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052184472Xc02 CUNY148/Severini May 24, 2005 2:29

2.5 Exchangeability 61

n
Let T denote a subset of the set of all permutations of (1,..., n). A function h :R → R m
is said to be invariant with respect to T if for any permutation τ ∈ T , h(x) = h(τx). Here τ
). If T is the set of all permutations,
is of the form τ = (i 1 ,..., i n ) and τx = (x i 1 , x i 2 ,..., x i n
h is said to be permutation invariant.

n
Example 2.21 (Sample mean). Let h denote the function on R given by
n
1
h(x) = x j , x = (x 1 ,..., x n ).
n
j=1
Since changing the order of (x 1 ,..., x n ) does not change the sum, this function is permu-
tation invariant.

Theorem 2.9. Suppose X 1 ,..., X n are exchangeable real-valued random variables and h
is invariant with respect to T for some set of permutations T . Let g denote a real-valued
function on the range of X = (X 1 ,..., X n ) such that E[|g(X)|] < ∞. Then
(i) The distribution of (g(τ X), h(X)) is the same for all τ ∈ T .
(ii) E[g(X)|h(X)] = E[g(τ X)|h(X)], with probability 1, for all τ ∈ T .

Proof. Since X 1 ,..., X n are exchangeable, the distribution of (g(τ X), h(τ X)) is the same
for any permutation τ.Part (i) now follows from the fact that, for τ ∈ T , h(τ X) = h(X)
with probability 1.
By Theorem 2.6, part (ii) follows provided that, for any bounded, real-valued function
f on the range of h,

E[E[g(τ X)|h(X)] f (h(X))] = E[g(X) f (h(X))].

Since
E[E[g(τ X)|h(X)] f (h(X))] = E[E[g(τ X) f (h(X))|h(X)]] = E[g(τ X) f (h(X))],

part (ii) follows provided that

E[g(τ X) f (h(X))] = E[g(X) f (h(X))];
the result now follows from part (i).

Example 2.22 (Conditioning on the sum of random variables). Let X 1 , X 2 ,..., X n
denote exchangeable, real-valued random variables such that

E[|X j |] < ∞, j = 1,..., n,

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and let S = X j . Since S is a permutation invariant function of X 1 ,..., X n ,it follows
j=1
from Theorem 2.9 that E[X j |S] does not depend on j. This fact, together with the fact that
n n

S = E(S|S) = E[ X j |S] = E[X j |S]
j=1 j=1
shows that E(X j |S) = S/n.