Maximal percentages in Polya's urn

0.16 MB·

by  Ernst Schulte-Geers

MAXIMAL PERCENTAGES IN PO´LYA’S URN
ERNSTSCHULTE-GEERSAND WOLFGANGSTADJE
4
1
Abstract. We show that the supremum of the successive percentages
0
2 ofredballsinPo´lya’surnmodelisalmostsurelyrational,givethesetof
values that are taken with positive probability and derive several exact
t
c distributional results for theall-time maximal percentage.
O 2010 Mathematics Subject Classification: 60C05.

9

]
R
1. Introduction
P
. The classical urn of Eggenberger-P´olya [3] contains initially r 1 red
h ≥
t balls and b 1 black balls. In the course of the drawing process in each
a ≥
drawone ball is taken from theurn(whereeach ballin the urnhasthe same
m
chance of being drawn), and this ball and another d 1 balls of the same
[ ≥
color are put into the urn. The theory of Po´lya urn schemes is expounded

4 in [8], where also an extensive bibliography can be found.
v
Let R , n 1, denote the number of red balls in the urn after the nth
5 n
≥
0 draw and set R0 = r. The ratio Zn = Rn/(nd+r+b) gives the percentage
8 of red balls among all balls in the urn at “time” n. Let
1
. S = supZ
1 r,b n
0 n≥0
4 be the supremum of all successive percentages of red balls during the entire
1
drawing process.
:
v InthispaperweshowthatS isattainedalmostsurelyandthatP(S is rational)=
r,b r,b
i
X 1, thereby settling in the affirmative a conjecture of Knuth (posed in the an-
r swer to problem 88 of pre-fascicle 5a to volume 4B of The Art of Computer
a
Programming [7]). The support of S (i.e., the set of values that are taken
r,b
with positive probability) is also given. Moreover, we derive exact formulas
for certain values of the distribution function of S , mainly in the case
r,b
d= 1. For general d 1 we show that
≥
∞
a+1
P(S > (t 1)/t) =
r,b
− n(t 1)+a+1
nX=0 −
nt+a B(n(t 1)+a+1+(r/d),n+(b/d))
− ,
(cid:18) n (cid:19) B(r/d,b/d)
Key words and phrases. Po´lya’surn;binomialrandomwalk;all-timemaximalpercent-
age; exact distribution.
1
2 ERNSTSCHULTE-GEERSANDWOLFGANGSTADJE
where (r +b)/r t N, a = (b(t 1) r)/d and B(α,β) is the Beta
≤ ∈ ⌊ − − ⌋
function. For d = 1 we have
1 ln2 for t = 2
−
P(S 1/t) =  1 t−2
1,t−1
≤  (t 1) 1 1 (t 2) for t > 2
− (cid:18) − t(cid:19) − −
h i.

and
2t 3 t 2
P(S = (t 1)/t) = − H(1 (1/t)) − H(1 (2/t))
1,1
− t − − t −
t 2
−
− t 1
−
P(S (t 1)/t) = (1 (1/t))H(1 (1/t)),
1,1
≤ − − −
where H(x) = Ψ(x+1)+γ. Here Ψ = Γ′/Γ denotes the Digamma function
and γ is Euler’s constant. Another exact result in form of an infinite series
is (again for d =1)
∞
(nt)! ((n+1)(t 1))!
P(S > (t 1)/t) = (t 1) − .
t−1,1
− − (nt n+1)! ((n+1)t)!
nX=0 −
In the course of our derivations we obtain a remarkable equidistribution
propertyofthesupremumM(p) = supn∈Nn−1(B1+ +Bn)ofthebinomial
···
random walk generated by i.i.d. Bernoulli variables B (i.e., P(B = 1) =
i i
1 P(B = 0) = p (0,1)). For t N with pt 1 it is shown that
i
− ∈ ∈ ≤
P(M(p) (1/(k+1),1/k]) = p/(1 p) for every k 1,...,t 1
∈ − ∈ { − }
P(M(p) (p,1/t]) = (1 tp)/(1 p).
