UNBIASED COIN TOSSING WITH A BIASED COIN
by
Wassily Hoeffding t
Gordon Simons*
Department of Statistics
University of North Carolina at Chapel Hill
Institute of Statistics Mimeo Series No. 618
APRIL 1969
•
t
Research partially sponsored by the Mathematics Division of the Air
Force Office of Scientific Research Grant No. AFOSR-68-l4l5; * and
the National Science Foundation Grant No. GU-2059.
UNBIASED COIN TOSSING WITH A BIASED COIN
by
Gordon Simons *
Wassily Hoeffdingt
University of North Carolina at Chapel Hill
O.
SUffi1.Uary.
PrQcedures are exhibited and analyzed for converting
a sequence of
iid
Bernoulli variables with unknown mean
Bernoulli variable with mean
1/2.
The
p
into a
efficiency of several pro-
cedures is studied.
1.
Coins probably have been flipped since the
Introduction.
dawn of their invention for the purpose of making impartial decisions.
That such decisions are in fact partial is widely accepted.
envision an idealized coin which produces
Xl' X , .•.
2
with unknown parameter
p ,
iid
<
0
Bernoulli variables
p
<
1.
(Such sequences
can be closely approximated using radio-active materials.)
and
M , M , •.•
2
l
sequence
is any predetermined sequence
E , E ,
2
l
is also a sequence of
the
defined by
iid
E
v
~ (M
v
We shall
of
If
zeros and ones, the
+ Xv) modulo 2
Bernoulli variables with mean
(v 6 1)
1/2.
M sequence represents a "message" in binary format, the
sequence represents an "unbreakable" encoded message.
M = (E
v
v
+ Xv) modulo 2
p = 1/2
If
E
The formula
(v 6 1) can be used for decoding.
Repeated
application of any of the procedures to be discussed will produce a
Bernoulli sequence with
p = 1/2
from a Bernoulli sequence with
One can not define a function of the
Xl s
p
E
wn.;i:chis':,a, 'Bernoulli
t Research partially sponsored by the Mathematics Division of the Air
Force Office of Scientific Research Grant No. AFOSR-68-l4l5; * and the
National Science Foundation Grant No. GU-2059.
(0,1).
Z
variable
Z with mean
1/2
using any (non-randomized) fixed sample
n ~ 1 , there are values of
size procedure since for any
An elementary sequential
p{X. = 0 , i=l,"',n} > l/Z.
such that
p (0 < p < 1)
~
procedure is
Ql:
Sample
time
sequentially in pairs and stop the first
Xl' X ,'"
2
Zm
for which
X
Zm
~
Zm- l '
Set
X
Z
= XZm
.
This procedure is described by von Neumann (1951) and has been rediscovered by others.
pointing it
o~t
The authors are indebted to N.L. Johnson for
to them, whfch led to the present investigation.
An extension of this procedure to Markov chains has recently been
considered by P. Samuelson [1].
Denoting the sample size by
N , one finds the expected sample
1
size to be
=
p
-1 -1
, 0
q
There are better procedures than
cedure
Q is better than procedure
sizes satisfy
N =< N' (N = N' =
00
p
<
Ql'
Q'
<
1 , q = 1 - p •
We shall say that a pro-
if the corresponding sample
is permitted) for all sample
sequences, with strict inequality for some sequence.
In Section Z,
we investigate a class called "even procedures".
belongs to this
class.
The best even procedure
size
(1)
=
Q
l
Q will require an expected sample
Z
3
In Section 3, we examine a non-even procedure
than
Q2'
We conjecture that
Q
3
that no better procedure exists.
Q which is better
3
is "weakly admissible" in the sense
In Section 4, we discuss admissibility
and show that there is no procedure which is better than (or as good as)
all others.
In connection with this, we obtain a lower bound for the
expected sample size and introduce a fourth (non-even and non-symmetric)
procedure
small
Q which has smaller expected sample size than
4
Q
3
for
p •
We tabulate the expected sample size for the range of
P
.4
.5
.6
Q1
4.17
4.00
4.17
Q2
3.5.7
3.40
3.57
Q3
3.28
3.10
3.28
Q4
3.50
3.38
3.55
3.17
3.00
3.17
Procedure
Lower Bound
The margin for improvement in
4%
one is
.4':::: P ~ ... 6 :
most likely to encounter,
size is less than
p
Q
3
in terms of the expected .sam'ph
over the entire range of
p , 0
<
p
<
1 •
n
and S - 0 .
