J Theor Probab (2015) 28:968–975
DOI 10.1007/s10959-013-0513-0
On Independence and Determination of Probability
Measures
Iddo Ben-Ari
Received: 18 March 2013 / Revised: 27 June 2013 / Published online: 13 September 2013
© Springer Science+Business Media New York 2013
Abstract We show that two probability measures defined on the same measure space
having the same pairs of independent events are either purely atomic or equal. In the
former case, either measures are trivial (taking values in {0, 1}) and singular, or equivalent. We also characterize nonatomic measures having the same events independent
of a fixed (nontrivial) event and provide sufficient condition on pairs on independent
intervals to determine a probability measure on the real line.
Keywords
Independence · Nonatomic
Mathematics Subject Classification (2010)
60A99
1 Introduction and Previous Results
The paper has two goals. The first is to publicize the fact that under appropriate conditions, independence of events uniquely determines a probability measure. We think
this is theoretically fundamental because most uniqueness results for probability measure are “measure-theoretic” (e.g., monotone class) or “analytic” (e.g., characteristic
functions, Stein’s equation), and the independence approach discussed here is—to
us—somewhat more elementary and genuinely “probabilistic.” One still has to show
that the independence approach is useful in applications, and perhaps the readers of
this work will... The second goal is to expand and complement the existing results.
This work was partially supported by NSA Grant H98230-12-1-0225.
I. Ben-Ari (B)
Department of Mathematics, University of Connecticut,
196 Auditorium Rd, Storrs, CT 06269-3009, USA
e-mail: [email protected]
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Throughout the discussion, we will fix a measurable space (Ω, F). In the sequel,
whenever we refer to a subset of Ω, we tacitly assume it is an event, that is, an
element of F, and all measures are probability measures on (Ω, F). Let P be a
probability measure. We say that events A and B are P-independent and write A ⊗ P B
if P(A∩ B) = P(A)P(B). We identify ⊗ P as a relation on F, that is, ⊗ P = {(A, B) ∈
F × F : P(A ∩ B) = P(A)P(B)}. An event A is P-null if P(A) = 0, and trivial if A
or Ac is P-null (equivalently, A is trivial if A ⊗ P Ac ). The measure P is trivial if every
event is P-trivial. An event is a P-atom if P(A) > 0, and for every B ⊂ A, B is P-null
or A − B is P-null (equivalently, P(A) > 0, and for every B ⊂ A, B ⊗ P (A − B)). An
event is P-nonatomic if it is not P-null and does not contain P-atoms. The measure P
is purely atomic if every event which is not P-null contains a P-atom (equivalently, P
is purely atomic if none of the events are P-nonatomic). The measure P is nonatomic
if it has no P-atoms. We may omit reference to P when there is no risk of ambiguity.
Finally, the probability measures P and Q are singular if there exists L with P(L) = 1
and Q(L) = 0 and are equivalent if their null events coincide. We begin with a review
of the results that motivated our work.
Theorem A ([3,9]) Suppose that P and Q are two probability measures at least one
of which is nonatomic. If they have identical independent events, then they coincide.
The theorem is a corollary to the main result of [9] (in Russian), predating [3]. Yet,
the derivation in the latter paper was done independently. We comment that [3] presents
two proofs. The second relies on Lyapunov’s convexity theorem for vector-valued
measures [1,4] and is short and elegant. That proof also yields the following corollary.
Corollary B ([3]) Suppose that P is nonatomic, and for some α ∈ (0, 1),
{B : P(B) = α} ⊂ {B : Q(B) = α}.
Then, P = Q.
In [3], it was chosen to state the result for α = 21 with equality between the sets,
although the proof gives the more general statement in the corollary. An alternative
proof is given in the “Appendix.”
In [7], the authors relaxed the assumption in Theorem A that one of the measures
is nonatomic. We quote the result below (with minor changes), but we need to fix
some additional notation. Recall the following key elementary fact. For every P, Ω
could be uniquely decomposed, up to P-null events, into a countable (possibly empty
or finite) union of disjoint atoms, and an event containing no atoms. We refer to the
latter event as the atomless part of P, and to its complement as the atomic part of P.
