Independence of Product Variables in Random Vectors

Consider a random vector $$U = (U_1,U_2)$$ where $$U_1$$ and $$U_2$$ are independent and identically distributed. Let $$X =(X_1,X_2)$$ with $$X_1$$ and $$X_2$$ being dependent random variables.

Suppose that $$X = (X_1,X_2)$$ and $$U = (U_1,U_2)$$ are independent. Define $$Z_1 = X_1 U_1$$ and $$Z_2 = X_2 U_2$$. The question at hand is to determine whether the variables $$Z_1$$ and $$Z_2$$ are independent.

Attempt: To investigate the independence of $$Z_1$$ and $$Z_2$$, we need to analyze their joint distribution. Let $$f_{Z_1,Z_2}(z_1,z_2)$$ denote the joint distribution of $$Z_1$$ and $$Z_2$$. We can express this joint distribution as:

$$f_{Z_1,Z_2}(z_1,z_2) = \int_{-\infty}^{\infty} f_{X_1,X_2}(z_1/u_1,z_2/u_2) \, f_{U_1,U_2}(u_1,u_2) \, du_1 \, du_2.$$

If this expression does not factorize into the product of the marginal distributions of $$Z_1$$ and $$Z_2$$, then $$Z_1$$ and $$Z_2$$ are not independent.

Is there a counterexample or is this true?

1 Answer

Certainly they're not independent in general. Here's a counterexample where the covariance of the $$Z$$'s will not be zero and hence they cannot be independent.

Suppose all the relevant expectations are finite:

$$\begin{eqnarray} \text{Cov}(X_1 U_1, X_2 U_2) &=& E(X_1 X_2 U_1 U_2) - E(X_1 U_1) E(X_2 U_2)\\ &=& E(X_1 X_2) E(U_1) E(U_2) - E(X_1) E(U_1) E(X_2) E(U_2)\\ &=& E(U_1)E(U_2) \text{Cov}(X_1,X_2) \end{eqnarray}$$

which is only zero when one of those terms is $$0$$, which we can readily construct so as not to be the case.

e.g. make $$U$$'s standard uniform and let $$X_1=X_2$$ also be standard uniform. We could proceed algebraically of course but all we really need is to double check that the above argument works, so a simulation will do:

 n=10000
u1=runif(n);u2=runif(n);x1=runif(n);x2=x1
cor(u1*x1,u2*x2)
[1] 0.4271108


Visually the dependence between $$Z_1$$ and $$Z_2$$ isn't super strong in this example (unless you know what you're looking for), but you can very clearly see it in their square roots:

When looking for counterexamples, often just thinking of very simple possibilities can be fruitful.