S - ~X \) , n = 1,2""
o
n
1
As one samples sequentially from
and plots S
versus
Xl' X2 ,
n
2.
Even Procedures
Let
...
n - S
n
, one generates a "path" in the Euc1idean.p1ane.
stopping set
S
is a set of points
(i, j)
with
i, j
A
nonnegative
4
integers, the stopping points.
(n - S
S)
n' n
smallest integer
N
=
00
to
points
(i, j)
The stopping variable
0 such that
Z,
Let
with
N is defined as the
(n - S , S ) E S
n
n
if one exists, and
denote the number of paths from (0,0)
m(i, j)
N~ i + j •
with
such points.
that
n
otherwise.
(i, j)
S
is in
Sampling is stopped as soon as
For mathematical convenience we include
min(i, j)
0
<
and note that
=0
m(i, j)
for
In order to make all stopping points reachable, we insist
m(i, j) > 0
for all
S , we call a point
(i, j) E S.
(i, j), not in
S
m(i, j) > 0
or
inaccessible point as
C
the corresponding sets by
(i, j) E I
(2)
~
(3)
N.
p E (0,1)
probability one for
=0
, respectively.
We denote
Thus
if, and only if,
Anyone of the three sets
and the stopping variable
a continuation point or
I.
and
With this restriction on
S, C and
=0
m(i, j)
.
I determines the other two
The condition that
N is finite with
is expressed by the identity
m(i, j) qi p
j
=
1,
0
<
P
<
1 .
(i,j)ES
If this condition is satisfied, an impartial procedure or, briefly,
procedure is determined by
(0,0)
on the paths from
which has mean
1/2
and, for each
(i, j)
from
(4)
(0,0)
to
(i, j)
and a Bernoulli variable
to the points of stopping
0 < P
for
E
S
S
<
1.
k(i, j)
on which
Z
~ k(i, j) qip j
(i,j hS
If
Z, .defined
(N - SN'
S~)
Z exists for a given
denotes the number of paths
=1
=
, we have
1/2,
0
<
P
<
1 .
,
S
5
Conversely, if
° S k(i,
is satisfied and there exist integers
(3)
j) ~ m(i, j) , such that
(4)
holds, we can define
(i, j) E S , the paths from
subdividing, for each
into two arbitrary subsets of
k(i, j)
paths and assigping the value
Z
=
k(i, j)
and
=
1 (Z
first (second) subset (or vice versa).
(0,0)
by
Z
to
(i, j)
m(i, j) - k(i, j)
0)
to each path in the
Two procedures defined on the
same stopping set will be considered as equivalent, and our main
concern will be to determine stopping sets for which at least one
impartial procedure exists and
EN
is small.
We introduce this
equivalence as a mat-ter gf convenience but do not wish to suggest that
the definition of
If
N
<
Z
is completely unimportant.
with probability one (0
00
even for every
p
<
(i, j) E S , we can satisfy
k(i, j)
(5)
We call such procedures
stopping set
<
81
-
m(i, j)/2
=
~
procedures.
1)
and
(4)
is
by setting
for every
Ql
m(i, j)
(i,j)
E
S •
is an even procedure with
{(i, j) : i, j = 1,3,5, ... } .
It may be instructive to note that an even procedure is equivalent
to a test of Neyman structure.
Indeed, a procedure of the type
studied in this paper may be viewed as a similar test of size 1/2
for testing the hypothesis that
with mean
p ,
considered.)
°
<
p < 1.
Xl' X ,
2
is a Bernoulli sequence
(Alternatives to the hypothesis are not
The very essence of the problem requires that the test
be nonrandomized, that is, that
Z
take only the values
For a given stopping rule, the random variable
sufficient statistic.
°
(N - Sw' SN)
and
1.
is a
In accordance with the usual definition,
Z
is
6
a nonrandomized test of Neyman structure if the conditional probability
of
Z = 1
(i, j)
given
S.
€
numbers
(N - Sw N) = (i, j)
is constant for all
Such a test exists (with constant 1/2) if and only if the
m(i, j)
are even for all
(i, j)
€
S .
