Supplementary Material. Dynamic Linear Panel Regression Models with Interactive Fixed Effects

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1 Supplementary Material Dynamic Linear Panel Regression Models with Interactive Fixed Effects Hyungsik Roger Moon Martin Weidner August 3, 05 S. Proof of Identification Theorem. Proof of Theorem.. Let Qβ, λ, f E Y β X λ f F λ 0, f 0, w, where β R K, λ R N R and f R T R. We have Qβ, λ, f { = E Tr [ Y β X λ f Y β X λ f ] } λ 0, f 0, w { [ λ = E Tr 0 f 0 λf β β 0 X + e λ 0 f 0 λf β β 0 X + e ] } λ 0, f 0, w [ ] = E Tr e e λ 0, f 0, w { [ λ + E Tr 0 f 0 λf β β 0 X λ 0 f 0 λf β β 0 X ] } λ 0, f 0, w. }{{} Q β,λ,f Department of Economics and USC Dornsife INET, University of Southern California, Los Angeles, CA moonr@usc.edu. Department of Economics, Yonsei University, Seoul, Korea. Department of Economics, University College London, Gower Street, London WCE 6BT, U.K., and CeMMaP. m.weidner@ucl.ac.uk.

2 [ ] In the last step we used Assumption IDii. Because E Tr e e λ 0, f 0, w is independent of β, λ, f, we find minimizing Qβ, λ, f is equivalent to minimizing Q β, λ, f. We decompose Q β, λ, f as follows Q β, λ, f { [ λ = E Tr 0 f 0 λf β β 0 X λ 0 f 0 λf β β 0 X ] } λ 0, f 0, w { [ λ = E Tr 0 f 0 λf β β 0 X Mλ,λ 0,w λ 0 f 0 λf β β 0 X ] } λ 0, f 0, w { [ λ + E Tr 0 f 0 λf β β 0 X Pλ,λ 0,w λ 0 f 0 λf β β 0 X ] } λ 0, f 0, w { [ β = E Tr high β 0,high ] } X high Mλ,λ 0,w β high β 0,high X high λ 0, f 0, w }{{} Q high β high,λ { [ λ + E Tr 0 f 0 λf β β 0 X Pλ,λ 0,w λ 0 f 0 λf β β 0 X ] } λ 0, f 0, w, }{{} Q low β,λ,f where β high β 0,high X high = K m=k + β m β 0 mx m. A lower bound on Q high β high, λ is given by Q high β high, λ { [ β min E Tr high β 0,high ] } X high M λ,λ,w β high β 0,high X high λ 0, f 0, w λ R N R+R+rankw = minn,t r=r+r+rankw [ β µ r {E high β 0,high X high β high β 0,high ]} λ X 0 high, f 0, w. S.. Because Q β, λ, f, Q high β high, λ, and Q low β, λ, f, are expectations of traces of positive semi-definite matrices we have Q β, λ, f 0, Q high β high, λ 0, and Q low β, λ, f 0 for all β, λ, f. Let β, λ and f be the parameter values that minimize Qβ, λ, f, and thus also Q β, λ, f. Because Q β 0, λ 0, f 0 = 0 we have Q β, λ, f = min β,λ,f Q β, λ, f = 0. This implies Q high β high, λ = 0 and Q low β, λ, f = 0. Assumption IDv, the lower bound S..,

3 and Q high β high, λ = 0 imply β high = β 0,high. Using this, we find Q low β, λ, f { [ = E Tr λ 0 f 0 λ f β low β 0,low X low λ 0 f 0 λ f ] } β low β 0,low λ X 0 low, f 0, w, { [ min E Tr λ 0 f 0 λf β low β 0,low X low λ 0 f 0 λf ] } β low β 0,low λ X 0 low, f 0, w f { [ = E Tr λ 0 f 0 β low ] } β 0,low X low M λ λ 0 f 0 β low β 0,low λ X 0 low, f 0, w, S.. where β low β 0,low X low = K l= β l β 0 l X l. Because Q low β, λ, f = 0 and the last expression in S.. is non-negative we must have { [ E Tr λ 0 f 0 β low ] } β 0,low X low M λ λ 0 f 0 β low β 0,low λ X 0 low, f 0, w = 0. Using M λ = M λm λ and the cyclicality of the trace we obtain from the last equality: { } Tr M λam λ = 0, where A = E [ λ 0 f 0 β low β 0,low X low λ 0 f 0 β low ] β 0,low λ X 0 low, f 0, w. The trace of a positive semi-definite matrix is only equal to zero if the matrix itself is equal to zero, so we find M λam λ = 0, This together with the fact that A itself is positive semi definite implies note that A positive semi-definite implies A = CC for some matrix C, and M λam λ = 0 then implies M λc = 0, i.e., C = P λc A = P λap λ, and therefore ranka rankp λ R. We have thus shown { [ rank E λ 0 f 0 β low β 0,low X low λ 0 f 0 β low ]} β 0,low λ X 0 low, f 0, w R. 3

