A Note on Joint Independence

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Some students asked about joint independence and factorization when there are more than 2 random variables. CB and my notes focus on the 2-variable case, so it may be useful to discuss the more general case.

Suppose we have random variables X_1, X_2, \ldots, X_k defined on the same probability space. Suppose they have a joint PDF/PMF f(x_1,x_2,\ldots,x_k), and marginal PDF/PMFs f_1(x_1), f_2(x_2),\ldots, f_k(x_k). The random variables are jointly independent if and only if

f(x_1,x_2,\ldots,x_k) = f_1(x_1) \cdot f_2(x_2) \cdots f_k(x_k)

for all x_1,x_2,\ldots,x_k.

(There is a formal definition of joint independence in terms of product measures, but the definition above will work for our purposes.)

In the midterm review questions, one of the questions had

f(x1,x2,x3) = exp( − (x1 + x2 + x3)).

You were asked to show (joint) independence. The easiest way is to directly solve for the marginal densities, and show that the joint equals the product of the marginals.

In this question, you might notice that we can write

f(x_1,x_2,x_3) = e^{-x_1} e^{-x_2} e^{-x_3}.

So, if a version of the factorization theorem holds for dimension greater than 2, then simply showing this factorization would be enough to conclude that the variables are jointly independent. However, our result in the notes only covers the 2 variable case. It turns out that it is possible to extend the factorization theorem to more than 2 variables (try proving this!), but you would have to show this yourself in order to answer this question, since we did not do so in class.

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