We turn our attention now toward characterizing even procedures.
We shall say that the variable
with certainty or not) if
(i, j)
coefficient
reaching
(i~j)
(i, j)
Theorem 1.
by
;L
divides
... ) .
= 2, 3,
S (k
€
k
N is "of type .k " (whether it stops
~
is the number of ways
( c(i, j)
.
(0,0)
The following
for every
In the sequel, we denote the binomial
c(i, j)
from
m(i, j)
c(i, j)
=
° if
min(i, j)
.
N is of
(b)
k
divides
m(i, j)
for every
(i, j)
€
S u
(c)
k
divides
c(i, j)
for every
(i, j)
€
S
(d)
k
divides
c(i, j)
for every
(i, j)
EO
S u
(e)
k
divides
c(i, j) - m(i, j)
Proof.
Observe that for all
k
c (i, j)
(6)
°
.)
equivalent.
(a)
~
<
ef
for every
r
.
r
.
.
(i, j)
.
(i, j)
=
c ( i-I, j) + c (i, j -1) ,
and
where
= m(i-l,j)I(i-l,
m(i, j)
(7)
I(i, j)
=1
or
°
Assume that at least one of
Certainly
(e)
holds for
as
(i, j)
(a) - (d)
(i, j)
E
j)
+ m(i, j-l)I(i, j-l) ,
C or
hold.
such that
sur
respectively.
We shall prove
i + j ~ 0.
Let
(e) .
n ~ 1
7
and assume that
Let
i + j
(e)
= n.
holds for all (i, j)
k
if
(8)
alone if
(i-I, j)
divides
(i-I, j)
S ,or
E
i + j
<
n .
By the induction assumption,
(8)
Using
for which
(8)
c(i-l, j) - m(i-l, j) •
E
and
C ,(8)
and anyone of
(2)
(i-I, j)
if
E
I
k
(9)
divides
(a) - (d)
1 , we conclude that
I
c(.:L-l.;- i) - m(i-rl.; j) I (i-;l., j ) .
Similarly,
k
(10)
divides
c(i, j-l) - m(i, j-l)I(i, j-l) .
Finally we complete the induction step by concluding
•
c(i, j) - m(i, j)
from
(6), (7), (9)
Conversely we shall derive
follows that
(e)
implies
(a)
Then there exists an
(11)
i + j
=
from
(a) - (d) .
does not.
n, (i, j)
n
E
Sand
(i+l, j-l)
(12)
and
j .
(i+l, j) .
c(i, j)
(e) .
With this, it easily
Suppose
(e)
(i, j)
m(i, j)
(i, j)
holds but
not divisible by
satisfying
We work with the three points
With
=
(6), (7)
and
(a)
with
(11)
c(i+l, j) - c(i+l, j-l)
and
(13)
divides
(10) .
and some
Let us assume we have chosen the poirit
the smallest value of
and
k
m(i+l, j) = m(i+l, j-l)I(i+l, j-l) .
(11)
k.
with
(i, j)
we conclude
8
From
(e), (11)
(lZ) , it follows that
and
c(i+1, j) - c(i+1, j-1) •
(14)
k
(Z), (13)
and
In turn,
does not divide
(14)
(15)
and
~l~),
k = Z
N<
for every
I(i+1, j-1) = 0 ,
= n , (i+1, j-1)
E:
S ,
is not divi,sibJ.ei.J;>y
m(i+1, j-1)
j
is the smallest value
U
it is clear that an even procedure exists if and
with probability one (0 < p < 1)
00
(i, j)
QZ with Sz
1
E:
S u 1.
and
c(i, j)
is even}.
Of course, no better
We note, in passing, that Theorem 1 directs us
to a class of procedures for defining a random variable
distinct values with the same probability
Z* which takes
11k.
For the remainder of this section we shall study procedure
It is helpful to observe Figure 1.
QZ'
The points indicated by the first
two symbols correspond to the noncontinuation points
Q2'
is even
In particular, there exists an even procedure
Z = {(i, j) : c(i,j)
even procedure exists.
k
}.:k, •
(11) •
Letting
only if
m(i+1, j) - m(i+1, j-1) •
contradicts the assumption that
satisfying
does not divide
implies
are compatible only if
(i+1) + <.1-1)
But
(e)
k
Sz u 1 Z of
Several easily proven facts become apparent.