4 We furthermore find { [ R rank E λ 0 f 0 β low β 0,low X low { [ rank M w E λ 0 f 0 β low β 0,low X low + rank λ 0 f 0 β low ]} β 0,low λ X 0 low, f 0, w P f 0 λ 0 f 0 β low β 0,low λ X 0 low Mw, f 0, w { [ P w E λ 0 f 0 β low β 0,low X low M f 0 λ 0 f 0 β low β 0,low λ X 0 low Pw, f 0, w rank [ ] M w λ 0 f 0 f 0 λ 0 M w { [ + rank E β low β 0,low X low M f 0 β low ]} β 0,low λ X 0 low, f 0, w. Assumption IDiv guarantees rank M w λ 0 f 0 f 0 λ 0 M w = rank λ 0 f 0 f 0 λ 0 = R, that is, we must have E [ β low β 0,low X low M f 0 β low ] β 0,low λ X 0 low, f 0, w = 0. ]} ]} According to Assumption IDiii this implies β low = β 0,low, i.e., we have β = β 0. This also implies Q β, λ, f = λ 0 f 0 λ f F = 0, and therefore λ f = λ 0 f 0. S. Examples of Error Distributions The following Lemma provides examples of error distributions that satisfy e = O p maxn, T as N, T. Example i is particularly relevant for us, because those assumptions on e it are imposed in Assumption 5 in the main text, i.e., under those main text assumptions we indeed have e = O p maxn, T. Lemma S... For each of the following distributional assumptions on the errors e it, i =,..., N, t =,..., T, we have e = O p maxn, T. i The e it are independent across i and t, conditional on C, and satisfy Ee it C = 0, and Ee 4 it C is bounded uniformly by a non-random constant, uniformly over i, t and N, T. Here C can be any conditioning sigma-field, including the empty one corresponding to unconditional expectations. 4

5 ii The e it follow different MA processes for each i, namely e it = ψ iτ u i,t τ, for i =... N, t =... T, S.. τ=0 where the u it, i =... N, t =... T are independent random variables with Eu it = 0 and Eu 4 it uniformly bounded across i, t and N, T. The coefficients ψ iτ satisfy τ max i=...n ψ iτ < B, max ψ iτ < B, i=...n τ=0 τ=0 for a finite constant B which is independent of N and T. S.. iii The error matrix e is generated as e = σ / u Σ /, where u is an N T matrix with independently distributed entries u it and Eu it = 0, Eu it =, and Eu 4 it is bounded uniformly across i, t and N, T. Here σ is the N N cross-sectional covariance matrix, and Σ is the T T time-serial covariance matrix, and they satisfy max j=...n N i= σ ij < B, max τ=...t T Σ tτ < B, S..3 for some finite constant B which is independent of N and T. In this example we have Ee it e jτ = σ ij Σ tτ. Proof of Lemma S.., Example i. Latala 005 showed that for a N T matrix e with independent entries, conditional on C, we have E e [ C c max E ] / [ e it C + max E ] / [ e it C + E ] /4 e 4 it C i j, t where c is some universal constant. Because we assumed uniformly bounded 4th conditional moments for e it we thus have e = O P T + O P N + O P T N /4 = O p maxn, T. Example ii. Let ψ j i = ψ j,..., ψ Nj be an N vector for each j 0. Let U j be an N T sub-matrix of u it consisting of u it, i =... N, t = j,..., T j. We can then write equation S.. in matrix notation as e = diagψ j U j = j=0 T diagψ j U j + r, j=0 t= i,t 5

6 where we cut the sum at T, which results in the remainder r = j=t + diagψ j U j. When approximating an MA by a finite MAT process we have for the remainder E r F = N i= T E r ij t= σ u N i= σ u N T σ u N T t= j=t + j=t + j=t + ψ ij max ψ ij i j max i ψ ij, where σ u is the variance of u it. Therefore, for T we have r F E 0, N which implies r F = O p N, and therefore r r F = O p N. Let V be the N T matrix consisting of u it, i =... N, t = T,..., T. For j = 0... T the matrices U j are sub-matrices of V, and therefore U j V. From example i we know V = O p maxn, T. Furthermore, we know diagψ j max ψij i. Combining these results we find e T diagψ j U j + r j=0 T j=0 [ j=0 max i ψ ij V + op N max i ψ ij ] O p maxn, T + o p N O p maxn, T, as required for the proof. Example iii. Because σ and Σ are positive definite, there exits a symmetric N N matrix φ and a symmetric T T matrix ψ such that σ = φ and Σ = ψ. The error term can then be generated as e = φuψ, where u is an N T matrix with iid entries u it such that Eu it = 0 and Eu 4 it <. Given this definition of e we immediately have Ee it = 0 and Ee it e jτ = σ ij Σ tτ. What 6

