2. LEARNING PROBABILITIES FROM RELATIVE
FREQUENCIES
Roger M Cooke
ShortCourse on Expert Judgment
National Aerospace Institute
April 15,16, 2008
1. Introduction
We focus on learning from observed relative frequencies. The following notation will
be used throughout.
11,....1n,...; indicator functions for events A1,...;
{0,1}n = 2n : set of sequences of length n containing only 0's and 1's.
For Q e 2n , #Q = number of 1's in Q.
wi(n) : probability that exactly i of A1,...An occur.
Sn = {p  Rn | pi ≥ 0, Σpi = 1}
(nj) = n!/(j!(n-j)!)
for any set A (finite or infinite), A! = the set of permutations of the elements of A.
2. The Frequentist Account
The frequentist interprets probability as limiting relative frequency. It is useful to
rehearse the frequentist account of learning from experience. The
weak law of large numbers reads:
Let Xi, i = 1,2,.. be independent and identically distributed variables taking values in
{0,1}, and let P{Xi = 1} = p. Then for all ε > 0
n
P{|(1/n)ΣXi – p| > ε} → 0 as n → ∞
i=1
1
The frequentist learning model works as follows. Suppose we have a variable X
taking values in {0,1} and we wish to know the probability that X = 1. If we can find
a sequence of independent variables Xi with the same distribution as X, then we can
observe a large number n of these variables. The relative frequency of "1"'s in
X1,...Xn is then unlikely to differ greatly from the
probability that X = 1. Two features of this account are noteworthy:
1. Probability exists. The person reasoning as above believes that there is some
objective quantity "p".
2. p is initially unknown, but can be measured to arbitrary precision, in a
probabilistic sense, by observing relative frequencies in large, finite
sequences of observations.
Note: Expressed in this way, we have to know that the variables Xi are independent
and identically distributed, which we could never know before measuring their
probabilities. This circularity in the naïve frequentist account is removed in more
sophisticated treatments, where probability is defined as limiting relative frequency
in a random sequence, and “random” is defined without reference to probability.
3.The Subjectivist Account
The subjectivist cannot accept this account of learning. "p" is not an objective
quantity, but a subjective degree of belief. It cannot be "observed" in the sense
implied by the frequentist account. It is a fact, nevertheless, that people attach
importance to relative frequencies. The subjectivist must offer an explanation of this
fact.
Expected Relative Frequency
Theorem 1: Let pi = p(Ai) = p(1i=1). Then the average of the probabilities pi equals
the expected relative frequency of occurrence of A1,...An, or
(1/n)Σpi = Σiwi(n) /n .
2
Proof: Rewriting the expected relative frequency of occurrence on the right hand
side, we have:
E(1/n)Σ1i = (1/n)ΣE(1i) = (1/n)Σpi.
Note that this theorem makes no independence assumptions. It also says nothing
about changing our beliefs in the light of observations.
Exchangeability
According to De Finetti, exchangeability provides the characterization of those belief
states in which we learn from relative frequencies.
Definition: The events A1,..An are exchangeable with respect to probability measure
p if for all Q, Q'  2n with #Q = #Q'; p(11,..1n = Q) = p(11,..1n = Q'). An infinite
sequence of indicators is exchangeable with respect to p if every finite subsequence
is exchangeable with respect to p.
Exercise: Suppose the events A1,...An are independent and identically distributed.
Are they exchangeable? What is wrn ?
Exercise: If A1,...An are exchangeable and #Q = r, then:
p(Q) = wr(n) / (nr) .
After observing a sequence of outcomes Q  2n , our probability for An+1 is no longer
p(An+1) but p(An+1|Q). We wish to calculate this quantity on the assumption that
A1,...An+1 are exchangeable with respect to p. Suppose #Q = r.
