20 1. Preparatory material
From the theory of product measure, we have the following equivalent
formulation of joint independence:
Exercise 1.1.11. Let (Xα)α∈A be a family of random variables, with each
taking values in a measurable space Rα.
(i) Show that the (Xα)α∈A are jointly independent if and only for every
collection of distinct elements α1,...,αk of A, and all measurable
subsets Ei Rαi for 1 i k , one has
P(Xαi Ei for all 1 i k ) =
P(Xαi Ei).
(ii) Show that the necessary and sufficient condition (Xα)α∈A being k-
wise independent is the same, except that k is constrained to be
at most k.
In particular, a finite family (X1,...,Xk) of random variables Xi, 1 i k
taking values in measurable spaces Ri are jointly independent if and only if
P(Xi Ei for all 1 i k) =
P(Xi Ei)
for all measurable Ei Ri.
If the are discrete random variables, one can take the Ei to be
singleton sets in the above discussion.
From the above exercise we see that joint independence implies k-wise
independence for any k, and that joint independence is preserved under
permuting, relabeling, or eliminating some or all of the Xα. A single random
variable is automatically jointly independent, and so 1-wise independence
is vacuously true; pairwise independence is the first non-trivial notion of
independence in this hierarchy.
Example 1.1.13. Let F2 be the field of two elements, let V F2
the subspace of triples (x1,x2,x3) F2
with x1 + x2 + x3 = 0, and let
(X1,X2,X3) be drawn uniformly at random from V . Then (X1,X2,X3)
are pairwise independent, but not jointly independent. In particular, X3 is
independent of each of X1,X2 separately, but is not independent of (X1,X2).
Exercise 1.1.12. This exercise generalises the above example. Let F be a
finite field, and let V be a subspace of
for some finite n. Let (X1,...,Xn)
be drawn uniformly at random from V . Suppose that V is not contained in
any coordinate hyperplane in
(i) Show that each Xi, 1 i n is uniformly distributed in F.
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