7 is left to show is e = O p maxn, T. From example i we know u = O p maxn, T. Using the inequality σ σ σ = σ, where σ = σ because σ is symmetric we find σ σ max j=...n N σ ij < L, i= and analogously Σ < L. Because σ = φ and Σ = ψ, we thus find e φ u ψ LO p maxn, T, i.e., e = O p maxn, T. S.3 Comments on Assumption 4 on the regressors Consistency of the LS estimator β requires the regressors not only satisfy the standard noncollinearity condition in assumption 4i, but also the additional conditions on high- and lowrank regressors in assumption 4ii. Bai 009 considers the special cases of only high-rank and only low-rank regressors. As low-rank regressors he considers only cross-sectional invariant and time-invariant regressors, and he shows that if only these two types of regressors are present, one can show consistency under the assumption plim N,T W > 0 on the regressors instead of assumption 4, where W is the K K matrix defined by W,k k = TrM f 0 X k M λ 0 X k. This matrix appears as the approximate Hessian in the profile objective expansion in theorem 4., i.e., the condition plim N,T W > 0 is very natural in the context of the interactive fixed effect models, and one may wonder whether also for the general case one can replace assumption 4 with this weaker condition and still obtain consistency of the LS estimator. Unfortunately, this is not the case, and below we present two simple counter examples that show this. i Let there only be one factor R = ft 0 with corresponding factor loadings λ 0 i. Let there only be one regressor K = of the form X it = w i v t + λ 0 i ft 0. Assume the N vector w = w,..., w N, and the T vector v = v,..., v N are such that the N matrix Λ = λ 0, w and and the T matrix F = f 0, v satisfy plim N,T Λ Λ/N > 0, and plim N,T F F/T > 0. In this case, we have W = TrM f 0 vw M λ 0 wv, and therefore plim N,T W = plim N,T TrM f 0 vw M λ 0 wv > 0. However, β is 7

8 not identified because β 0 X + λ 0 f 0 = β 0 + X wv, i.e., it is not possible to distinguish β, λ, f = β 0, λ 0, f 0 and β, λ, f = β 0 +, w, v. This implies that the LS estimator is not consistent both β 0 and β 0 + could be the true parameter, but the LS estimator cannot be consistent for both. ii Let there only be one factor R = f 0 t with corresponding factor loadings λ 0 i. Let the N vectors λ 0, w and w be such that Λ = λ 0, w, w satisfies plim N,T Λ Λ/N > 0. Let the T vectors f 0, v and v be such that F = f 0, v, v satisfies plim N,T F F/T > 0. Let there be four regressors K = 4 defined by X = w v, X = w v, X 3 = w + λ 0 v +f 0, X 4 = w +λ 0 v +f 0. In this case, one can easily check plim N,T W > 0. However, again β k is not identified, because 4 k= β0 kx k + λ 0 f 0 = 4 k= β0 k + X k λ 0 + w + w f 0 + v + v, i.e., we cannot distinguish between the true parameters and β, λ, f = β 0 +, λ 0 w w, f 0 + v + v. Again, as a consequence the LS estimator is not consistent in this case. In example ii, there are only low-rank regressors with rankx l =. One can easily check assumption 4 is not satisfied for this example. In example i the regressor is a low-rank regressor with rankx =. In our present version of assumption 4 we only consider low-rank regressors with rankx =, but as already noted in a footnote in the main paper it is straightforward to extend the assumption and the consistency proof to low-rank regressors with rank larger than one. Independent of whether we extend the assumption or not, the regressor X of example i fails to satisfy assumption 4. This justifies our formulation of assumption 4, because it shows in general the assumption cannot be replaced by the weaker condition plim N,T W > 0. S.4 Some Matrix Algebra including Proof of Lemma A. The following statements are true for real matrices throughout the whole paper and supplementary material we never use complex numbers anywhere. Let A be an arbitrary n m matrix. In addition to the operator or spectral norm A and to the Frobenius or Hilbert-Schmidt 8

9 norm A F, it is also convenient to define the -norm, the -norm, and the max-norm by n m A = max A ij, A = max A ij, A max = max max A ij. j=...m i=...n i=...n j=...m i= j= Lemma S.4. Some useful inequalities. Let A be an n m matrix, B be an m p matrix, and C and D be n n matrices. Then we have: i A A F A rank A /, ii AB A B, iii AB F A F B A F B F, iv TrAB A F B F, for n = p, v Tr C C rank C, vi C Tr C, for C symmetric and C 0, vii A A A, viii A max A nm A max, ix A CA A DA, for C symmetric and C D. For C, D symmetric, and i =,..., n we have: x µ i C + µ n D µ i C + D µ i C + µ D, xi µ i C µ i C + D, for D 0, xii µ i C D µ i C + D µ i C + D. Proof. Here we use notation s i A for the ith largest singular value of a matrix A. i We have A = s A, and A F = ranka i= s i A. The inequalities follow directly from this representation. ii This inequality is true for all unitarily invariant norms, see, e.g., Bhatia 997. iii can be shown as follows AB F = TrABB A = Tr[ B AA A B I BB A ] B TrAA = B A F, where we used A B I BB A is positive definite. Relation iv is just the Cauchy Schwarz inequality. To show v we decompose C = UDO singular value decomposition, where U and 9

