of the $Y_i.$ Note that the $(X_i,Y_i)$ have different distributions, though, because

$$P_1((-1,0)) = 1/6 \ne 0 = P_2((-1,0)).$$

Thus the $X_i$ are exchangeable, the $Y_i$ are exchangeable, the $X_iY_i$ are exchangeable, but the $(X_i,Y_i)$ are not exchangeable.


No. Suppose that the sample space consists of three equally likely outomes for which $X$ takes values of $$(1,0,0), (0,1,0), (0,0,1)$$ and for which $Y$ takes values of $$(1,0,0), (0,0,1), (0,1,0).$$ Then $X_1,X_2,X_3$ are exchangeable and so are $Y_1,Y_2,Y_3$. But the corresponding values of $Z=(X_1 Y_1,\dots, X_3 Y_3)$ are $$ (1,0,0),(0,0,0),(0,0,0) $$ so clearly $Z_1,Z_2,Z_3$ are not exchangeable.


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