(i)
N is even.
Z
(ii)
m(i, j) = 0, 1; or Z as
(iii)
For each
11.' ne
V' ~ 0
i +. J' -- Z\)
1 , £2 ,~.or :...·SZ:·2respecl=ive1y.
2
the only continuation points (i, j) on the
(i, j)
are th e t·wo
.'
E:
':t1t
pc~
s
and
\)
(0, Z ) •
•
9
v ~ 1 , if one shifts the origin from
For each
(iv)
(ZV, 0)
to
(0, Zv) , the stopping variable which results
or
does:~not.;dltffer.from:~:N
are made on
(v)
(0,0)
Unless: 2Y; or m017e.obs.ervations
2
X.
A recursive definition for
N may be roughly stated as
Z
"Stop the first time one reaches a point with an even number
(1ttVo)of not previously stopped paths to it."
From
(ii) , it follows that a legitimate definition for
Z
is
given by
(16)
S
Z
modulo Z •
NZ-1
Now let
an
(17)
-
P{N
Z
>
,
n}
n
00
(18)
r
.-:- r=O a r t
l:
A(t)
n = 0, 1, Z,
A (t)
n
-
t
r=O
a t
r
r
,
...
n = 0, 1, Z,
... .
We have from (iii) ,
and from
(ZO)
1
=
(19)
(iv) ,
am+n
m =
=
Hence for
=
=
Zl ,..,
Z2 •••
.
10
Letting
m-+-oo ,
00
A(t)
(22)
Setting
v~O {1 + (pt)2
=
00
=
Higher moments of
N
2
(24)
EN(N
Z
IT
{1 + p2
v=O
- l)···(N
defined by
(Neither
€
C
2
N nor
Z
= kA(k-1) (1) .
Under procedure
2
for which, with certainty,
,
(16)
v
+ q2 }
- k+1)
Q
A better procedure than
(i, j)
v
can be obtained from
222
points
v
+ (qt)2 } •
t = 1 , we have (in agreement with (1))
(23)
3.
v
there are
N = i + j + 1.
With
Nth ob servat~on.
.
there is no need to take the
depend on it.)
Q2
This suggests procedure
Q3
given by:
S3 u 1 3
-
{(i, jc) : c(i, j)
is even or both
c(i+1, j)
c(i, j+1)
and
,
are even}
and
Z
-
SN-Ci.
(or [because
S3 u 1
modulo 2 , Ci. = 0
3
3
N
2
c(N3-SN' SN) is odd or even
3
3
is even] equivalently as N is odd or even).
3
1
as
is denoted in Figure 1 by all .three symbols.
b
(25)
:~
n
P{N
00
(26)
or
B(t)
-
~
b t
r=O r
r
3
>
n}
,
n = 0, 1, 2,
n
B (t)
n
Let
-
~
r=O
b t
r
r
,
n = 0, 1, 2,
... .
11
Procedure
Q
3
has similar counterparts to properties
of procedure
m=2 1 , 22 ,.•••
and one finds for
(27)
b ill-:: l
(28)
bm+n
-.=
p
m-1
+
b b·.
=
mn
q
m-l
b
n =
m
O~
(: ~~)
(:Ui) and
that
= p m + qm ;
m-2 •
1,
By direct computation, one can show
n
(29)
Corresponding to
(21)
_ (t)]
2m 1
-1
B _ (t)
2m 1
Upon summing over
(21)
=
m = 2 1 , 22 ,
m= 2 1 , 2 2 ,
one can obtain for
Using the extremes of
[A
= m-1 ,
and the fact that
b
m
= a
m
,we find
[Am-l ( t.)] -1Bm- L~()
t - [A2m_1(t)]-1(pq)m~lt2m-l
m and using the fact
=
A (t) = B (t) , we conclude
1
1
m = 2 1 , 22 ,
1 - D (t)
m
•••
where
~
D (t)
r =21 , ••• , m[AZr _ 1 (t)]
m
Letting
m -+
00
-1
r-l 2r-1
(pq): t
,
B(t)
(30)
=
A(t)(l - D(t))
where
D(t)
00
2s -1 2s +'1-1 s
2v
2v
t .