We also say that P and Q have the same conditionally independent events if for any
L with P(L)Q(L) > 0, ⊗ P(·|L) = ⊗ Q(·|L) . This of course implies ⊗ P = ⊗ Q as seen
by taking L = Ω.
Theorem C ([7]) Suppose that P and Q are not singular. If P and Q have the same
collection of conditionally independent events, then the atomless part of P and Q
coincide, and the atomic parts of P and Q vanish outside of the union of a countably
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many pairwise disjoint events, which are atoms of P as well as of Q. The condition
on P and Q to be not singular can be omitted when the atomless part of P (or Q)
does not vanish.
We close this section mentioning several related publications. The problem of triviality or nontriviality of the independence relation was addressed by several authors.
In [6,8], the authors provide sufficient conditions for having infinitely many pairs of
nontrivial independent events. In [5], the authors provide a sufficient conditions for
nonexistence of pairs of nontrivial independent events. We also mention [2] which
discusses independence of events under uniform distribution on finite sample spaces.
2 Main Results
Before stating our results, we recall the following well-known result which is our main
tool.
Lemma 1 Let P be nonatomic. Then, the range of P is [0, 1].
Note that this is a corollary to Lyapunov’s convexity theorem, but can be derived
independently. In order to keep this note self-contained, we include a proof in the
“Appendix.” Observe that if P is a probability measure and C is not P-null and Pnonatomic, then P( · |C) is a nonatomic probability measure and it follows from
Lemma 1 applied to this measure that the restriction of P to subsets of C has range
[0, P(C)].
Our main results are Theorems 1 and 2 below. Theorem 1 characterizes the relation
between two probability measures having identical pairs of independent events. This
extends and combines Theorem A and Theorem C. Theorem 2 characterizes the relation between two probability measures P and Q with a weaker assumption on the pairs
of independent events: There exists some P nontrivial event such that the set of events
that are P-independent of it is contained in the set of events that are Q-independent of it.
2.1 On Probabilities with Identical Independent Events
Theorem 1 Suppose that ⊗ P = ⊗ Q . Then, either
1. P and Q are singular and trivial; or
2. P is not purely atomic, and then P = Q; or
3. P is purely atomic, and then Q is purely atomic and every P-atom is a Q-atom.
Independence determines whether P is trivial or not. Indeed, P is trivial if and only
if every event is P-independent of itself. When P is trivial, then only 1. or 3. can hold,
but since all events are trivial, dependence does not determine whether it is 1. or 3.
that holds (of course, under assumption of triviality, 3. is equality of P and Q). When
P is not trivial, A is a P-atom if and only if A is not P-independent of itself, and every
B ⊂ A is P-independent of A − B. This determines whether 2. or 3. holds.
The proof of the theorem will be obtained through a sequence of lemmas. We begin
with a simple dichotomy.
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Lemma 2 Suppose that ⊗ P = ⊗ Q . Then, either
1. P and Q are singular and trivial; or
2. Q and P are equivalent.
Note that the first case complements Theorem C and gives an answer to what can
happen when P and Q are singular: They are also trivial measures, having an atom of
full measure. In particular, in this case, ⊗ P = ⊗ Q = F × F.
Proof Observe that under any probability measure, an event A is trivial if and only
if it is independent of its complement. Since ⊗ P = ⊗ Q , we conclude that the set
of P-trivial events is equal to the set of Q-trivial events. Thus, either Q and P are
equivalent, or, for some A, P(A) = 1, Q(A) = 0. In the latter case, let B ⊂ A.
Since B is Q-trivial, it is Q-independent of its complement, and as a result, B is
P-independent of its complement. Thus, B is P-trivial. Thus, A is a P-atom.
Lemma 3 Suppose that C is P-nonatomic, and D ⊂ C is not P-null. Then, there
exist A, B ⊂ D with A ⊗ P B and P(A)P(B) > 0.
