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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. 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