SUPPLEMENT TO “EXACT EXPONENT IN OPTIMAL RATES FOR
CROWDSOURCING”
Chao Gao, Yu Lu and Dengyong Zhou
A
Proof of Corollary 3.1
Proof. Under the assumption that mI(⇡) ! 1, the upper bound is a special case of Theorem
3.1. Note that
I(⇡)
m
=
1 X
min
log p1i t (1
0t1 m
pi )t + p1i t (1
pi ) t
i=1
=
m
⇣ p
1 X
log 2 pi (1
m
pi )
i=1
= I(p).
⌘
We focus on the proof of the lower bound, which involves weaker assumptions than that of
Theorem 3.1. Using a similar analysis as (15)-(16), we have
sup EL(ŷ, y)
inf
ŷ y2{1,2}n

n
X
1
n
inf
j=1
ŷj
1
1
P1 {ŷj = 2} + P2 {ŷj = 1} .
2
2
Following the proof of Theorem 3.1 with the confusion matrix ⇡ (i) replaced by (5), we have

1
1
inf P1 {ŷj = 2} + P2 {ŷj = 1}
ŷj
2
2
exp ( mI(p)) e
where Sm =
P
Therefore, Sm
L
Q(0 < Sm < L),
and under the distribution Q,
◆
✓
◆
1
1 pi
1
pi
1
Qi Wi = log
= Qi Wi = log
= .
2
pi
2
1 pi
2
p
has a symmetric distribution around 0. Letting L = 2 VarQ (Sm ), we have
i2[m] Wi ,
✓
1
1 VarQ (Sm )
Q(Sm L)
2
2
L2
Finally, we need to show that L = o(mI(p)). We claim that
◆
m
m ✓
X
1X
1 pi 2
VarQ Wi =
log
4
pi
Q(0 < Sm < L)
i=1