Proof Indeed, let A ⊂ D satisfy P(A) = P(D)/2, and choose B1 ⊂ A with P(B) =
P(D)2 /4, and B2 ⊂ D − A1 with P(B2 ) = P(D)/2 − P(D)2 /4 < P(D − A1 ) =
P(D)/2. Letting B = B1 ∪ B2 completes the claim.
Lemma 4 Suppose that C is P-nonatomic and that ⊗ P and ⊗ Q coincide on subsets
of C. Then P(C) = Q(C).
Proof Since C is nonatomic, C is not P-independent of itself. Therefore, C is not Qindependent of itself. Thus, Q(C) < 1. Now, if A is P-null, then A ⊗ P A; therefore,
Q(A) ∈ {0, 1}, but since Q(A) ≤ Q(C) < 1, Q(A) = 0. Conversely, if A ⊂ C is Qnull, then it is P-independent of itself, and since P(C) < 1, P(C) = 0. Summarizing,
Q(C) ∈ (0, 1), and the measures Q(· ∩ C) and P(· ∩ C) are equivalent. Let f = dd Q
P.
To complete the proof, we need to show that on C, { f = 1}, P-a.s. It is enough to
show that on C, { f ≤ 1} P-a.s, because the same argument, with roles of P and Q
interchanged, shows that on C, { 1f ≤ 1} Q-a.s., hence P-a.s. To prove the claim, we
argue by contradiction and assume that there exists t0 > 1 satisfying P( f ≥ t0 ) > 0.
n
n+1 } ∩ C. Then, by
0
Let := 1+t
2 , and for n ∈ Z+ , let C n := {t0 ≤ f < t0 assumption, P(Cn 0 ) > 0 for some n 0 ∈ Z+ . By Lemma 3, we have A, B ⊂ Cn 0
which are not P-null and satisfy A ⊗ P B. This gives the following:
Q(A ∩ B) =
f d P ≤ t0 n 0 +1 P(A ∩ B) = t0 n 0 +1 P(A)P(B)
A∩B
= t0 n 0 +1
A
1
dQ
f
B
t0 n 0 +1
1
dQ ≤
Q(A)Q(B)
f
(t0 n 0 )2
< Q(A)Q(B) < Q(A)Q(B).
t0
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Lemma 5 Suppose that P is not purely atomic. If ⊗ P = ⊗ Q , then P = Q.
Proof Lemma 4 guarantees that P and Q coincide on all P-nonatomic events. Since
every event is a countable union (possibly empty) of P-atoms or a countable union
of P-atoms and a P-nonatomic event, it is sufficient to prove that Q(L) = P(L)
for all P-atoms L. Fix such L and let R ⊂ L c be P-nonatomic. Choose R0 ⊂ R
with P(R0 ) ∈ (0, P(R)). Let A = L ∪ R0 and observe that P(A) ∈ (0, 1) since
0 < P(R0 ) < P(R) ≤ 1. Since both R0 and R − R0 are P-nonatomic, we can find
D0 ⊂ R0 satisfying 1−P(A)
P(A) P(D0 ) = P(R − R0 ). Let B = D0 ∪ (R − R0 ). We claim
that A ⊗ P B. Indeed, A ∩ B = D0 , while
P(A)P(B) = P(A)(P(D0 ) +
1 − P(A)
P(D0 )) = P(D0 ).
P(A)
Therefore, A ⊗ Q B. Since A = L ∪ R0 , the independence statements with respect to
P and Q could be rewritten as
(P(L) + P(R0 ))P(B) = P(D0 ) and (Q(L) + Q(R0 ))Q(B) = Q(D0 ).
However, since R0 , B, D0 ⊂ R, and R is P-nonatomic, it follows from Lemma 4 that
Q(R0 ) = P(R0 ), Q(B) = P(B) and Q(D0 ) = P(D0 ), and all are strictly positive.