10 O are n rankc that satisfy U U = O O = I and D is a rankc rankc diagonal matrix with entries s i C. We then have O = U = and D = C and therefore TrC = TrUDO = TrDO U rankc = η ido Uη i i= rankc i= D O U = rankc C. For vi let e be a vector that satisfies e = and C = e Ce. Because C is symmetric such an e has to exist. Now choose e i, i =,..., n, such that e i, i =,..., n, becomes a orthonormal basis of the vector space of n vectors. Because C is positive semi definite we then have Tr C = i e ice i e Ce = C, which is what we wanted to show. For vii we refer to Golub and van Loan 996, p.5. For viii let e be the vector that satisfies e = and A CA = e A CAe. Because A CA is symmetric such an e has to exist. Because C D we then have C = e A CAe e A DAe A DA. This is what we wanted to show. For inequality ix let e be a vector that satisfied e = and A CA = e A CAe. Then we have A CA = e A DAe e A D CAe e A DAe A DA. Statement x is a special case of Weyl s inequality, see, e.g., Bhatia 997. The inequalities xi and xii follow directly from ix because µ n D 0 for D 0, and because D µ i D D for i =,..., n. Definition S.4.. Let A be an n r matrix and B be an n r matrix with ranka = r and rankb = r. The smallest principal angle θ A,B [0, π/] between the linear subspaces spana = {Aa a R r } and spanb = {Bb b B r } of R n is defined by cosθ A,B = max 0 a R r max 0 b R r a A Bb Aa Bb. Lemma S.4.3. Let A be an n r matrix and B be an n r matrix with ranka = r and rankb = r. Then we have the following alternative characterizations of the smallest principal angle between spana and spanb sinθ A,B = min 0 a R r = min 0 b R r 0 M B A a A a M A B b. B b

11 Proof. Because M B A a + P B A a = A a and sinθ A,B + cosθ A,B =, we find proving the theorem is equivalent to proving cosθ A,B = min 0 a R r P B A a A a = min 0 b R r P A B b A b. This last statement is theorem 8 in Galantai and Hegedus 006, and the proof can be found there. Proof of Lemma A.. Let The theorem claims S Z = min f,λ Tr [Z λf Z fλ ], S Z = min f TrZ M f Z, S 3 Z = min λ TrZ M λ Z, S 4 Z = min λ, f TrM λ Z M f Z, T S 5 Z = µ i Z Z, S 6 Z = i=r+ N i=r+ µ i ZZ. S Z = S Z = S 3 Z = S 4 Z = S 5 Z = S 6 Z. We find: i The non-zero eigenvalues of Z Z and ZZ are identical, so in the sums in S 5 Z and in S 6 Z we are summing over identical values, which shows S 5 Z = S 6 Z. ii Starting with S Z and minimizing with respect to f we obtain the first-order condition λ Z = λ λ f.

12 Putting this into the objective function we can integrate out f, namely Tr [ Z λf Z λf ] = Tr Z Z Z λf = Tr Z Z Z λλ λ λ λf = Tr Z Z Z λλ λ λ λλ Z = Tr Z M λ Z. This shows S Z = S 3 Z. Analogously, we can integrate out λ to obtain S Z = S Z. iii Let M λ be the projector on the N R eigenspaces corresponding to the N R smallest eigenvalues of ZZ, let P λ = I N M λ, and let ω R be the R th largest eigenvalue of ZZ. We then know the matrix P λ[zz ω R I N ]P λ M λ[zz ω R I N ]M λ is positive semi-definite. Thus, for an arbitrary N R matrix λ with corresponding projector M λ we have { 0 Tr P λ [ZZ ω R I N ]P λ } M λ [ZZ ω R I N ]M λ Mλ M λ = Tr { P λ[zz ω R I N ]P λ + M λ[zz } ω R I N ]M λ Mλ M λ = Tr [Z M λ Z] Tr [ Z M λ Z ] [ + ω R rankmλ rankm λ ], and because rankm λ = N R and rankm λ N R we have Tr [ Z M λ Z ] Tr [Z M λ Z]. This shows M λ is the optimal choice in the minimization problem of S 3 Z, i.e., the optimal λ = λ is chosen such that the span of the N-dimensional vectors λ r r =... R equals to the span of the R eigenvectors that correspond to the R largest eigenvalues of ZZ. This shows S 3 Z = S 6 Z. Analogously one can show S Z = S 5 Z. iv In the minimization problem in S 4 Z we can choose λ such that the span of the N- dimensional vectors λ r r =... R is equal to the span of the R eigenvectors that correspond to the R largest eigenvalues of ZZ. In addition, we can choose f such that the span of the T -dimensional vectors f r r =... R is equal to the span of the R eigenvectors that correspond to the R + -largest up to the R-largest eigenvalue of Z Z. With this choice of λ and f we actually project out all the R largest eigenvalues of Z Z