∈ − −
2. Used facts and related work
In Po´lya’s urn scheme let X = 1 if a red ball is drawn in the nth draw
n
and X = 0 otherwise, and let X = (X ,X ,...) be the full sequence of
n 1 2
these red ball indicators.
The following facts about Po´lya’s urn are well-known [2, 4].
(1) Z converges almost surely to a random variable Z, which has a
n
Beta(r/d,b/d) distribution on (0,1). (Here and in the sequel Beta(α,β)
denotes the Beta distribution with parameters α,β > 0).
(2) Conditionally on Z = z , X ,X ,... are independent, 0,1 -valued
1 2
{ }
random variables with P(X = 1) = z.
i
Thusthepercentage Z of redballs eventually tendstoarandomvalueZ,
n
and if Z “exits” to Z = z the red ball indicators behave like independent
n
0 1variableswith“successprobability”z. Onecanthereforetrytoaverage
−
known results for the Bernoulli sequence to obtain results for Po´lya’s urn.
MAXIMAL PERCENTAGES IN PO´LYA’S URN 3
Forasequenceof 1,1 -valuedBernoullivariablesU ,U ,...withP(U =
1 2 i
{− }
1) = 1 P(U = 1) = p (0,1) the supremum
i
− − ∈
M = supn−1(U +...+U )
1 n
n≥1
of the averages of their partial sums was considered in [9]. It was shown
that this supremum is attained and that
(1) P(M (2p 1,1] Q)= 1
∈ − ∩
(2) P(M =x) > 0 for each x (2p 1,1] Q.
∈ − ∩
Moreover, explicit formulas for the distribution of M were derived. Note
that M is one of the rare examples of a naturally occurring random variable
taking every rational number in some interval with positive probability.
In the sequel we combine these results to study the all-time maximal
percentage of red balls in Po´lya’s urn.
3. Existence and Possible Values of Maxima
LetX bethePo´lyaredballindicatorsequence. Wehavealreadyremarked
that one may view this sequence as a randomized Bernoulli sequence. To
make this note self-contained we include a short proof.
Theorem 3.1. Let Z and Y = (Y ,Y ,...) be defined on some probability
1 2
space such that: (a) Z is Beta(r/d,b/d)-distributed.
(b) Conditionally on Z = z , (Y ,Y ,...) is an i.i.d. sequence of 0,1 -
1 2
{ }
valued random variables with P(Y = 1) = 1 P(Y = 0) = z.
i i
−
d
Then Y = X.
Proof. Let (y ,...,y ) 0,1 n, s = y +...+y . Then
1 n n 1 n
∈ { }
P(Y = y ,...,Y = y Z = z) = zsn(1 z)n−sn.
1 1 n n
| −
Therefore,
P(Y = y ,...,Y = y )
1 1 n n
1 1
= z(r/d)−1+sn(1 z)n+(b/d)−1−sndz
B(r/d, b/d) Z −
0
B((r/d)+s , n+(b/d) s )
n n
= − .
B(r/d, b/d)
Hence a straightforward calculation (using the elementary properties of the
Beta function) yields P(Y = 1) = r/(r+b) and
1
r+s d
P(Y = 1 Y = y ,...,Y = y ) = n ,
n+1 1 1 n n
| nd+r+b
so that the finite-dimensional distributions of Y coincide with those of X.
Thus, Y =d X. (cid:3)
Since the red ball indicators in Po´lya’s urn scheme are a randomized
Bernoulli sequence they have similar properties.
4 ERNSTSCHULTE-GEERSANDWOLFGANGSTADJE
Proposition 3.2. In the situation of Theorem 3.1 let S = n Y and
n i=1 i
Zn = (r+dSn)/(nd+r+b). Thenlimn→∞Zn = Z a.s. andlimsuPpn→∞(Sn−
nZ)= a.s.