P(D0 )
0)
Thus, Q(L) = Q(D
Q(B) − Q(R0 ) = P(B) − P(R0 ) = P(L).
Proof of Theorem 1 From Lemma 2, we conclude that either 1. holds or that P and Q
are equivalent. In the latter case, 2. follows from Lemma 5. To prove 3., assume that
1. and 2. do not hold. Then, P and Q are equivalent, and P is purely atomic. Let A be
a P-atom. Then, for B ⊂ A, B ⊗ P (A − B). Consequently, B ⊗ Q (A − B). It follows
that Q(B) = 0 or Q(A − B) = 0. This shows that A is Q-null or a Q-atom. Since P
and Q share the same null events, and since A is a P-atom, P(A) > 0, which implies
Q(A) > 0; thus, A is a Q-atom. Finally, since Ω is a countable union of disjoint
P-atoms, it is also a countable union of disjoint Q-atoms, proving that Q is purely
atomic.
2.2 On Independence with Respect to a Fixed Event
Theorem 2 Let P be nonatomic, and let L be an event. Then,
{A : A ⊗ P L} ⊂ {A : A ⊗ Q L} F,
(1)
if and only if Q(L) ∈ (0, 1) and Q = Q(L)P( · |L) + Q(L c )P( · |L c ).
The equivalent condition to (1) could be expressed in terms of the Radon-Nykodim
Q(L)
Q(L c )
c
derivative as Q ∼ P and dd Q
P = P(L) 1 L + P(L c ) 1 L .
Corollary 1 Let P be nonatomic and suppose that L satisfies (1). Then, P = Q if
P(L) = Q(L). In particular, P(L) = Q(L) if there exist A, B ⊂ L which are not
P-null and satisfy A ⊗ P B as well as A ⊗ Q B
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Proof of Corollary 1 If Q(L) = P(L), then equality between P and Q follows immediately from the conclusion of Theorem 2. Assume now that there exist A, B ⊂ L
which are not P-null and satisfy A ⊗ P B and A ⊗ Q B. Let ρ = Q(L)
P(L) . Then, ρ > 0
because Q(L) > 0. Observe that Q(A ∩ B) = ρ P(A ∩ B), and Q(A)Q(B) =
ρ 2 P(A)P(B). Since P(A ∩ B) = P(A)P(B) > 0 and A ⊗ Q B, we have ρ = ρ 2 .
Since ρ > 0, it follows that ρ = 1.
Proof of Theorem 2 Observe that a A ⊗ P L if and only if P(A|L) = P(A|L c ).
“⇐” Assume that Q = ρ P(·|L) + (1 − ρ)P(·|L c ) and let A satisfy A ⊗ P L.
Clearly, Q(L) = ρ. Furthermore, Q(A ∩ L) = ρ P(A|L) and Q(A ∩ L c ) = (1 −
ρ)P(A|L c ) = (1 − ρ)P(A|L). Adding both equalities, we obtain Q(A) = P(A|L).
Thus, Q(A ∩ L) = ρ P(A|L) = Q(L)Q(A), that is, A ⊗ Q L.
“⇒” Observe that L is not Q-trivial because the set of events which are Qindependent of L is a proper subset of F. Next, we show that if B1 , B2 ⊂ L c satisfy P(B1 ) = P(B2 ), then Q(B1 ) = Q(B2 ). Since P is nonatomic, there exists
P(B2 )
1)
L 0 ⊂ L with P(L 0 ) = P(B
P(L c ) P(L) = P(L c ) P(L). Let A1 = B1 ∪ L 0 and let
P(B1 )
c
0)
A2 = B2 ∪ L 0 . Observe that P(A1 |L) = P(L
P(L) = P(L c ) = P(A1 |L ), with a similar statement for A2 . Therefore, A1 ⊗ P L and A2 ⊗ P L, and from our assumption,
we conclude that A1 ⊗ Q L and A2 ⊗ Q L. But rewriting these statements, we obtain
(Q(B1 ) + Q(L 0 ))Q(L) = Q(L 0 ), and (Q(B2 ) + Q(L 0 ))Q(L) = Q(L 0 ). Since
Q(L) > 0, it follows that Q(B1 ) = Q(B2 ).