13 and ZZ. This shows that S 4 Z S 5 Z. This result is actually best understood by using the singular value decomposition of Z. We can write M λ Z M f = Z Z, where Z = P λ Z M f + Z P f. Because rankz rankp λ Z M f + rankz P f = R + R = R, we can always write Z = λf for some appropriate N R and T R matrices λ and f. This shows that S 4 Z = min λ, f TrM λ Z M f Z min TrZ ZZ Z { Z : rank Z R} = min f,λ Tr [Z λf Z fλ ] = S Z. Thus we have shown here S Z S 4 Z S 5 Z, and this holds with equality because S Z = S 5 Z was already shown above. S.5 Supplement to the Consistency Proof Appendix A Lemma S.5.. Under assumptions and 4 there exists a constant B 0 > 0 such that for the matrices w and v introduced in assumption 4 we have w M λ 0 w B 0 w w 0, v M f 0 v B 0 v v 0, wpa, wpa. Proof. We can decompose w = w w, where w is an N rankw matrix and w is a rankw K matrix. Note w has full rank, and M w = M w. By assumption i we know λ 0 λ 0 /N has a probability limit, i.e., there exists some B > 0 such that λ 0 λ 0 /N < B I R wpa. Using this and assumption 4 we find for any R vector a 0 we have M v λ 0 a λ 0 a = a λ 0 M v λ 0 a a λ 0 λ 0 a > B B, wpa. 3

14 Applying Lemma S.4.3 we find b w M min λ0 w b 0 b R rankw b w w b = min 0 a R R a λ 0 M w λ 0 a a λ 0 λ 0 a Therefore we find for every rankw vector b that b w M λ 0 > B B, wpa. w B/B w w b > 0, wpa. Thus w M λ 0 w B/B w w > 0, wpa. Multiplying from the left with w and from the right with w we obtain w M λ 0 w B/B w w 0, wpa. This is what we wanted to show. Analogously we can show the statement for v. As a consequence of the this lemma we obtain some properties of the low-rank regressors summarized in the following lemma. Lemma S.5.. Let the assumptions and 4 be satisfied and let X low,α = K l= α lx l be a linear combination of the low-rank regressors. Then there exists some constant B > 0 such that Xlow,α M f 0 X low,α min > B, wpa, {α R K, α =} Mλ 0 X low,α M f 0 X low,α min M λ 0 > B, wpa. {α R K, α =} Proof. Note M λ 0 X low,α M f 0 X low,α M λ 0 X low,α M f 0 X low,α, because M λ 0 =, i.e., if we can show the second inequality of the lemma we have also shown the first inequality. We can write X low,α = w diagα v. Using Lemma S.5. and part v, vi and ix of Lemma S.4. we find M λ 0 X low,α M f 0 X low,α M λ 0 = Mλ 0 w diagα v M f 0 v diagα w M λ 0 B 0 M λ 0 w diagα v v diagα w M λ 0 B 0 K Tr [M λ 0 w diagα v v diagα w M λ 0] = B 0 K Tr [v diagα w M λ 0w diagα v ] B 0 K v diagα w M λ 0w diagα v B 0 K v diagα w w diagα v B 0 K Tr [v diagα w w diagα v ] = B 0 Tr [ X K low,α X low,α]. 4

15 Thus we have M λ 0 X low,α M f 0 X low,α M λ 0 / B 0 /K α W low α, where the K K matrix W low W and thus W low above. is defined by W low,l l = Tr X l X l, i.e., it is a submatrix of W. Because converges to a positive definite matrix the lemma is proven by the inequality Using the above lemmas we can now prove the lower bound on the consistency proof. Remember [ S β, f = Tr λ 0 f 0 + K β 0 k β k X k M f λ 0 f 0 + k= k= S β, f that was used in ] K β 0 k β k X k P λ 0,w. We want to show under the assumptions of theorem 3. there exist finite positive constants a 0, a, a, a 3 and a 4 such that S β, f a 0 β low β 0,low β low β 0,low + a β low β 0,low + a a 3 β high β 0,high a 4 β high β 0,high β low β 0,low, wpa. Proof of the lower bound on S β, f. Applying Lemma A. and part xi of Lemma S.4. 5