i=1
8 max (| log(pi )| _ | log(1
1im
m
X
i=1
⇣ p
log 2 pi (1
⌘
pi ) .
pi )| _ 2)
1
.
4
This is becase when pi 2 [1/16, 15/16], we have log 1 pipi
2
 6(2pi 1)2 
When pi 2 (0, 1/16)[(15/16, 1), log 1 pipi  2 log (4pi (1
2| log(1 pi )|. Therefore, under the assumption that
max (| log(pi )| _ | log(1
1im
6 log (4pi (1
pi )).
pi )) and log 1 pipi  2| log(pi )|_
pi )|) = o(mI(p)),
L = o(mI(p)) holds, and the proof is complete.
B
Proof of Lemma 6.1
P
(i)
(i)
0
Let f (t) = m
i=1 log Bt (⇡1⇤ , ⇡2⇤ ). Then we have f (t0 ) = 0 by its definition. First, we are
going to prove 0 < t0 < 1. The concavity of logarithm gives us xt y 1 t  tx + (1 t)y for
non-negative x, y and t 2 [0, 1], which implies
!
k ⇣
⌘
X
X
X
(i)
(i)
(i)
(i)
f (t) =
log Bt (⇡1⇤ , ⇡2⇤ ) 
log
(1 t)⇡1h + t⇡2h
= 0.
i2[m]
h=1
i2[m]
(i)
(i)
For t 2 (0, 1), the equality holds if and only if ⇡1h = ⇡2h for all h 2 [k] and i 2 [m]. As there
is at least one non-spammer, we must have f (t) < 0 = f (0) = f (1) for t 2 (0, 1). Hence the
minimizer t0 2 (0.1).
Now we are going to show the uniqueness of t0 by proving that
f 00 (t) = Var(Sm ) > 0, 8t 2 (0, 1)
where Sm =
⇣
⌘
(i) 1
pih = ⇡1h
(i)
(i)
P
Bt (⇡1⇤ , ⇡2⇤ ) =
i2[m] Wi .
t⇣
(i)
⇡2h
P
⌘t
To simplify the notation, let us define wih = t log
for all i 2 [m] and h 2 [k]. Now Qi (Wi = wih ) = pih /
d
Notice that dt
pih = pih wih , we have
X P pih wih
X
d
h
P
f (t) =
=
EWi = ESm ,
dt
h pih
h2[k] pih .
i2[m]
and
✓
◆
and
h pih
and
(i)
⇡2h
(i)
⇡1h
P
(19)
i2[m]
X P pih w2 P pih (P pih wih )2
X
d2
h
h
h
ih
f
(t)
=
=
Var(Wi ) = Var(Sm ).
P
dt2
( h pih )2
i2[m]
(20)
i2[m]
Since the set A↵ is non-empty, there is at least one Var(Wi ) > 0. Thus, f 00 (t) = Var(Sm ) > 0.
C
Proof of Lemma 6.2
From (19), we know ESm = f 0 (t0 ) = 0. Since t0 > 0 by lemma 6.1, we can rescale Wi
p
by Wi /( t0 log ⇢m ) and the value of Sm / Var(Sm ) will not change. Let us define Vi =
2
Pm
Wi /( t0 log ⇢m ) and Rm =
i=1 Vi . Then we have |Vi |  1. To prove a central limit
theorem of Sm , it is sufficient to check the following Lindeberg’s condition [? ], that is, for
any ✏ > 0,
m
X
1
E (Vi
Var(Rm )
EVi )2 I{(Vi
EVi )2
✏2 Var(Rm )} ! 0 as m ! 1.
i=1
(21)
Note that for a discrete random variable X who takes value xa with probability pa for a 2 [N ],
Var(X) =
X
pa
a
!
X
Then, for any i 2 A↵ , we have
Var(Vi ) =
1
log2 ⇢m
1
log2 ⇢m
1
2
pa x2a
a
!
X
a
⇣
⌘
(i) (i) 1
X ⇡1a ⇡1b
⇣
=
X
t ⇣ (i) (i) ⌘t
⇡2a ⇡2b
(i)
pa pb (xa
xb ) 2 .
a,b
(i)
Bt2 (⇡1⇤ , ⇡2⇤ )
a,b
⇣
pa x a
!2
(i) (i)
log2
⌘ ⇣
⌘
(i) (i) 1 t
(i) (i) t
⇡22 ⇡21 log2
⌘ ⇣
⌘
(i) (i) 1 t
(i) (i) t
⇡22 ⇡21 log2
log ⇢m
⇢2m
4 log2 (1 + ↵)
log2 ⇢m
⇢2m
min{↵2 , 1}.
log2 ⇢m
(i) (i)
⇡1a ⇡2b
!
(i) (i)
⇡22 ⇡11
⇡12 ⇡11
⇡12 ⇡11
⇡2a ⇡1b
!
(i) (i)
⇡12 ⇡21
(1 + ↵)2
(i)
(i)
Here the second inequality is due to the assumption that for any i 2 A, ⇡aa ⇡ab (1 + ↵) for
(i)
any b 6= a. We have used the assumption that ⇡ab
⇢m for the third inequality. The last
inequality is because log(1 + ↵) ↵/(1 + ↵) min{↵/2, 1/2} for positive ↵. Take a sum of
Var(Vi ) over i 2 A↵ ,
Var(Rm ) =
X
Var(Vi )
i2[m]
|A↵ |
for some constant c 2 (0, 1). Since (Vi
I{(Vi
⇢2m
min{↵2 , 1}
log2 ⇢m
cm
⇢2m
min{↵2 , 1},
log2 ⇢m
EVi )2  2Vi2 + 2(EVi )2  4, we will have
EVi )2
✏2 Var(Rm )} = 0
when 4 log2 ⇢m < ✏2 |A↵ |⇢2m min{↵2 , 1}. Notice that E(Vi EVi )2  4, we apply the Dominated
Convergence Theorem to conclude
E (Vi
EVi )2 I{(Vi
EVi )2
✏2 Var(Rm )} ! 0.
Thus, the Lindeberg condition holds when | log ⇢m | = o(⇢m |A0.01 |1/2 ).
3
D
Proof of Lemma 6.3
We are first going to show
0

2 log ⇢m . Recall that
f( ) =
m ⇣
Y
(1
pi )e
/2
+ pi e
i=1
/2
⌘
.
Q
For all
2 log ⇢m , we have f ( ) > ⇢mm m
pi )
1, and f (0) = 1. Thus, the
i=1 (1
minimizer of f ( ) must be in the interval (0, 2 log ⇢m ].
Again, we are going to prove the central limit theorem of Sm by checking the following
Lindeberg’s condition. For any ✏ > 0,

m
X
1
lim
EQ (Wi
m!1 VarQ (Sm )
2
EQ Wi ) I |Wi
i=1
When
0
⇢
E Q Wi | > ✏
q
VarQ (Sm )
=0
(22)
2 (0, 2 log ⇢m ], a lower bound of VarQ (Wi ) is given by
VarQ (Wi ) =
2
0
pi (1
(1
pi )e
0 /2
pi )
+ pi e
2
0 /2
2
0e
pi (1
pi )
2 2
0 ⇢m pi (1
P
P
2 ⇢2
Therefore, VarQ (Sm ) = m
0 m
i=1 VarQ (Wi )
i2[m] pi (1 pi ). Notice that |Wi
|Wi | + E|Wi | = , for any fixed ✏ > 0, we will have
⇢
q
I |Wi EQ Wi | > ✏ VarQ (Sm ) = 0
pi )
EQ Wi | 
P
when ⇢2m i2[m] pi (1 pi ) ! 1 as m ! 1. Since VarQ (Wi )/ 20  1/4, the Dominated
Convergence Theorem implies the desired Lindeberg’s condition (22).
4