∞
Proof. Let z (0,1). We only need to prove the assertion conditionally on
∈
Z = z. SinceY is, conditionally on Z = z, asequenceofi.i.d. 0 1variables
−
with probability z for the value 1, we have
P(Z z,limsup(S nz)= Z = z) =1.
n n
−→ − ∞ |
n→∞
Note that P(Z z Z = z) = 1 follows from the strong law of large
n
−→ |
numbers, and P(limsup (S nz)= Z = z)= 1 follows (e.g.) from
the law of the iterated long→a∞rithmn−. ∞ | (cid:3)
Let M = sup Z . We now use a variant of the proof of the corre-
r,b n≥1 n
sponding property of M in [9] to show that this supremum is almost surely
attained.
Proposition 3.3. (a) The supremum in the definition of M is almost
r,b
surely attained.
(b) P(M is rational) = 1.
r,b
Proof. Again we condition on Z = z. Since limsup (S nz) > b+1
n→∞ n −
a.s. we have that a.s. infinitely often (r+dS ) (nd+r+b)z > d(b+1)+
n
−
r(1 z) bz 1+(r+b)(1 z) > 0, i.e., Z > z for infinitely many n a.s.,
n
− − ≥ −
and since Z z a.s. the supremum is a.s. attained. It is then obviously
n
rational. −→ (cid:3)
Define
r+i d
j
Q = max n N,i ,...,i N 0 ,
1 n
n1≤j≤n r+b+jd (cid:12) ∈ ∈ ∪{ }
(cid:12)
(cid:12)
i j for all j 1,...,n ,0 i i ... i .
j 1 2 n
≤ ∈ { } ≤ ≤ ≤ ≤
o
Clearly, Q is the set of all possible maximal percentages for finite sequences
ofdrawsfromtheurn. Itfollows fromProposition3.3thatP(M Q)= 1.
r,b
∈
Now we show
Proposition 3.4. P(M = q)> 0 for all q Q.
r,b
∈
Proof. Fix an element q of Q and corresponding m N and nonnegative
∈
integers i i ... i such that i j for all j 1,...,m and
1 2 m j
≤ ≤ ≤ ≤ ∈ { }
r+i d
j
q = max .
1≤j≤mr+b+jd
Let the maximum be attained at n 1,...,m . Then we have
∈{ }
r+i d r+i d
n j
q = = max .
r+b+nd 1≤j≤nr+b+jd
MAXIMAL PERCENTAGES IN PO´LYA’S URN 5
Let E be the event that R = i d for j = 1,...,n and M = q (thus on
j j r,b
this event the percentage after the nth draw is the largest among the first
n ones). It remains to show that P(E) > 0.
Write P(E Z = z) in the form
|
P(E Z = z) = P R = i d for j = 1,...,n and
j j
|
(cid:16)
r+i d+(Y′+...+Y′)d r+i d
sup n 1 k n Z = z
k∈N r+b+nd+kd ≤ r+b+nd (cid:12) (cid:17)
(cid:12)
= P(R = i d for j = 1,...,n Z = z) (cid:12)
j j
|
r+i d+(Y′+...+Y′)d r+i d
P sup n 1 k n ,
× (cid:16)k∈N r+b+nd+kd ≤ r+b+nd(cid:17)
where Y′,Y′,... is an i.i.d. sequence with P(Y′ = 1) = 1 P(Y′ = 0) = z
1 2 i − i
which is independent of Y and Z. The first factor on the righthand side is
obviously positive. Next, the inequality
r+i d+(Y′+...+Y′)d r+i d
sup n 1 k n
k∈N r+b+nd+kd ≤ r+b+nd
holds if and only if
Y′+...+Y′ r+i d
sup 1 k n .
k∈N k ≤ r+b+nd
By [9], this latter inequality occurs with positive probability if
P(Y′ = 1) < (r+i d)/(r+b+nd).
1 n
It follows that P(E Z = z) > 0 for z (0,(r+i d)/(r+b+nd)). Hence,
n
| ∈
min[1,(r+ind)/(r+b+nd)]
P(E) P(E Z = z)β (z) dz > 0,
r/d,b/d
≥ Z |
0
where β (z) is the density of Beta(r/b,b/d). The proof is complete. (cid:3)
r/b,b/d
Remark 3.5. Proposition 3.4 can also be formulated in the following form.
