Pólya urn model
Matija Vidmar
December 8, 2012
Problem 1 (Pólya urn model). Let an urn contain black and white balls, B0 ≥ 1 (resp. W0 ≥ 1)
being the initial number of black (resp. white) balls (non-random). Draw from the urn a ball,
uniformly at random, then return the ball so drawn, together with a ∈ N additional balls of the same
colour. Repeat. Let Bn (resp. Wn ) denote the number of black (resp. white) balls in the urn after
the n-th draw-and-replacement. Show that the sequence ρ := (Bn /(Bn + Wn ), n ≥ 0) converges a.s.
to a random variable ρ∞ and find the distribution of ρ∞ !
Solution. Let for n ∈ N, Xn be an indicator random variable, 1 if a black ball was drawn on the
n-th draw, 0 otherwise. Next allow Fn := σ(Xi : 1 ≤ i ≤ n) to be the natural filtration of this
sequence, n ∈ N0 (F0 being trivial). We denote this filtration F. Note that conditionally on Fn ,
Xn+1 has distribution Ber(Bn /(Bn + Wn )).
We show ρ is an F-martingale. Indeed, certainly ρ is adapted and integrable, and
E[ρn+1 |Fn ]
=
=
=
=
=
=
Bn+1
|Fn ]
Bn+1 + Wn+1
Bn+1
E[
|Fn ]
B0 + W0 + (n + 1)a
1
E[Bn + aXn+1 |Fn ]
B0 + W0 + (n + 1)a
1
Bn
Bn + a
B0 + W0 + (n + 1)a
Bn + Wn
Bn + Wn + a
Bn
B0 + W0 + (n + 1)a Bn + Wn
ρn ,
E[
as required. Being a nonnegative supermartingale, ρ → ρ∞ , a.s. (and hence in distribution). It
remains to find the law of ρ∞ . In fact we know ρ∞ ∈ [0, 1] a.s. Observe that, while the sequence X
is not an independency, nevertheless the probability of drawing k black balls in the first n draws,
0 ≤ k ≤ n, does not depend on the precise sequence of the colours so drawn (but only on their
cumulative number). Hence:
P(k black balls in first n draws) =
i.e.
n B (B + a) · · · (B + (k − 1)a)W (W + a) · · · (W + (n − k − 1)a)
0
0
0
0
0
0
,
(B0 + W0 )(B0 + W0 + a) · · · (B0 + W0 + (n − 1)a)
k
n β(B0 /a + k, W0 /a + n − k)
P(k black balls in first n draws) =
.
k
β(B0 /a, W0 /a)
1
It follows that
φρn (t)
=
=
E[eitρn ]
n X
n
k=0
=
=
k
e
it B
B0 +ka
0 +W0 +na
B0
B0
Z
eit B0 +W0 +na
1
pk (1 − p)n−k
0
Z
eit B0 +W0 +na
Z
n 1X
a
n
pB0 /a−1 (1 − p)W0 /a−1
(peit B0 +W0 +na )k (1 − p)n−k
dp
k
β(B0 /a, W0 /a)
0 k=0
1
it B
(pe
a
0 +W0 +na
+ (1 − p))n
0
Z
→
0
1
pB0 /a−1 (1 − p)W0 /a−1
dp
β(B0 /a, W0 /a)
pB0 /a−1 (1 − p)W0 /a−1
dp
β(B0 /a, W0 /a)
pB0 /a−1 (1 − p)W0 /a−1
eitp
dp
β(B0 /a, W0 /a)
by the dominated convergence theorem. Lévy’s continuity theorem then implies that the law of ρ∞
is absolutely continuous w.r.t. Lebesgue and with Radon-Nikdoým density:
pB0 /a−1 (1 − p)W0 /a−1
1(0,1) (p) .
p 7→
β(B0 /a, W0 /a)
Put in other words ρ∞ ∼ Beta(B0 /a, W0 /a).
J
2