/ vllO (1 + (pt)
+ (qt) )}
s~l {(pq)'
,
12
In particular, since
(31)
B(l) = EN
3
and
A(l) = EN
2
'
13
~.
5
+
3
'lor
I""
~r~
...
~
",Lor
J
jllIl.
41"
lIp1f
. ,..
... ~
2
Lor
I
0
I""
.1...
JI""
"11'J"
41 lit
.pIf
....
... ~
lIl5"
... "
;Ill'"
~
.
... '
~
iH
"'~
..
111'"
• I'
.. r~
•
'"
....
~
I•
j~
lIl'"
i•
1111"
.
~
I"
"lit
lIpif
....
4llpif
•
~1It
JJIIr
"'"
... ...
'
~
..
~
lIpif
"
o
.....
...
I~
~
...
~
,
.ti
"
~Iit
.... .
4111'"
1115"
.....
..
-i. ,..
,~
~
~"
...
~
~
;lIl
.....
4111'"
~
....
~.
"11'"
...... '
~
;lIl1""
2
3
..
~
•
••
• •*
~
•
.•
.. ~
-.I.
... ~
,~
..,
~
..~
.,.
... .......
...
~
4~
....
,~
['
JJIIr
5
.till.
1111'"
...
"
.,.. ,I...' ,
r~
·... ,...
.....
~
.. I.
"115"
• Lot
JIIr
.. La
lIpif
....•
.. ..
.'.l.
" I'"
... '
~
~
['
T
O~
8
9
~
..
~
tilt
• fit
1I
lIpt
"lit
lIpIf
I"
41"
lIl'"
jllit
4IIp1f
.J1t.
jllit.
1I1'J"
"1"
Lor
...
.. ..
.... .... -.I"
jilt.
.til It.
4111"
"Of
41"
.. lit
lIpif
.. Lor
.tllit
.pIf
I"
.tllit.
:Jl1'L
.,.
~
tI I'"
~It.
JI""
...
.. lit.
t"
"lit
1I"
..
~
...~
... ~
NONCONTINUATION POINTS OF
Q1
NONCONTINUATION POINTS OF
Q2
NONCaNT INUAT ION POINTS OF
Q3
Lor
•
4
"I'"
t~
.,.
~
jllit.
Jllit.
11 I'"
1115"
,~
r, ~
"
TAilS
.
.
jilt.
.'
...
II
Lor
11"
1I1'"
•... [r..'
.,.
.,. iH If
11 I'"
,~
.J1t.
1111'"
111'"
....
lor
~
"1p1f
111 pt
41
•
1111'"
4llpIf
.
.
jllit.
41"
lIl'"
4llpIf
~ It.
lor
i~
Jill..
.... ".
J
• •
. ...
lit
111'"
·... ,...
I~
~
~
II pt
['
Lor
NUMBER
•
41 l.
-JIIr
6
...
lIp1f
"..
....
i.
.,.. I ' -ill-
~
:Jl
jl ~
.pt
....
,
iH H
.... ..
1I I'"
"11'"
~
.. l.
-it
.
.,.
jlllt
"lJllr
lIIIt"
... ....• ....I
tII~
"1p1f
jilt.
r~
-
•
1111""
I'
+
,
~~
111'"
['
.. ..
"111"
1111'"
jl ~
'11 I'"
i.
~
... ,
41"
• lit
'11 I'"
lIp1f
111'"
1111"
... ...•
111'"
.. r~
.... ,
~1It
.. It
.. "
~
lit"
'111'"
... ... ....
... ...
... La
lIIp1f
...
111'"
...
it
.1'
Lor
.
111 I'"
..
til'"
.. ~
~,'~
"'"
"
..
w
-lit"
~
U.s
•
...
....
II.
.3
15
---
n-Sn
14
4.
Problems of existence and admissibility.
n 6 2
for which
min(n-S , S ) = 1.
n
n
NS be the first
Let
There is no procedure that stops
NS for the same reason (given in Section 1) there is no
sooner than
fixed sample size procedure. Hence;,fol::.an'a.rbitrary. p-ro'cedurewith stppping variable
N there is a general lower bound
EN
(32)
-1 -1
>
P
q
-1, 0
<
p
<
1 .
We raise the question whether there isa procedure with stopping
variable
N
S
ever equality.