Let supp(U) denote the support of the random variable U. Let q (i,n) =
r,b
(r+id)/(r+b+nd) for i,n N 0 . Then
∈ ∪{ }
supp(S )= q (i,n) 0 i n,q (i,n) r/(r+b) (3.1)
r,b r,b r,b
{ | ≤ ≤ ≥ }
and
supp(M ) = supp(S ) supp(S ). (3.2)
r,b r,b+d r+d,b
∪
(3.2) is obvious, and (3.1) is proved in the same way as Proposition 3.4
(considersequencesofdrawsbeginningwithX = ... = X = 0,X =
1 n−i n−i+1
...,X = 1).
n
In particular, for d =1 we obtain supp(S ) = [r/(r+b),1) Q.
r,b
∩
6 ERNSTSCHULTE-GEERSANDWOLFGANGSTADJE
4. Some Closed-Form Results on Maximal Percentages for the
Binomial Random Walk
In this section we only consider the case d = 1. We start with two exact
results for the binomial random walk, which may be of independent interest.
Although they are consequences of the t-ballot theorems they apparently
have not been formulated in this form before.
Inthesequelfix0 < p < 1, q = 1 pandletY ,Y ,...bei.i.d. 0 1random
1 2
− −
variableswithP(Y = 1) = p. IntroducetheirpartialsumsS = Y + +Y
i n 1 n
···
and define, for r,b N 0 ,
∈ ∪{ }
r+S
n
M (p)= sup and M(p)= M (p)
r,b 0,0
r+b+n
n≥1
r+S
n
L (p) = inf and L(p) = L (p).
r,b 0,0
n≥1r+b+n
Since P(M (p) x) = P(L (q) 1 x) for 0 x 1 the distribution
r,b b,r
≤ ≥ − ≤ ≤
function of L (q) can be obtained from that of M (p).
b,r r,b
Clearly M (p) x if and only if S nx bx (1 x)r for all n N.
r,b n
≤ − ≤ − − ∈
Thusifx Q (0,1),sayx = s/tforpositiveintegerssandtwithgcd(s,t) =
∈ ∩
1, the value of P(M (p) s/t) depends on s,t only through the integer
r,b
≤
value m = sb (t s)r.
− −
For m Z we define a = P(tS ns m for all n N).;
+ m n
∈ − ≤ ∈
Remark 4.1. For two integers s,t satisfying (s,t) = 1, p < s/t < 1 the
following can be shown (along the lines of [9]):
(a) The sequence a ,a ,a ,... has a rational generating function; it is
0 1 2
given by q( s−1a zj)/(pzt zs +q). This function is regular for
j=0 j −
z < 1. P
| |
(b) The denominator f(z)= pzt zs+q has only simple roots: exactly
−
s 1 roots z ,...,z inside the unit disk, z = 1, and exactly t s
1 s−1 s
− −
roots z ,...,z outside the unit disk.
s+1 t
Thus a can be written as a linear combination of 1 and the (m + 1)th
m
negative powers of the roots of f(z) outside the unit disk. However, explicit
expressions are hard to come by.
Let t 2 be an integer. In the sequel we give expressions for the proba-
≥
bilities P(M(p) (t 1)/t) and P(M(p) 1/t).
≤ − ≤
For the first case we need the t ary tree function T (z). Its power series
t
−
is given by
∞ nt zn
T (z) = .
t
(cid:18)n(cid:19)n(t 1)+1
Xn=0 −
This power series converges for z < 1(1 1)t−1 and for z = 1(1 1)t−1
| | t − t t − t
and represents there the unique solution of the implicit equation T (z) =
t
1 + zT (z)t. (Thus 1/T is the inverse function of y yt−1(1 y) for
t t
7→ −
MAXIMAL PERCENTAGES IN PO´LYA’S URN 7
1 y < 1/t.) We define
| − |
R (p)= pT (qpt−1).
t t
We recall the following fact from path counting combinatorics (“t-ballot
numbers”). (See e.g. [6], Problem 26 of Section 7.2.1.6., for an equivalent
statement.)