This in particular implies that if B1 ⊂ L c is P-null, then since P(B1 ) = P(∅) = 0,
we also have Q(B1 ) = 0. Thus, Q(· ∩ L c ) is absolutely continuous with respect to
P(· ∩ L c ). Let f denote the Radon-Nykodim derivative. We want to show that f is
constant P-a.s. Otherwise, there exist disjoint B1 , B2 ⊂ L c , with P(B1 )P(B
2) > 0
and supx∈B1 f < inf x∈B2 f (x). But this implies Q(B1 ) = B1 f d P < B2 f d P =
Q(B2 ), a contradiction. Therefore, f is some constant β, P-a.s. This implies Q(L c ) =
β P(L c ), hence β = Q(L c )/P(L c ).
3 Additional Results
In this section, we present two additional simple results on characterizations of measures through independence of events. The first provides a mechanism for finding
events with rational probabilities from independence. The second discusses equality of
probability measures on the real line through independence of certain pairs of intervals.
Proposition 1 Let A be an event and let k ∈ N. Suppose that there exists P-nontrivial
B and a partition of B, B0 , . . . , Bk−1 with B0 = B ∩ A, such that for every j =
0, . . . , k − 1, ((A − B) ∪ B j ) ⊗ P B. Then, P(A) = k1 .
Applying the proposition for k = 2, we obtain that P(A) = 21 provided A is not
P-trivial, and for some P-nontrivial B, we have A ⊗ P B and (AB) ⊗ P B, where
AB is the symmetric difference (A − B) ∪ (B − A).
Proof We have P(B j ) = (P(A − B) + P(B j ))P(B). That, is P(B j )(1 − P(B)) =
P(A − B)P(B). Since P(B) ∈ (0, 1), it follows that P(B j ) is independent of j, but
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then P(B) = k P(B0 ) > 0 and the independence between A = (A − B) ∪ B0 and B
implies P(B0 ) = P(A ∩ B) = P(A)P(B0 )k = P(A)P(B).
In analogy with monotone class results, one may ask whether independence of
events in a “small” (and usually easily identifiable) subset of F determines P. The
following result provides a positive answer for distributions on R, under a mild regularity condition.
Proposition 2 Let X be a random variable with distribution function F. If, for some
differentiability point b of F, F (b) > 0, then F is uniquely determined by the pairs of
intervals of the form (a, c) and (b, d) where a < b < c < d, which are independent
under the Borel measure induced by X .
The proof is essentially a restatement of the fact that the probability of an interval
with one end point equal to b is proportional to its length as the length tends to 0, and
so conditional probabilities involving intervals around b reduce to ratios of lengths of
the intervals. Also, as the proof shows, the condition in the theorem could be relaxed
to the following: There exists a dense A with the property that for every d ∈ A and
d > b, there exist sequences an < b < cn satisfying limn→∞ (cn − an ) = 0 and
(an , cn ) ⊗ (b, d), and similarly, that for every a ∈ A with a < b, there exist sequences
b < cn < dn with limn→∞ dn − b = 0 such that (a, cn ) ⊗ (b, dn ).
Proof of Proposition 2 Let P denote the Borel probability measure induced by X .
That is, P(A) = P(X ∈ A). Fix d > b. Observe that for a < b < c, (a, c) ⊗ (b, d) if
P((b,c))
and only if P((b, d)) = P((a,c))
. However, P((b, c)) = F(c)− F(b) = F (b)(c−b)+
o(c−b), and similarly, P((a, c)) = P((b, c))+ P((a, b)) = (F(c)− F(b))+(F(b)−
F(a)) = F (b)(c−b)+o(b−c)−(F (b)(a−b)+o(a−b)) = F (b)(c−a)+o(c−a).