Theorem 2.
or, equivalently, whether the inequality in
a
= 0,
Ss = {(i, j):
There is E£ procedure with stopping set
or
j=l, i ~ I} .
By equation (4) , we must show that there do not exist numbers
or
1,
2,
= o
(33)
or
1 , ~
=
o
or
1 ,
k = 1, 2, ...
such that
apq
(34)
+
=
Clearly we can not have
for all
is
The answer is no as the following theorem shows.
i=l, j ~ 1
Proof.
(32)
k
.
~
k
= 0
a = 2 , then
Hence i f
which is not true for
= n
p = 1/2
a similar contradiction.
.
for all
(34)
o
1/2 ,
k
nor
1/2
implies
a = 1
.
= nk = 1
~
The assumption '"a'·=';O
Thus we must have
p < 1.
<
>
2pq
leads to
Since
15
equation
(34)
with
a
00
k~1{(2~-1)p
(35)
Let
f(p)
=1
k
is equivalent to
k
=
+ (2nk -l)q}
0, 0
denote the left-hand side of
k(k-l)···(k-r+l)
denoted by
the r-th derivative of
f(p)
k(r)
(k(O)
<
p
(35) •
=1
or
<
= l-p
1 , q
•
With
0
as
k
>
= 0)
or
is
and
Hence, for
= 0,
r
1, 2, ...
f(2r)(1/2)
(36)
f(2r+l) (1/2)
(37)
where
(f(0)(1/2) _ f(1/2)
=
=
and)
k
(38)
Since
f(p)
= 1,
2, ••..
is identically zero, it follows that all its derivatives
are identically zero and, in particular, that the expressions in
and
(37)
are zero.
The numbers
f(2r)(1/2)
c
k
=1
We first show that
c
=0
k
are restricted to the values
=0
for
r
= 6,
k.
k
-1, 0, 1 .
= 1,
2,
Since
1, •.• , itf61lows that we can not have
for all sufficiently large
sufficiently large
for
(36)
f(0)(1/2)
k
=~
nor
c
k
= -1
for all
is equivalent to
,
16
00
k~12
-k
If not all
~ = 0 •
such that
c
m
~
ck
=0
and
m denotes the first integer
0 , then we must have the strict inequality
't' 2-k
I
= Ik=m+1
ck
2-m
a contradiction.
With all
~:ck
00
2:
<
k=m+1
2-
k
=
-m
2
,
= 0 , it follows that the sequence
{d } is restricted to the values
k
( 39)
Furthermore, the conditions
(40)
d
or
-1
=
k = 1, 2, .•..
1 ,
f(2r+1)(1!2) = 0
imply that
changes' sign infinitely often.
k
Now consider the equation
f(1)(1/2) = 0 , which is equivalent to
(41)
Let
=
( 42)
We now show that equation
(41) , subject to conditions
(40) , is satisfied (if and) only if the
as follows.
d
1
= 11
d
k
if
1
(43)
d
k
=
1
or
-1
if
if
-1
First note, by
(41)
and
(42)
00
IDkl
=
Iv=k+1
~ 2-v
vd v I
,
~
2
D_ < 0
k 1
Dk - 1 = O'
Dk - 1 > 0
.
that
00
<
2:
v=k+12
-v
and
are recursively defined
k
is arbitrary, and for
(39)
v
=
2-k (k+2)
,
17
with the strict inequality.
integer.
Clearly
It is seen by
that
(42)
Hence
d
l
is arbitrary.
If
d , "', d - (k
k l
1
=
~
2)
are given, then
<
Equivalently,
-(1 + dk )k/2
_ 1/
~
2
2k - l n
~
k-l
or, since
(1 - dk )k/2 + 1/ 2
are integers,
-(1 + d )k/2
k
k l
2 - n
k-1
<
=
This immediately implies that
d
k
<
must satisfy
(43).
(It is easy
{d } defined by
(43)
satisfies
We complete the proof by showing that
(43)
is incompatible with
to see that any sequence
f(3) (1/2) = O.
By
k
(40)
and
(37) ,
(45)
where
n =
(46)
Now
( 47)
<
'f (k)2-k
k=n 3
=
4, 5, . . . .
(41).)