Lemma4.2. Leta N 0 . Thenumberofpaths, withstepsin (0,1),(1,0) ,
∈ ∪{ } { }
from (0,0) to (n,n(t 1) + a) whose points lie on or below the line y =
−
a+x(t 1) is the coefficient [zn]T (z)a+1. We have
t
−
a+1 nt+a
[zn]T (z)a+1 = .
t
n(t 1)+a+1(cid:18) n (cid:19)
−
Proposition 4.3. If m = b(t 1) r 0, we have
− − ≥
P(M (p) (t 1)/t) = 1 R (p)m+1 (4.1)
r,b t
≤ − −
(4.2)
p R (p)+qR (p)t−1 for m = 0
t t
−
P(M (p) = (t 1)/t) =  (4.3)
r,b
−  1 R (p) R (p)m for m 1
t t
− ≥
Proof. (1) Consider P(M (p) >((cid:0)t 1)/t).(cid:1)Start a lattice path R =
r,b n
−
r + S ,B = b + n S in (r,b) by stepping (1,0) if Y = 1 , (0,1) if
n n n i
−
Y = 0. Since M (p) > (t 1)/t, there must be a smallest j b such that
i r,b
− ≥
R = j(t 1)+1,B = j and (t 1)B R for i ℓ = jt r b+1.
ℓ ℓ i i
− − ≥ ≤ − −
Equivalently, j b is the first position where the lattice path (n S ,S )
n n
− −
starting at (0,0) steps at time ℓ for the first time above the line y = (t
−
1)x+m. Call this event A . Clearly the last step is Y = 1, appended to a
j ℓ
path from (0,0) to (j b,j(t 1) r)= (j b,(j b)(t 1)+(t 1)b r)
− − − − − − − −
of the type considered in the lemma above. Since each path from (0,0) to
(j b,j(t 1) r+1) has the same probability pj(t−1)−r+1qj−b, we get
− − −
P(A ) = [zj−b]T (z)m+1pj(t−1)−r+1qj−b
j t
= pm+1[zj−b]T (z)m+1p(j−b)(t−1)qj−b.
t
Since M (p)> (t 1)/t is the disjoint union of the A , the first assertion
r,b j
{ − }
follows.
(2) For k 0 let b = P(tS < n(t 1)+k) for all n N). Conditioning
k n
≥ − ∈
with respect to Y then shows that b = qa and b = a for j 1.
1 0 t−2 j j−1
≥
Since P(M (p) = (t 1)/t) = a b the second assertion follows from
r,b m m
the result in (1). − − (cid:3)
Remarks 4.4. (a)ThepathcountingargumentaboveisduetoKnuth,who
used it to determine P(S > (t 1)/t) in Po´lya’s urn (see next section).
1,1
−
Alternatively the results could be obtained using the theorem from [9] given
below, but in a more laborious way.
(b) In the derivation no restrictions on p (other than 0 < p < 1) were
imposed, the formulas above are thus also valid for p (t 1)/t. In fact, by
≥ −
8 ERNSTSCHULTE-GEERSANDWOLFGANGSTADJE
the remark before Lemma 4.2 we have R (p) = 1 for p (t 1)/t and the
t
≥ −
corresponding probabilities in Proposition 4.3 evaluate to 0 (as they must
do in view of the strong law of large numbers).
(c) R (p) p for t . More precisely, lim q−1p−t[R (p) p]= 1.
t t→∞ t
−→ −→ ∞ −
Examples 4.5. (a) Let t = 2, p < 1/2. Then R (p) = min(1,p/q) and we
2
recover the well-known facts that for m = 0
P(M(p) > 1/2) = p/q and P(M(p) = 1/2) = (q p)p/q.
−
and that for m 1
≥
P(M (p) 1/2) = (p/q)m and P(M (p) = 1/2) = (p/q)m(q p)/q.
r,b r,b
≥ −
(b) Let p = 1/2. Then R (p) = 1 and
2
R (p) 0.618034,R (p) 0.543689,R (p) 0.518790
3 4 5
≈ ≈ ≈
R (p) 0.50866,R (p) 0.504138.