P(An+1 | Q) = P(An+1  Q)/p(Q) =
[wr+1(n+1) / (n+1r+1)]
────────────
[wr(n) / (nr)]
3
= [wr+1(n+1) / wr(n)] × [(r+1)/(n+1)]
To reduce this note that
wr(n) / (nr) = [wr+1(n+1) / (n+1r+1)] + [wr(n+1) / (n+1r)]
Putting s = n - r and simplifying:
wr(n)
wr+1(n+1) (r+1) + wr(n+1) (s+1)
= ──────────────────── ;
n+1
Since r+1 = n+2 - (s+1). This yields:
wr+1(n+1) (r+1)
p(An+1|Q) = ───────────────────
wr(n+1) (s+1) + wr+1(n+1) (r+1)
r+1
= ────────────────────
(r+1) + (s+1) [wr(n+1) / wr+1(n+1) ]
r+1
= ──────────────────────
(n+2) + (s+1) [wr(n+1) / wr+1(n+1) -1]
4
If the probability of r occurrences in A1,...An+1 is close to the probability of r+1 such
occurrences, then the term in brackets is close to one and
p(An+1|Q) ~ (r+1)/(n+2).
We have proved the following result.
Theorem: Let A1, A2,...be exchangeable, and suppose there is a constant K such that
for all n:
Max {r = 0...n } |[wr+1(n+1) / wr(n) ] – 1| < K/n
then max Q2ω |p(An+1|Qn) - #Q/n| → 0 as n →∞.
Notice that this theorem does NOT say that the relative frequency of 1's converges to
some limit as n goes to infinity.
Exercise: Consider an Urn with n (n even) black and white balls drawn without
replacement. Let Ai denote the event that the i-th ball is black. Suppose the Ai's are
exchangeable. Consider three cases: (1) we are told that there are exactly n/2 black
balls in the urn; (2) we are told that there are either n/2 - 1 or n/2 + 1 black balls in
the urn, each option having probability 1/2; and
(3) we are told that for r = 0,...n the urn contains r black balls with probability 2-n(nr)
• Calculate P(An) in all cases
• Calculate the possible values p(An|Qn-1) in all cases.
(hint: think before calculating)
4. DeFinetti's representation theorem
Instead of speaking of exchangeable sequences A1,...An, we may just as well speak of
exchangeable measures on the field of events generated by A1,...An.
5
DeFinetti's theorem states that an exchangeable measure can be uniquely represented
as probabilistic mixtures of certain "extreme" measures. Any measure can be
represented as a mixture of other measures via conditionalization. For example, if
{Bi}i=1,...n is a partition, then
p(A) = Σp(A|Bi)p(Bi)
represents the measure p(.) as a mixture of the measures pi = p(P|Bi), with mixing
weights p(Bi). Exchangeability is preserved under mixing, (see exercise), hence if
events are exchangeable under each pi, then they are
exchangeable under p as well. The question of representing exchangeable sequences
can be translated into the question of representing exchangeable
probability measures.
DeFinetti's representation theorem for finite exchangeable sequences of events:
(1) Any exchangeable measure on P(2n) is uniquely determined by the probabilities
wj(n) , j = 0,...n, and (2) any q  Sn+1 determines an exchangeable measure by setting
wj(n) = qj+1; j=0...n.
Proof: (1) Partition P(2n) into sets Bi = {Q2n | #Q = i}, i = 0,...n. For any
exchangeable probability measure p on P(2n),
p = Σp(•|Bi)p(Bi).
p(•|Bi) is uniform on P(2n) and p(Bi) = wi(n). Hence if p' is another exchangeable
measure and p'(Bi) = p(Bi), then p'=p.
(2) For fixed i, the probability μi on P(2n) which assigns Bi probability one and for
which P(•|Bi) is uniform, is exchangeable; and for any qSn+1,the measure p = Σμiqi
is also exchangeable (see exercise).
Consider drawings from an urn containing r black and n-r white balls. The
probability of drawing j black and m-j white balls, with m < n, j < r is given by the
hypergeometric distribution:
(mj) (n-mr-j)
wj(m) = ──────────.
