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Some Extensions of a Lemma of Kotlarski
Halbert Whitey; z
UC San Diego
Kirill Evdokimov
Princeton University
July 25, 2011
Abstract
This note demonstrates that the conditions of Kotlarski’s (1967) lemma can be
substantially relaxed. In particular, the condition that the characteristic functions of
M , U1 , and U2 are non-vanishing can be replaced with much weaker conditions: the
characteristic function of U1 can be allowed to have real zeros, as long as the derivative
of its characteristic function at those points is not also zero; that of U2 can have a
countable number of zeros; and that of M need satisfy no restrictions on its zeros.
We also show that Kotlarski’s (1967) lemma holds when the tails of U1 are no
thicker than exponential, regardless of the zeros of the characteristic functions of U1 ,
U2 , or M .
1
Introduction
This note provides new regularity conditions ensuring that the conclusion of Kotlarski’s
(1967) lemma holds. Kotlarski’s result may be explained as follows. Suppose one observes
the joint distribution of two noisy measurements (Y1 ; Y2 ) = (M + U1 ; M + U2 ) of a random variable M , where random variables U1 and U2 are measurement errors. Kotlarski
showed that when (M; U1 ; U2 ) are mutually independent, E[U1 ] = 0, and the characteristic functions of M; U1 ; and U2 are non-vanishing, it is possible to recover the unknown
distributions of M , U1 , and U2 from the joint distribution of (Y1 ; Y2 ):
Kirill Evdokimov: Department of Economics, Princeton University, Fisher Hall 001, Princeton, NJ
08544; email: [email protected]
y
Halbert White: Department of Economics 0508, UCSD, La Jolla, CA 92093-0508; email:
[email protected]
z
We thank Stéphane Bonhomme, Susanne Schennach, the Editor, the Co-editor, and two anonymous
referees for helpful comments. The …rst author gratefully acknowledges the support from the Gregory C.
Chow Econometric Research Program at Princeton University.
1
Kotlarski’s lemma has been applied to identify and estimate a wide variety of models
in economics, such as measurement error models (e.g., Li and Vuong 1998, Schennach
2004), auction models (e.g., Li et al. 2000, Krasnokutskaya 2011), panel data models (e.g.,
Arellano and Bonhomme 2009, Evdokimov 2008, 2010), and in various labor economics
applications (e.g., Bonhomme and Robin 2010, Kennan and Walker 2011).
Kotlarski’s lemma requires that the characteristic functions of the random variables
M , U1 , and U2 do not have real zeros. This is restrictive; the characteristic functions
of many standard distributions have zeros (e.g. the uniform, the truncated normal, and
many discrete distributions). Thus, it is important to consider identi…cation when the
characteristic functions may have real zeros.1 Our aim here is to provide less restrictive
alternative conditions for Kotlarski’s conclusions to still hold.
Instead of requiring that the characteristic functions of M , U1 , and U2 are non-vanishing,
we require that the sets of zeros of the characteristic function of U1 and its derivatives have
empty intersection and that the real zeros of the characteristic function of U2 are isolated.
We impose no restrictions on the zeros of the characteristic function of M .
We also show that the conclusion of Kotlarski’s lemma holds when U1 has tails that
are no thicker than exponential. This alternative result imposes strong restrictions on
the tails of one of the measurement errors, but does not require any assumptions on its
characteristic function, aiding economic interpretability. Further, the distributions of M
and U2 are completely unrestricted, apart from a …rst moment restriction.
Thus, we not only relax the assumption of nonvanishing characteristic functions of
the errors U1 and U2 , but we also provide conditions that may have a direct economic
interpretation and that may thus be more appealing to researchers than those previously
imposed.
2
Let
i
Main Results
X
p
denote the characteristic function of X ;
1: We write
0
X
following assumption:
(@[email protected])
X;
and let
X (s)
E [exp(isX )] ; s 2 R; where
denote Lebesgue measure. We impose the
Assumption A: (i) M , U1 , and U2 are mutually independent; and Y1
2
M + U1 and
Y2
0
U1
M + U2 ; (ii) E [jY1 j + jY2 j] < 1 and E(U1 ) = 0 ; (iii) the real zeros of
are disjoint; and (iv)
U2
U1
and
has only isolated real zeros.