6 7
≈ ≈
For example we have
S 6 S 2
n n
P sup > 0.504138 and P sup > 0.381937
n 7 ≈ n+1 3 ≈
(cid:16)n≥1 (cid:17) (cid:16)n≥1 (cid:17)
Let us now turn to the second probability P(M(p) 1/t). We use the
≤
following result of [9].
Theorem 4.6. Let s,r Z 0 , s > 0, r < s. The polynomial
∈ \{ } | |
pz2s zs+r +q (4.4)
−
has exactlys r simple roots z ,...,z outside the unitdisk. If p < r+s,
− s+r 2s−1 2s
we have
2s−1
r+s
P(M(p) ) = (1 z−1) (4.5)
≤ 2s 1 − i
− i=Yr+s
2s−1 2s−1
r+s
P(M(p) = ) = (1 z−1)+p (1 z ). (4.6)
2s 1 − i − i
− i=Yr+s i=Yr+s
Remarks 4.7. (a) In [9] this theorem is proved under the additional as-
sumption (r,s) = 1, but it remains valid if (r,s) = k > 1. To see this
let u = zk. As a function of u the polynomial (4.4) has (s r)/k simple
−
roots u outside the unit disk, say u ,...,u , by the theorem above. If
i 1 (s−r)/k
η is a primitive kth root of unity, then as a function of z the polynomial
(4.4) has exactly the s r simple roots z = ηj u 1/k,j = 0,...k 1 and
i,j i
− | | −
i= 1,...,(s r)/k,outsidetheunitdisk,andsince k−1(1 z−1) = (1 u−1)
− j=0 − i,j − i
and k−1(1 z ) = (1 u ) the formulas above rQemain valid.
j=0 − i,j − i
(b) IQn [9] also formulas for the corresponding a in terms of the roots of
m
(4.4) were provided, but we do not use them here.
MAXIMAL PERCENTAGES IN PO´LYA’S URN 9
Proposition 4.8. Let p < 1/t. If m = b (t 1)r 0,...,t 1 , we have
− − ∈ { − }
P(M (p) 1/t) = (1 tp)q−(m+1) (4.7)
r,b
≤ −
P(M (p) = 1/t) = (1 tp)q−(m+1)p. (4.8)
r,b
−
For m t the a can be computed by the recursion
m
≥
qa = a pa . (4.9)
k k−1 k−t
−
For p 1/t the probabilities in (4.7) and (4.8) are 0.
≥
Proof. (1) Let m = 0 and set s = t, r = (t 2) in Theorem 4.6. The
− −
polynomial pz2t z2+q has 2t 2 roots z ,...,z outside the unit disk,
2 2t−1
− −
anda = 2t−1(1 1). Equivalently, since +z , z are roots, we have a =
0 i=2 −zi i − i 0
t−1(1 Q1)wherey ,...,y aretheroots of pyt−1 y+q outsidetheunit
i=1 −yi 1 t−1 −
dQisk. Let y = 1/(1 u). Then we can write a = t−1u where u ,...,u
− 0 i=1 i 1 t−1
are the non-zero roots of the polynomial g(u) =Qp (1 u)t−1 +q(1 u)t.
− − −
Thus qa = g′(0) = 1 pt. Furthermore, since 2t−2z = q/p we find
0 − − i=2 i −
from the second formula P(M(p) = 1/t) = a0 qaQ0 = pa0.
−
(2) By conditioning on Y we find that a = qa for k = 0,...,t 2, and
1 k k+1
a = pa +qa . The rest is straightforward. − (cid:3)
k k−t+1 k+1
Remarks 4.9. (a) In the case m = 0 and p 1/t, we find a curious
≤
equidistribution property: the maximum M(p) lies in any of the intervals
(1/(k+1),1/k] (for 1 k t 1) with the same probability p/q (and it lies
≤ ≤ −
in the interval (p,1/t] with probability 1 (t 1)p/q).