( n r)
6
If m = n, then wj(m) = 1 if j = r and = 0 otherwise. The probability on P(2n) induced
by drawing from such an urn is exchangeable and corresponds to the measure μr in
the above proof. Hence any exchangeable measure can be
represented uniquely as a mixture of urn drawing measures. The probability p(Bi)
corresponds to the probability of "drawing" an urn with i black balls, i
= 0,...n; and the probabilities p(P|Bi) correspond to drawings from the urn with i
black balls.
DeFinetti's representation for infinite sequences of exchangeable events
For large n, drawing a few balls without replacement from an urn with r black and nr white balls is "almost" like flipping a coin with probability r/n of "black" and (nr)/n of "white". One can prove, letting n→ ∞ keeping the ratio r/n constant, that the
hypergeometric distribution converges to the binomial distribution, that is
wj(m) → (mj)(r/n)j ((n-r)/n)m-j .
For infinite sequences of exchangeable events, DeFinetti's theorem states that any
exchangeable probability measure may be uniquely represented as a mixture of
binomial measures. For Q2n , #Q= r, the probability of Q under the binomial
measure with parameter q  (0,1) is
P(Q) = qr (1-q)n-r .
Taking mixtures of such measures corresponds to integrating the above over q 
(0,1), with respect to some measure dp(q) on the interval (0,1). Every
exchangeable probability corresponds to a unique measure dp(q), and we may write:
P(Q) = ∫[0,1] q#Q (1-q)n-#Q dp(q)
where Q is a sequence of r 1's and n-r 0's. This notation is a bit ambiguous, as the P
on the left hand side is a probability on 2ω ; the p under the integral is a probability
measure on (0,1). The quantity dp(q) may be thought of as the probability of
"drawing" a binomial probability with parameter q.
7
The key to applying DeFinetti's theorem for infinite exchangeable sequences lies in
finding a class of measures dp(q) for which the above integral can be
evaluated easily.
5. The subjectivist resolution of the Ellsberg Paradox
The following "paradox" was proposed by Daniel Ellsberg to argue that people's
degrees of belief cannot be represented as (subjective) probability measures.
Consider two cases:
Case 1: A subject is offered the opportunity to bet on the outcomes of tosses with a
coin which the subject believes to be fair (P(heads) = 0.5, tosses independent).
Case 2: A subject is offered the opportunity to bet on the outcomes of tosses of a coin
about which he believes that the probability of heads could with equal probability be
any number r  (0,1).
On the subjectivist account the subject's beliefs are modeled by, Q  2n:
p(Q) = ∫q#Q (1-q)n-#Q dp(q).
In case 1, the subject is certain that q = 0.5, hence his beliefs are modeled by setting
dp(q) = δ(q-0.5); where δ(x) is the "Dirac delta measure" assigning unit mass to the
point x=0; hence for any function F(q);
∫F(q) δ(q-θ) = F(θ).
In case 2, dp(q) = dq. Hence
p(heads on one toss) = ∫qd(q) = 0.5.
The "paradox" is that subjects in case 1 will accept even money bets on heads,
whereas in case 2 they supposedly will not, even though the subjective probabilities
are the same in both cases. Hence there must be a difference in the beliefs in these
two cases, even though P(heads on one toss) is the same.
On the subjectivist account the difference in these two cases only appears when we
consider the subject's willingness to revise his/her betting rates on the basis of
observations of outcomes of previous tosses.
Exercise:Let Hi denote the event "heads on toss i".
8
(i) Suppose that the subject believes the tosses are independent with P(Hi) = r; i =
1,2,... . Show
that P(H2|H1) = r.
(ii) Suppose that dp(q) is not concentrated at one point; use Jensen's inequality to
show that P(H2|H1) > p(H1). (iii) Suppose dp(q) = dq, use the Beta integral:
∫qr (1-q)n-rdq = r!(n-r)!/(n+1)!
to derive Laplace's rule:
P(Hn+1|r heads on n tosses) = (r+1)/(n+2).
In case (ii) the subject's beliefs are modeled as a mixture of "independent coin
tosses", with mixing measure dp(q); the mixture of independent measures is not
independent; hence the subject in case (ii) exhibits willingness to learn from previous
experience whereas the subject in case 1 does not learn from experience.
9