16 we find [ S β, f µ R+ λ 0 f 0 + [ = µ R+ λ 0 f 0 + K β 0 k β k X k P λ 0,w λ 0 f 0 + k= K l= ] K β 0 k β k X k k= K β 0 l β l w l v l λ 0 f 0 + β 0 l β l w l v l K K + λ 0 f 0 + β 0 l β l w l v l P λ 0,w β 0 m β m X m l= m=k K K + β 0 m β m X mp λ 0,w λ 0 f 0 + β 0 l β l w l v l m=k l= ] K K + β 0 m β m X mp λ 0,w β 0 m β m X m m=k m=k [ K µ K R+ λ 0 f 0 + β 0 l β l w l v l λ 0 f 0 + β 0 l β l w l v l l= l= K K + λ 0 f 0 + β 0 l β l w l v l P λ 0,w β 0 m β m X m l= m=k ] K K + β 0 m β m X mp λ 0,w λ 0 f 0 + β 0 l β l w l v l m=k l= K l µ K R+ λ 0 f 0 + β 0 l β l w l v l λ 0 f 0 + β 0 l β l w l v l= l= l= a 3 β high β 0,high a4 β high β 0,high β low β 0,low, wpa, where a 3 > 0 and a 4 > 0 are appropriate constants. For the last step we used part xii of Lemma S.4. and the fact that K K β 0 m β m X mp λ 0,w λ 0 f 0 + β 0 l β l w l v l m=k l= K β high β 0,high max X m λ 0 f 0 m + K β low β 0,low max w l v l. Our assumptions guarantee the operator norms of λ 0 f 0 / and X m / are bounded from above as N, T, which results in finite constants a 3 and a 4. We write the above result as S β, f µ R+A A/ + terms containing β high, where we defined A = λ 0 f 0 + K l= β0 l β l w l v l. We also write A = A + A + A 3, with A = 6 l

17 M w A P f 0 = M w λ 0 f 0, A = P w A M f 0 = K l= β0 l β l w l v l M f 0, A 3 = P w A P f 0 = P w λ 0 f 0 + K l= β0 l β l w l v l P f. We then find A A = A A + A + A 3A + A 3 and A A A A a / A 3 + a / A a / A 3 + a / A = [A A a A 3A 3 ] + a A A, where for matrices refers to the difference being positive definite, and a is a positive number. We choose a = + µ R A A / A 3. The reason for this choice becomes clear below. Note [A A a A 3A 3 ] has at most rank R asymptotically it has exactly rank R. The non-zero eigenvalues of A A are therefore given by the at most R non-zero eigenvalues of [A A a A 3A 3 ] and the non-zero eigenvalues of a A A, the largest one of the latter being given given by the operator norm a A. We therefore find µ R+ A A µ [ ] R+ A A a A 3A 3 + a A A min { a A, µ R [A A a A 3A 3 ] }. Using Lemma S.4.xii and our particular choice of a we find Therefore µ R [A A a A 3A 3 ] µ R A A a A 3A 3 = µ RA A. µ R+A A { µ RA A min, where we used A A 3 and A A. A µ R A A A + µ R A A, } A A 3 + µ R A A Our assumptions guarantee there exist positive constants c 0, c, c, and c 3 such that A λ0 f 0 + µ R A A A = µ K l= β 0 l β l w l v l c 0 + c β low β 0,low, wpa, = µ R f 0 λ 0 M w λ 0 f 0 c, wpa, [ K ] K β 0 l β l w l v l M f 0 β 0 l β l v l w l l = c 3 β low β 0,low, wpa, l = 7

18 were for the last inequality we used Lemma S.5.. We thus have µ R+ A A Defining a 0 = c c 3, a c = c 0 c and a = c c µ R+ A A c 3 β low β 0,low + c c0 + c β low β 0,low, wpa. we thus obtain a 0 β low β 0,low β low β 0,low + a β low β 0,low, wpa, + a i.e., we have shown the desired bound on S β, f. S.6 Regarding the Proof of Corollary 4. As discussed in the main text, the proof of Corollary 4. is provided in Moon and Weidner 05. All that is left to show here is the matrix W = W λ 0, f 0, X k does not become singular as N, T under our assumptions. Proof. Remember W = TrM f 0 X k M λ 0 X k. The smallest eigenvalue of the symmetric matrix W λ 0, f 0, X k is given by a W a µ K W = min {a R K, a 0} a = min {a R K, a 0} a Tr = min {α R K, ϕ R K α 0, ϕ 0} [ Tr [ M f 0 M f 0 K k = a k X k M λ 0 K k = a k X k ] X low,ϕ + X high,α Mλ 0 X low,ϕ + X high,α ] α + ϕ, where we decomposed a = ϕ, α, with ϕ and α being vectors of length K and K, respectively, and we defined linear combinations of high- and low-rank regressors: X low,ϕ = K l= ϕ l X l, X high,α = 8 K m=k + α m X m.