Given A(i); the moment condition A(ii) implies EjM j < 1 and EjU2 j < 1: Given
A(i) and E(U1 ) = 0; it su¢ ces for E [jY1 j + jY2 j] < 1 that E jY2 j < 1; but we write the
condition as we do to avoid obscuring the moment requirements on Y1 and Y2 .
Let Z0 denote the set of real zeros of
Y1 ;Y2 (s1 ; s2 ) = E
(@[email protected] ) Y1 ;Y2 ( ; )
)
Y1 ;Y2 ( ;
Y1 Y2 .
Also, de…ne the characteristic function
[exp(is1 Y1 + is2 Y2 )], the set of singular points S0
fs 2 Z0 : lim sup
= 1g, and the function
(s)
(
(@[email protected] )
Y1 ;Y2 (s;
Y1 ;Y2 (s;
s)
s)
!s
, if s 62 Z0 ;
0, if s 2 Z0 .
Below we show that A implies that all elements of Z0 are isolated (and hence are
countable). Since S0 is a subset of Z0 , we can enumerate all positive elements of S0 :
Placing these in increasing order, for k > 0 we let s0 (k) be the kth smallest positive
element of S0 . Similarly, for k < 0, we let s0 (k) be the
kth largest negative element of
S0 . Thus, S0 = f: : : ; s0 ( 2) ; s0 ( 1) ; s0 (1) ; s0 (2) ; : : :g ; and s0 (k) < s0 (l) for all k < l.
In addition, for notational convenience, denote s0 (0) = 0. For all s
largest k such that s0 (k)
0 let k 0 (s) be the
s. Thus k 0 (s) = 0 for all s 2 [0; s0 (1)), k 0 (s) = 1 for all
s 2 [s0 (1) ; s0 (2)) and so on. We extend Kotlarski’s (1967) lemma as follows.
Lemma 1
(a) Let (L; V1 ; V2 ) be random, and let (Z1 ; Z2 )
(L + V1 ; L + V2 ); with (V1 ; V2 ) dis-
tributed identically to (U1 ; U2 ): If A(i) holds for both (M; U1 ; U2 ) and (L; V1 ; V2 ); then Z0
is also the zero set of
Z1 Z2 .
If A(i; ii) hold for (M; U1 ; U2 ) and
(Z0 ) > 0; then A (iii)
or A (iv) fail for (M; U1 ; U2 ) and there exist (L; V1 ; V2 ) such that
L
6=
=
Y1 ;Y2
but
M:
(b) if A(i)
U1 (s)
Z1 ;Z2
(iv) hold, then, with
2
= exp[ is
1]
lim 4( 1)k0 (s)
"&0
1
Y
E(Y1 ); for all s 2 R+ nS0
(Z
exp
s0 (k) "
s0 (k 1)+"
0<k k0 (s)
( )d
)
(Z
exp
s
( )d
s0 (k0 (s))+"
(1)
3
)3
5:
A similar formula holds2 for all s 2 R nS0 . Then
Y1 ;Y2
(s; s) =
U1 (s)
(s) =
M
Y1
for all s 62 Z0 . Moreover, the functions
(s) =
M(
U1 (s)
);
U1 (
and
U2
), and
( s) =
U2 (
) are
continuous on R and hence can be uniquely extended from RnZ0 to R.
Proof: We begin with some simple but useful Facts:
(1) Given A(i); we have
Y1 ;Y2 (s1 ; s2 )
= E[exp(i(s1 + s2 )M + is1 U1 + is2 U2 )]
=
Letting s1 = s and s2 =
U1 (s) U2 (s);
M (0)
as
s gives
= 1 and
union of the zero sets Z01 of
zeros of
U2 ;
M (s1
Y1 ;Y2 (s;
U2 (
U1
+ s2 )
U1 (s1 ) U2 (s2 ):
s) =
s) =
U2 (s):
and Z02 of
U2 :
Y1 Y2 (s)
=
M (0) U1 (s) U2 (
Thus, the zero set Z0 of
As the zeros of
U2
Y1 Y2
s) =
is the
are identical to the
say Z02 ; we have Z02 = Z02 : Thus, A(i) implies Z0 = Z01 [ Z02 :
(2) A(i; ii) imply E [jM j + jU1 j + jU2 j] < 1; which in turn implies that the functions
Y1 ;Y2 ,
U1 ,
U2 ,
(3) A (i)
and
M
are continuously di¤erentiable.