P((b,c))
As a result for any > 0, the ratios { P((a,c))
: a < b < c, |c − a| < } are dense
in [0, 1]. In particular, there exist sequences dn d, an b and cn b such that
(an , cn ) ⊗ (b, dn ). In particular,
P((b, dn )) =
cn − b
F (b)(cn − b)(1 + o(1))
=
(1 + o(1)).
F (b)(cn − an )(1 + o(1))
cn − an
By letting dn d, P((b, dn )) → P((b, d]) = F(d) − F(b). Therefore,
these differences are uniquely determined by the pairs (an , cn ) ⊗ (b, dn ). By letting d → ∞, F(b) is determined as well. Therefore, F is determined for all x ≥ b.
The remainder of the argument is similar. First, observe that for any > 0, the ratios
P((b,c)
: c > d > b, d − b < } are dense in [0, 1], because the ratio is asymptot{ P((b,d))
c−b
ically equivalent to d−b
as d, c b. As a result, given any a < b, then there exist
sequences an a, dn > cn with dn b, such that
P((b, cn ))
− P([b, cn )) = P((an , b)),
P((b, dn ))
that is, (a, cn ) ⊗ (b, dn ). Since all terms on the right-hand side are already determined,
and the right-hand side converges to F(b−) − F(a) = F(b) − F(a), it follows that
F(a) is determined as well.
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Acknowledgments I would like to thank R. A. Vitale for introducing me to the topic as well as for making
useful comments and suggestions. I would also like to thank and A. P. Yurachkovskiı̆ and A. L. Rukhin
for sending me copies of [9] and [7], and an anonymous referee for comments and suggestions that greatly
improved the manuscript.
Appendix
Proof of Lemma 1 First, we observe that any event A with P(A) > 0 has an event
with arbitrarily small but strictly positive measure. Indeed, let such A be given. Then,
since P is nonatomic, there exists A1 ⊂ A with P(A1 ), P(A − A1 ) > 0. Without
loss of generality, P(A1 ) ≤ P(A)/2. Continue inductively. Given An ⊂ A with
0 < P(An ) ≤ P(A)/2n , there exists An+1 ⊂ An , with 0 < P(An+1 ) < P(An )/2.
We turn to the main claim. Fix α ∈ (0, 1), and let A0 be such that P(A0 ) ∈ (0, α).
Continue inductively, assuming P(An ) ∈ (0, α). Let In = {B ⊂ Acn : P(B) ≤
α − P(An )}. Choose Bn+1 ∈ In that satisfies P(Bn ) ≥ (1 − n1 ) sup B∈In P(B), and
let An+1 = An ∪ Bn+1 . Note that the events (Bn : n ∈ N) are disjoint. Therefore,
P(Bn ) → 0. Consequently, sup B∈In P(B) → 0. Let A∞ = ∪n∈N An . We continue
the proof arguing by contradiction, assuming that P(A∞ ) < α. Then, by the first
paragraph, there exists B∞ ⊂ Ac∞ satisfying 0 < P(B∞ ) < α − P(A∞ ). Since
Ac∞ ⊂ Acn for all n ∈ N, it follows that B∞ ∈ In for all n. But this contradicts the fact
that sup B∈In P(B) → 0.
Proof of Corollary B Let = min(α, 1 − α). It is enough to show that if P(A) =
P(B) ≤ , then Q(A) = Q(B). Without loss of generality, A ∩ B = ∅; otherwise, we
will consider the pair A − (A ∩ B) and B − (A ∩ B) instead of A and B. Since A ⊂ B c
and P(B c ) ≥ α, by Lemma 1, there exists an event L ⊂ B c , with P(L) = α and
A ⊂ L. But since P((L − A) ∪ B) = P(L) − P(A) + P(B) = P(L) = α, it follows
that Q((L − A) ∪ B) = α. However, Q((L − A) ∪ B) = Q(L) − Q(A) + Q(B) =
α − Q(A) + Q(B). Thus, Q(A) = Q(B).
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