19 Here, as in assumption 4 the components of α are denoted α K +,..., α K to simplify notation. We have M λ 0 = M λ 0,w + P Mλ 0 w, where w is the N K matrix defined in assumption 4, i.e., λ 0, w is an N R + K matrix, whereas M λ 0w is also an N K matrix. Using this we obtain µ K W = min {ϕ R K, α R K ϕ 0, α 0} = min {ϕ R K, α R K ϕ 0, α 0} { Tr [ M ϕ + α f 0 + Tr [ M f 0 X low,ϕ + X high,α Mλ 0,w X low,ϕ + X high,α ] X low,ϕ + X high,α PMλ0 w X low,ϕ + X high,α ] } { Tr [ ] M ϕ + α f 0 X high,α M λ 0,w X high,α + Tr [ M f 0 X low,ϕ + X high,α PMλ0 w X low,ϕ + X high,α ] }. S.6. We note there exists finite positive constants c, c, and c 3 such that Tr [ M f 0 Tr [ ] M f 0 X high,α M λ 0,w X high,α c α, X low,ϕ + X high,α PMλ0 w X low,ϕ + X high,α ] 0, wpa, Tr [ ] M f 0 X low,ϕ P Mλ 0 w X low,ϕ c ϕ, wpa, Tr [ ] M f 0 X low,ϕ c 3 P Mλ 0 w X high,α ϕ α, wpa, Tr [ ] M f 0 X high,α P Mλ 0 w X high,α 0, S.6. and we want to justify these inequalities now. The second and the last equation in S.6. are true because, e.g., Tr [ M f 0 X high,α P M λ 0 w X high,α ] = Tr [ Mf 0 X high,α P M λ 0 w X high,α M f 0], and the trace of a symmetric positive semi-definite matrix is non-negative. The first inequality in S.6. is true because rankf 0 + rankλ 0, w = R + K and using Lemma A. and assumption 4 we have α Tr [ ] M f 0 X high,α M λ 0,w X high,α α µ [ ] R+K + Xhigh,α X high,α > b, wpa, i.e., we can set c = b. The third inequality in S.6. is true because according Lemma S.4.v 9

20 we have Tr [ M f 0 X low,ϕ P Mλ 0 w X high,α ] K X low,ϕ X high,α K X low,ϕ F X high,α F K K K ϕ α max k =...K X k F max k =K +...K X k F c 3 ϕ α, where we used that assumption 4 implies X k / < C holds wpa for some constant C F as, and we set c 3 = K K K C. Finally, we have to argue that the third inequality in S.6. holds. Note X low,ϕ P M λ 0 w X low,ϕ = X low,ϕ M λ 0 X low,ϕ, i.e., we need to show Using part vi of Lemma S.4. we find Tr [ M f 0 X low,ϕ M λ 0 X low,ϕ ] c ϕ. Tr [ ] M f 0 X low,ϕ M λ 0 X low,ϕ = Tr [ ] M λ 0 X low,ϕ M f 0 X low,ϕ M λ 0 Mλ 0 X low,ϕ M f 0 X low,ϕ M λ 0, and according to Lemma S.5. this expression is bounded by some positive constant times ϕ in the lemma we have ϕ =, but all expressions are homogeneous in ϕ. Using the inequalities S.6. in equation S.6. we obtain µ K W min {ϕ R K, α R K min ϕ 0, α 0} c, c c c + c 3 { c α + max [ 0, c ϕ + α ϕ c 3 ϕ α ]}, wpa. Thus, the smallest eigenvalue of W is bounded from below by a positive constant as N, T, i.e., W is non-degenerate and invertible. S.7 Proof of Examples for Assumption 5 Proof of Example. We want to show the conditions of Assumption 5 are satisfied. Conditions i-iii are satisfied by the assumptions of the example. 0

21 For condition iv, notice Cov X it, X is C = E U it U is. Because β 0 < and sup it Ee it <, it follows N T i= t,s= Cov X it, X is C = = N i= t,s= N T E U it U is T i= t,s= p,q=0 β 0 p+q E e it p e is q <. For condition v, notice by the independence between the sigma field C and the error terms {e it } that we have for some finite constant M, = = N T i= t,s,u,v= N T i= t,s,u,v= N M T T = M T T = M T T i= t,s,u,v= p,q=0 t,s,u,v= p,q=0 s t,u,s,v= k= l= T s,v= min{s,v} k= Cov e it Xis, e iu Xiv C Cov e it U is, e iu U iv β 0 p+q E eit e is p e iu e iv q β 0 p E eit e is p β 0 q E eiu e iv q β 0 p+q [I {t = u} I {s p = v q} + I {t = v q} I {s p = u}] v β 0 s k+v l I {t = u} I {k = l} + M β 0 s+v k + M T T s,u= s u 0 β 0 s u T T T v,t= v t 0 T s,u= s u 0 β 0 s u T β 0 v t. T v,t= v t 0 β 0 v t

22 Notice and T T = T = T = = T s,v= T min{s,v} k= s v s= v= k= T s= v= β 0 T β 0 = O, T s,u= s u 0 β 0 s u = T s β 0 s v Therefore, we have the desired result N β 0 s+v k β 0 s v+v k + T l=0 s= v= T β 0 l + T T s= T s s= k= β 0 l l=0 T s β 0 s v + β 0 β 0 l l + T β 0 T i= t,s,u,v= l= s= u= T β 0 s k s β 0 s u T = β 0 l l = O. T l=0 Cov e it Xis, e iu Xiv C = Op. Preliminaries for Proof of Example Although we observe X it for t T, here we treat Z it = e it, X it as having an infinite past and future. Define Gτ t i = C σ {X is : τ s t} and Hτ t i = C σ {Z it : τ s t}. Then, by definition, we have Gτ t i, Hτ t i Fτ t i for all τ, t, i. By Assumption iv of Example, the time series of {X it : < t < } and {Z it : < t < } are conditional α-mixing conditioning on C uniformly in i. Mixing inequality: The following inequality is a conditional version of the α-mixing inequality of Hall and Heyde 980, p. 78. Suppose X it is a F t -measurable random