(iii) imply that Z01 has no limiting points; hence, all elements of Z01
are isolated (in R) and Z01 is a countable set. To prove this, suppose there exists a
sequence of points f k g1
k=1 , such that
for all k. By Fact (2), the function
limk!1
U1
( k ) = 0; and
0
U1
6=
k
U1
0
for all k,
0
= limk!1
k,
and
is continuously di¤erentiable. Then
( 0 ) = limk!1
U1
( k)
U1
( 0) = (
0)
k
U1
( k) = 0
U1
( 0) =
= 0, which
contradicts A (iii).
(4) By Fact (3) and A (iv) ; Z0 = Z01 [ Z02 is countable. The Lebesgue measure of a
countable set is zero, so Assumption A implies (Z0 ) = (Z01 ) = (Z02 ) = 0:
We are now ready to prove the lemma:
(a) Because (U1 ; U2 ) and (V1 ; V2 ) are identically distributed,
so the zero sets of
V1
and
V2
U1
=
V1
and
U2
=
V2 ;
are Z01 and Z02 ; respectively. Given this and A(i); Fact (1)
ensures that Z0 is the zero set of both
Y1 Y2
and
Z1 Z2 .
If A(i; ii) hold and (Z0 ) > 0
then A(iii) or A (iv) must fail due to Fact (4). The proof is completed by the example
of Kotlarski (1967, p.72), which speci…es two random triplets having the given properties
with (Z0 ) > 0 and with
Z1 ;Z2
=
Y1 ;Y2
but
4
L
6=
M:
(b) (1) By Fact (2); (@[email protected] )
(@[email protected] )
Y1 ;Y2 (s1 ; s2 )
(@[email protected] )
Y1 ;Y2 (s;
=
s) =
exists and
Y1 ;Y2
0
M (s1
+ s2 )
U1 (s1 ) U2 (s2 )
0
M (0) U1 (s) U2 (
s) +
+
M (s1
0
U1 (s) U2 (
+ s2 )
0
U1 (s1 ) U2 (s2 );
so
s):
Suppose s 62 Z0 : Then
(s) =
(@[email protected] )
Y1 ;Y2 (s;
Y1 ;Y2 (s;
s)
s)
0
M (0) U1 (s) U2 (
s) + 0U1 (s)
U1 (s) U2 ( s)
=
U2 (
s)
=i
1+
0
U1 (s)
U1 (s)
:
(2)
Note that for s 62 Z0 we can write
s 2 S0 ; lim sup
U2
!s j
U1
Y1 ;Y2 (s;
( )j is in…nite, which implies that
0
U1
is bounded and the function
Thus
(s) = (@[email protected] ) ln
U1
s). Also note that for
(s) = 0; because the function
is locally bounded away from both zero and in…nity.
(s) = 0 for all s 2 S0 .
The proof now proceeds by induction. First, for all s 2 [0; s0 (1)), i.e., all s such that
k 0 (s) = 0, the right hand side of expression (1) simpli…es to
Z s
exp[ is 1 ] lim exp
( )d
"&0
= exp[ is
=
=
where ln
0 for all
U1 (
"
Z
1 ] lim exp
"&0
lim exp[ i"
"&0
s
i
"
1 ] exp
U1 (s),
Z
"
s
1+
@
ln
@s
2 [0; s0 (1)) and
U1
(s0 (k
U1 (
d
) d
U1 (
) 6=
(0) = 1. Hence, formula (1) is shown to hold for all
Now suppose (1) holds for all s such that k 0 (s)
k K
)
U1 ( )
) is the principal value of the logarithm and is well de…ned since
s 2 [0; s0 (1)).