− −
(b) The resultfor a can also beshown to follow from a path counting result
0
(Barbier’s theorem): the number of paths with step set (0,1),(1,0) from
{ }
(0,0) to (k,n) with n > tk that never touch the line y = tx except at (0,0)
is equal to (n tk) n+k /(n+k).
− n
(cid:0) (cid:1)
5. Distributional results for Po´lya’s urn
Returning to Po´lya’s urn, recall that S = sup Z = max r , M
r,b n≥0 n {r+b r,b}
and let I = inf Z = min r , L . We now give some formulas for
r,b n≥0 n {r+b r,b}
the distribution function of S and I for special values.
r,b r,b
We introduce the generalized harmonic number function
1 1
H(x) = = Ψ(x+1)+γ, x R 1, 2, 3,...
n − n+x ∈ \{− − − }
nX≥1(cid:16) (cid:17)
where Ψ is the Digamma function (the logarithmic derivative of the Γ-
function) and γ is Euler’s constant. For positive integers p,q with p < q
Gauss has shown (see e.g. Problem 19 of Section 1.2.9 in [5]) that
q π p 2pn n
H(p/q) = cot π ln(2q)+2 cos π ln sin π .
p − 2 (cid:18)q (cid:19)− (cid:18) q (cid:19) (cid:18)q (cid:19)
X
1≤n<q/2
Thus H(p/q) can be expressed in terms of finitely many elementary func-
tions.
10 ERNSTSCHULTE-GEERSANDWOLFGANGSTADJE
Let us first consider the distribution function of S at (t 1)/t. The
r,b
−
obvious route to results is to condition on Z, use the formulas for the bino-
mial random walk and integrate the power series of Tm term-by term, using
t
the Beta integrals. The general solution is given in the following proposi-
tion. Note that in the case d > 1 we have to determine the probabilities
a = P(tdS nds m for all n N). Since clearly a = a , (for
m,d n m,d ⌊m/d⌋
− ≤ ∈
r R, r denotes the largest integer not exceeding r) this can be reduced
∈ ⌊ ⌋
to the computation of the a , i.e., the case d = 1, which we dealt with in
m
Section 4.
Proposition 5.1. Let m = b(t 1) r 0 and a = m/d . Then
− − ≥ ⌊ ⌋
∞
a+1
P(S > (t 1)/t) =
r,b
− n(t 1)+a+1
nX=0 −
nt+a B(n(t 1)+a+1+(r/d),n+(b/d))
− .
(cid:18) n (cid:19) B(r/d,b/d)
Ford= 1onegetsseriesofrationalfunctionsofn,whichcan(inprinciple)
be evaluated in closed form in terms of the Digamma function Ψ. We only
give two examples for the results of these calculations:
Proposition 5.2. For d= 1 and 2 t N we have
≤ ∈
2t 3 t 2
P(S = (t 1)/t) = − H(1 (1/t)) − H(1 (2/t))
1,1
− t − − t −
t 2
−
− t 1
−
P(S (t 1)/t) = (1 (1/t))H(1 (1/t)).
1,1
≤ − − −
Essentially this has already been shown by Knuth ([7], Problem 88).
The situation for b = 1,r = t 1 is of special interest because in this case
−
p (t) = P(I (t 1)/t) (the probability that the urn content process
+ t−1,1
≥ −
staysabovetheliney = (t 1)x)aswellasp (t) =P(S (t 1)/t)(the
− t−1,1
− ≤ −
probability that it stays below the line y = (t 1)x) can both be considered.
−
Let q (t) =1 p (t) = P(S > (t 1)/t). One finds that
− − t−1,1
− −
Proposition 5.3. For d= 1,
∞
(nt)! ((n+1)(t 1))!
q (t) = (t 1) − .
−
− (nt n+1)! ((n+1)t)!
nX=0 −
Special values are
q (2) = ln2 0.6931
−
• 4 ≈
q (3) = π√3 0.8061
−
• 27 ≈
9 27
q (4) = ln2+ π 0.8576
−
• 32 128 ≈
q (5) 0.8874,q (6) 0.9068,...,q (20) 0.9726.
− − −
• ≈ ≈ ≈