18
IA2 ,9 1
and; in particular,
(43)
and
(46),
IA1 ,9 1
~ 65/128.
By direct computation from
= 79/128 ~
This contradicts
(45)
and
proves the theorem.
It is an interesting fact that while no procedure exists for
N
5
Q
there is a procedure
N
N
4
{
-
1
N
Q exists.
4
Proof that
s*
$**
==
for a slight modification, namely
4
5
if
(N 5 - SN ' SN )
_. {w
,
(1, 2)
otherwise.
Let
w = (i,l)
{w
=
5
5
w = (i,l)
or
or
(l,j); i=2,4,6,··· ;j=4,6,8,···}
(l,j);i,j,= 4,8,12,···} ,
and set
(48)
1
for
w
€
s**
o
for
w
€
s* - s** .
g(w)
Consider the condition
(49)
Z = g(w)
If we can find a (even) procedure
Q1 *
satisfies the restrictions imposed by
then we can define
stopping in
It is
with
Sl
Q
and
~ufficient
(5), (48) and
w, w
for every path which continues through
4
by assigning
Z = g(w)
(5)
Z
with
according to
Z
which.~
and
Q1 * for paths
S* .
can be defined in acoordance
on the paths to points
s* .
,,
Sl
S = Sl ' (48)
for paths stopping in
to show that
(49)
for the stopping set
€
(i, j)
€
Sl
(49),
19
i, j ~ 3
with
.
Let
(i, j)
m(i, j) = m + m + m
2
3
1
paths to
(i, j)
points in
to
be any of these points •
m , m
1
2
where
m
3
and
that pass through the point
(i, j)
on which
k
1
and through
Considering the paths
Z = 1 , we must have
k(i, j) = 1/2 m(i, j) = k
where
denote the numbers of
(1,2)
s** and s* - s** , respectively.
Then
+ m2
1
is the number of these paths which pass through
Thus we must show that the integer
k
1
(1,2) •
defined by
=
a
satisfies the inequalities
~ k
or, equivalently, that
1
<
=
(50)
Let
m(i, j;
(i, j)
m~(i,
and
i~)
m~(i,
which pass through
j;
j~)
= m(j, i;
j~)
j,
j~)
denote the number of paths to
and
(l,j~)
(i~,l)
and
m =
1
m~(i,
, respectively.
Then
j, 2) , m = 1 + 1 '
2
2
1
m3 = 1 + 1 4 ' where
3
1
1
1
1
1
2
3
4
-
E
m(i
(i~,l)€S**
E
( ...o{,j~)"'S**m
""
'
~ (i ,
J','
i~)
' J''')
J;
=
v=4, 8,E . • • m(i , j; v)
=
E
("
v= 4 ,8,'" m J,
=
-
~;
v)
v=2J, •.• m(i, j; v)
-
An elementary combinatorial' a1:"gumen·t shows that
m(i,j;v)
=
2c«i-v-1)/2, (j-3)/2)
for
(i,j)
€
Sl' v=2,4,'"
•
20
It follows that
=
m(i, j; 2)
(51)
m(j, i; 2) ,
and
m(i, j; v)
( 52)
(50)
and
quickly follows from
m(j, i; v)
(51)
and
are non-increasing in
(52).
EN
We shall not derive the formula for
difficult and one finds that
Q
3
This means that
EN. for
mize
EN
4
<
EN
This completes the proof.
here, but it is not
4
for small values of
3
v •
p > 0 •
is not uniformly optimal (i.e., does not mini-
0 < p < 1 ) among all procedures.
Our:.last theorem concerns the subject of weak admissibility.
Recall that a procedure
Q is weakly admissible if there is no
better procedure (i.e., no procedure which stops as soon as and sometimes sooner than
For every procedure
Theorem 3.
procedure
Q.)
Q*
Q there is
~
weakly admissible
Q.
which is better than E.E. equivalent to
We shall need the following lemma.
Lenuna.
Let
variables
M , M , ••• , respectively.
2
1
v
-+
~
00
where
for all .p
~
variable
R , R , •..
2
1
Moo
E
M00
is
~
be
~
sequence of procedures with stopping
If
-
Mv
-+
M
00
in law for all pE (0,1) as
---
----
stopping variable which is finite with probability
(0,1) , then there exists i! procedure
Roo
with stop-
21
Rather. than giving a constructive proof, we
Proof of Theorem 3.
shall use Zorn's lemma and some of the terminology commonly used with it.