[1
0
U1 (
K, i.e., it holds for all s 2
1) ; s0 (k)). We now show that this implies that (1) also holds for all
s 2 (s0 (K) ; s0 (K + 1)). Write s^
(s0 (K) + s0 (K
5
1))=2. Since
s)
U1 (^
6= 0 and (1)
holds for s^, for any s 2 (s0 (K) ; s0 (K + 1)) we can write the right hand side of (1) as
"
(Z
)
(Z
)#
s0 (K) "
s
( )d
exp
( )d
s) exp[ i (s s^) 1 ] lim ( 1) exp
U1 (^
"&0
s^
s0 (K)+"
(s0 (K) ")
s)
U1 (^
0
")
U1 (s0 (K)
U1 (s) lim exp[ 2i" 1 ]
"&0
0
U1 (s0 (K) + ")
=
lim exp[ 2i"
s)
U1 (^
=
"&0
=
U1 (s) lim exp[ 2i"
=
U1
where
1
1] (
1]
"&0
because
0
U1
0
U1
U1 (s)
(s
0 (K) + ")
U1
U1
1)
( 1) "
( 2) "
(s) ;
2 (s0 (K)
U1
"; s0 (K)),
2
2 (s0 (K) ; s0 (K) + "). The second equality holds
(s0 (K)) = 0 and because
s)
U1 (^
can be cancelled out, since
s)
U1 (^
6= 0 by
construction of s^. The third equality follows from the mean value theorem applied to
U1
0
U1
(s0 (K)) 6= 0 and the continuity of
from Fact (2). Thus, we have shown that (1) holds for k 0 (s) = K + 1, i.e. for all
s 2 [1
k K+1 (s0 (k
1) ; s0 (k)). The proof by induction is therefore complete.
The above establishes identi…cation of
(s) =
Y1
(s) =
continuity of
M;
M
0
U1
(s0 (K)) = 0, and the last equality holds by
U1
(s) and
U1 ,
and
U2
U2
(s) =
U1
Y2 Y1
(s) for all s 2 RnZ0 . Then we also identify
(s) =
U1
( s) for all s 2 RnZ0 . Finally, the
implies the uniqueness of their continuous extension from
RnZ0 to R.
Remark 1: When the characteristic function Y1
Z s
( )d
= exp[ is
U1 (s) = exp[ is 1 ] exp
0
has no zeros, eq. (1) becomes
Z s
(@[email protected] ) Y1 ;Y2 ( ; )
d
1 ] exp
)
0
Y1 ;Y2 ( ;
Y2
,
(3)
which is exactly the expression obtained in Evdokimov (2008), who assumes that the characteristic functions
U1
and
U2
are nonvanishing. Similar to Evdokimov (2008), Lemma
1 relaxes Kotlarski’s condition that the characteristic function
Remark 2: In Lemma 1, we essentially recover
equals the ratio
tion
U1
0
U1
(s) =
(0) = 1. When
U1
M
is nonvanishing.
(s) by observing
(s)
i
1,
which
U1
(s) = (@[email protected]) ln
U1
(s) is nonzero, solving the di¤erential equation (2) immedi-
U1
(s) , and by imposing the initial condi-
ately yields eq. (3). Nevertheless, we run into obvious problems when
6
U1
(s0 ) = 0 for
some s0 . Here, A(iii) is very important; for a small " > 0 we can write
(s0
U1
0
U1
") + 2
B
0
A (1)
(s) = (1
=
A (1)
0
U1
s) j1
sj for s
0
B
=
(1) =
B
(s0 ) 6= 0. For example, the functions
0
A (s) = A (s)
=
1
0
B
=
Remark 3: If the zeros of
U1
(s) =
B
0
U1
( 0 ) = 0, but
moment), and that
00
U1
00
U1
(=
0
(s) = 2= (s
(s) = (s) because for both functions
1) for s
0.
are not disjoint, identi…cation may be obtained by
considering higher-order derivatives, say
0
U1
s)2
= (1
(1) = 0 and thus violate A(iii). Indeed, one cannot
and
(2)
U1 )
A (s)
0 (although not proper characteristic functions) have
distinguish between these two functions based on
(s)
(s0 + ") =
(s0 ) " + o (") and hence "jump" through the singular point s0 . This ex-
pression is uninformative unless
and
U1
(n)
U1 ;
n > 1: For example, suppose that
U1
( 0) =
exists and is continuous (so that U1 has …nite second
( 0 ) 6= 0, so that the zeros of
00
U1
a similar argument delivers identi…cation. If
0
U1
and
00
U1
are disjoint at
0.
