1
APPENDIX S1
1
2
Result 1 Suppose that Y1 ,, Yr is a random sample of size r from a distribution
3
Geometric P  1  (1  p) . Let h( x) 
4
P

k

k
2
and let Tr


r
Y
i 1 i
r
. Then for
2k
2k

 1 
r h(Tr )  h 
 P 

5
6
1  1 / x 2 / k 1 1 / x 2
2
d


N (0,1)
1
h   r2
P
that is,

1
1 
Tr
h(Tr )  
7
8
1 P  1 
where  
, h  
rP 2
P
9
Proof
2
r
2 / k 1



k2
1  P 2 / k 1 P 2
k2
1

 Tr
2


2
 ~ N h 1 , h 1   2 
      r
  P   P 

P3  2  k
 1  1 1  P 

P  1.
and h  

2
/
k

1
2
 P  k 1  P 

P  2k
2 (1 / k 1)
1  1 / x 2 / k 1 1 / x 2
10
1 P 
1
Note that Tr ~ N  ,  r2 
. Then, since  r2  0 if r   , h( x) 
2 
P
rP


11
1  1 1  P 
P3  2  k


P  1  0
is differentiable with respect to x  (0,1) and h  

2
/
k

1
2
 P  k 1  P 

P  2k
k2
2 (1 / k 1)
12
13
14
  1    1  2 2 
2k
for P 
, then using the delta method we get, h(Tr ) ~ N h ,  h'     r  .
2k
  P    P  

2
15
Result 2 Suppose that Y1 ,, Yr is a random sample of size r from a distribution
16
Geometric P  1  (1  p) k . Let h( x) 
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the smallest integer value rm such that



P 2Z1 / 2


18
1  1 / x 2 / k 1 1 / x 2
k
2
1  Pˆ  Pˆ 
2 / k 1
2
rm k 2
1
and let Pˆ  
Tr
1
i1Yi / r
is approximately
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



rm  



21
Proof
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  1    1  2 2 
1 P
From Result 1 we have that h(Tr ) ~ N h ,  h'     r  where  r2m 
,
rm P 2
  P    P  

23
1
h  
P
24
 quantile of the standard normal distribution. Therefore,
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1  P 
k
2
P
2

1  P 2 / k 1 P 2
k

P 2 Z1 / 2


2
1  P 
2 / k 1
k2

P
2
1 (1  P)
P 2k
 1 P 

Z
P  1

2

Z1 / 2
k 1  P 2 / k 1 P 2  2k
 P 



Z1 / 2


2 (1 / k 1)
2
P3
 1  1 1  P 
and h  
 P  k 1  P 2 / k 1 P 2
2 (1 / k 1)
1  Pˆ  Pˆ 
2 / k 1
rm k 2
2
. Then,

   


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2 / k 1
r




    P





2
3
2k
2k

P  1 for P 
. Let Z  the

2k
 2k


1 
 h 
2 Z1 / 2
P 

 
2
2
  1   (1  P )
  1   (1  P ) 
 h'   
 h'   
2
2 
  P   rm P
  P   rm P 
1 1
h   h 
 Pˆ   P 
 rm
3


 rm
1 

 h 

2 Z1 / 2
P 
 P Z 
  
2

  1   (1  P ) 
 h'   

2 
  P   rm P 

26

27

2Z1 / 2

28

2
  1   (1  P )
 h'   
2
  P   rm P
2
that
Eq.
30
x  rm , a 
31
x
(1.S1)
 Z
  1   (1  P)
0
 h'   
2
  P  P
2Z1 / 2
Note
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2 Z1 / 2
1
 h 
P
1
 1  (1  P)
rm  h  rm  Z  h 
0
P2
P
P
29
32
1
rm  h  rm  Z 
P
 rm
has
a
quadratic
(1.S1)
form
ax 2  bx  c  0 ,
with

1
 1  (1  P)
, with two solutions given by
, b  h , and c   Z  h 
2Z1 / 2
P2
P
P
 b  b 2  4ac
 b  b 2  4ac
. Taking x 
for fixed  , the desired rm is given by
2a
2a
2
 1
 h   h 1   2 Z h 1  1  P

2
 P
Z1 / 2
P
P P

rm 



Z1 / 2

we obtain the result and Eq. 9.
2



1
1
 , in which after replace h  and h 
P
P