(See, for instance, [2, page 40].)
Let
of complements of continuation sets
C'
a procedure
Q'
tinuation set of
with continuation set
Q.)
P
P
be the class composed
C such that there exists
c
C'.
(C
denotes the
con~
is partially ordered by set inclusion.
Our lemma verifies that every completely ordered subclass has an
upper bound since the number of elements in the subclass is necessarily
coub.1:ahle.
P
Zorn's lemma implies that there exists a maximal set in
C*.
whose complement we denote by
weakly admissible procedure
Q*
Consequently, there exists a
which (has continuation set
C*
and) is better than or equivalent to
Q.
Proof of Lemma.
be an ordering of all points
(i, j) .
Let
wI' w ' •••
2
By taking a subsequence of
if necessary, we can assume
{R }
\)
without loss of generality that the classification of the point
w
or.
as a continuation, stopping or inaccessible point remains the same
and
m(wor.)
is constant for procedures Ror.' Ror.+l' .•. , for every
This is because
P{M\) = n} , when expressed as a linear combination
of the linearly independent functions
qp
n-l
coefficients which must be integer valued.
wI
is a stopping point for
•.• ,q n-l p, has
Starting with
R , R , •.• , we let
1
2
Z = 1, is a constant, say
stopping point then let
R , R ,
1I
12
Proceedin,g ,recursively., -we; l.e.t.
k*(w ) •
I
De
wI' if
R , R , ••.
11
12
a subsequence for which the number of paths stopping at
on·.which~
or.
If
the original
wI " and
is not a
sequence~
be
22
equal
if
Ra - 1 ,1' Ra - 1 ,2'
point and let it be some subsequence of
w
is not a stopping
a
R
R
a-1,1' a-1,2
on which
the number of paths stopping at
constant, say
k*(w ) , i f
w
a
a
is a stopping point,
Working with
of the paths to
paths to
w
a
w
a
R
R00
a = a
O
k*(w)
a
We claim that this
w
a
with stopping variable
e: > 0
This is because
M00
such
there is an
Z = 1
' the probabilities that
differ by less than
'€Xa!
to
= 0 to the remaining m(w a ) - k*(w a )
p , 0 < p < 1 , and fixed
>
that for
Z
Z = 1
Moo' we assign
for each stopping point
defines a procedure
for fixed
and
a > 1 .
under
and
R00
e: •
Remarks.
1)
Because of Theorem 2 , the
non-continuation; set:co.:L
stopping point of
non~equiva1ent
2)
Q4'
can not be added to the
Since the point
(1,2)
is a
weakly admissible procedures exist.
The lemma (used to prove Theorem 3) and Theorem 2 can be used to
p
-1 -1
q
-1
(32)
is not "tight" in the sense that
is not the greatest lower bound for
EN.
We conjecture that there is no uniformly optimal procedure, that
is, a procedure which minimizes
4)
(1,2)
Q ' Theorem 3 guarantees us that at least two
3
show that inequality
3)
poin~
Let
Sand
EN
for all
S' c S be two stopping sets.
fn~ ~topping variable
N'
associated with
p e: (0,1) •
Suppose that
S'
is finite with
23.
probability one (0
S.
stopping set
Q'
procedure
5)
Either
\.Q
procedure
3
<
p
<
1)
and there exists a procedure
Q with
Then (it is easily shown) there exists a
with stopping set
S' .
- ,is weakly admissible or (because of the way
Q' , referred to in remark 4, is constructed) there
exists a procedure which is equivalent to
Q2 but violates
Showing that a procedure is weakly admissible seems to be a
difficult· task.
(5)
.~
24
REFERENCES
[1]
Samuelson, Paul A. (1968). Constructing an unbiased random
sequence. J. Amer. Statist. Assoc. 63. 1526-1527.
[2]
Taylor, Angus E. (1958).
Wiley, New York.
[3]
von Neumann, John. (1951). Various techniques used in connection
with random digits. Monte Carlo Method, Applied Mathematics
Series, No. 12. 36-38. u.S. National Bureau of Standards,
Washington D.C.
Introduction to Functional Analysis.
\
\
--