If so,
( 0 ) = 0, one can consider the next higher
derivative, and so on. That is, identi…cation continues to hold, given that the characteristic
function
U1
is su¢ ciently continuously di¤erentiable and its higher-order derivatives have
suitably disjoint zeros. The ( 1) factor in equation (1) appears only when n is even, with
(n+1)
( 0)
U1
6=
(n)
U1 ( 0 )
= 0: A su¢ cient (but not necessary) condition for A(iv) is the
disjointness of the zeros of
U2
and
0
U2 ,
since
0
U2
exists by Fact (2). This holds by the
argument of Fact (3). Just as for U1 ; the properties of higher-order derivatives of
U2
can
also ensure A(iv):
Remark 4: When A(i) holds, the assumptions E [jY1 j + jY2 j] < 1, A(iii), and A(iv) can
be checked for any given
Y1 ;Y2 .
Remark 5: Although the uniform distribution is not a common measurement error distribution, it nicely illustrates the power of A(iii). If U1
the functions
U1
and
0
U1
U [ a; a] then for any value of a > 0
have real zeros, but these zeros never coincide. Thus, the original
result of Kotlarski (1967) as well as the lemmas of Li and Vuong (1998), Schennach (2004),
and Evdokimov (2008) do not apply, yet our Lemma 1 does guarantee identi…cation.
The assumptions of Lemma 1 are weak and hold for all standard probability distributions. However, they are stated in terms of characteristic functions. Economic models
rarely impose restrictions on characteristic functions; hence any assumptions stated in
7
terms of characteristic functions might lack an economic interpretation. To address this
issue, we introduce an alternative assumption and identi…cation lemma.
Assumption B: A(i) and A(ii) hold, and (iii) there exist positive constants c1 and c2
such that the density of U1 satis…es fU1 (u) < c1 exp ( c2 juj) for large u:
Lemma 2 Let Assumption B hold. Then the distributions of M; U1 ; and U2 are identi…ed.
Proof: The characteristic functions
U1
and
Thus, there is an s > 0 such that
Uj
(s) > 1=2 for all s 2 [ s; s] and j = 1; 2. Then
Y1 ;Y2
(s; s) =
U1
(s)
U2
U1
are continuous, and
U1
(0) =
U2
(0) = 1.
( s) > 1=4 for all s 2 [ s; s] : Thus, B ensures that equation
(3) applies for all s 2 [ s; s] ; identifying
B(iii) implies that
U2
U1
(s) on this interval.
is analytic on R; see page 3 of Paley and Wiener (1934). Then,
by the properties of analytic functions,
U1
is identi…ed not only on the interval [ s; s]
but also on the whole real line. Moreover, functions analytic on R may only have isolated
real zeros, and hence
M
(s) =
Y1
(s) =
countable number of the isolated zeros of
whole real line. We identify
U2
U1
(s) for all points s 2 R; except for at most a
U1 .
Then, by continuity,
in a similar way as
U2
(s) =
Y2 Y 1
M
is identi…ed on the
(s) =
U1
( s).
Remark 6: Here, Assumption B(iii) replaces A(iii); and it makes A(iv) unnecessary.
Clearly, B(iii) is strong for U1 . The advantage of this assumption is its potential economic
interpretability; in a variety of economic applications, researchers may have some intuition
or economic model that implies that one of the measurement errors, U1 ; has thin tails
(or even bounded support). Apart from the requirement that U2 has …nite …rst moment
(implied by B(ii)), its distribution is completely unrestricted.
Remark 7: The key property of
U1
ensured by B(iii) is its analyticity on R: Although
economically interpretable conditions are more compelling, any other condition ensuring
this analyticity can replace B(iii) to deliver the same conclusion.
Remark 8: Lemma 2 is not a corollary of Lemma 1, as B(iii) (or analyticity of
not imply A(iii); and it says nothing about
U2 :
8
U1 )
does
Remark 9: An interesting topic for future research is whether our approach can be applied
or adapted to models identi…ed using generalized functions and their Fourier transforms,
as in Schennach (2007) and Zinde-Walsh (2010).
Notes
1 Note
that when the distribution of the error term is known, deconvolution can be
performed even when the characteristic function of the distribution of the error has real
zeros; see Devroye (1989) and Carrasco and Florens (forthcoming).
2 For
all s
0 let k 0 (s) be the smallest k such that s0 (k) s. Then for all s 2 R nS0 ;
2
(Z
)
(Z
)3
s0 (k)+"
s
Y
k0 (s)
4
( )d
exp
exp
( ) d 5:
U1 (s) = exp[ is 1 ] lim ( 1)
"&0
s0 (k+1) "
k0 (s) k<0
s0 (k0 (s)) "
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9
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