23 variable with E X it max{p,q} C X it C,p = E X it p C /p. Then, for each i, we have <, where p, q > with /p + /q <. Denote Cov X it, X it+m C 8 X it C,p X it+m C,q α p q m i. S.7. Proof of Example. Again, we want to show the conditions of Assumption 5 are satisfied. Conditions i-iii are satisfied by the assumptions of the example. For condition iv, we apply the mixing inequality S.7. with p = q > 4. Then, we have N i= t,s= N i= T Cov X it, X is C T T t t= m=0 = 6 N T T m i= m=0 t= 6 sup X it C,p i,t O p, Cov X it, X it+m C = X it C,p X it+m C,p α m i p P m=0 α p P m N T T m i= m=0 t= Cov X it, X it+m C where the last line holds because sup i,t X it C,p = O p for some p > 4 as assumed in the example, and m=0 α p P m = p m=0 m ζ P For condition v, we need to show N T i= t,s,u,v= = O because of ζ > 3 4p 4p Cov e it Xis, e iu Xiv C = Op. and p > 4. Notice = N T i= t,s,u,v= N T i= t,s,u,v= N T i= t,s,u,v= = I + II, say. Cov e it Xis, e iu Xiv C E e it Xis e iu Xiv C E e it Xis C E e iu Xiv C E e it Xis e iu Xiv C + N N T i= T t,s= E e it Xis C 3

24 First, for term I, there are a finite number of different orderings among the indices t, s, u, v. We consider the case t s u v and establish the desired result. The other cases can be shown analogously. Note N + N N i= N i= T N i= T T t T k t= T T t= k=0 T l l=0 m=0 0 l,m k 0 k+l+m T t T t= 0 k,m l 0 k+l+m T t E e it Xit+k e it+k+l Xit+k+l+m C E e it Xit+k e it+k+l Xit+k+l+m C [ E e it Xit+k e it+k+l Xit+k+l+m ] C E e it Xit+k C E e it+k+l Xit+k+l+m C + N + N N i= N i= T T T t= T t= = I + I + I 3 + I 4, say. 0 k,m l 0 k+l+m T t 0 p,l m 0 k+l+m T t E e it Xit+k C E e it+k+l Xit+k+l+m C E By applying the mixing inequality S.7. to X it+k e it+k+l Xit+k+l+m, we have E e it Xit+k e it+k+l Xit+k+l+m C [ e it Xit+k e it+k+l Xit+k+l+m C] E e it Xit+k e it+k+l Xit+k+l+m C with eit and 8 e it C,p Xit+k e it+k+l Xit+k+l+m C,q α p q k 8 e it C,p Xit+k C,3q e it+k+l C,3q Xit+k+l+m C,3q α p q k i, i where the last inequality follows by the generalized Holder s inequality. Choose p = 3q > 4. 4

25 Then, I 8 N 8 N i= T T t= sup e it C,p i,t 8 sup e it C,p i,t O p, 0 l,m k 0 k+l+m T t sup i,t sup i,t X it+k C,p C,p e it C,p Xit+k e it+k+l C,p Xit+k+l+m α 4p k i C,p X it+k C,p T k=0 T t= α 4p k 0 l,m k 0 k+l+m T t k α 4p k where the last line holds because we assume in example that sup i,t e it C,p sup i,t X it+k O p for some p > 4,, and p > 4. m=0 m α 4p m = 4p m=0 m ζ 4p By applying similar arguments, we can also show I, I 3, I 4 = O p. C,p = O because of ζ > 3 4p 4p and = S.8 Supplement to the Proof of Theorem 4.3 Notation E C and Var C and Cov C : In the remainder of this supplementary file we write E C, Var C and Cov C for the expectation, variance and covariance operators conditional on C, i.e., E C A = EA C, Var C A = VarA C and Cov C A, B = CovA, B C. What is left to show to complete the proof of Theorem 4.3 is that Lemma B. and Lemma B. in the main text appendix hold. Before showing this, we first present two further intermediate lemmas. 5

26 Lemma S.8.. Under the assumptions of Theorem 4.3 we have for k =,..., K, a P λ 0 Xk = o p, b X k P f 0 = o p, c P λ 0eX k = o p N 3/, d P λ 0eP f 0 = O p. Proof of Lemma S.8.. # Part a: We have P λ 0 Xk = λ 0 λ 0 λ 0 λ 0 Xk λ 0 λ 0 λ 0 λ 0 Xk λ 0 λ 0 λ 0 λ 0 Xk F = O p N / λ 0 Xk F, where we used part i and ii of Lemma S.4. and Assumption. We have [ ]} R T N E {E C λ 0 Xk F = E E C λ 0 X ir k,it r= t= i= { R T N } = E λ 0 ir E C X k,it = R r= r= T t= = O p, t= i= N E [ λ 0 ir Var C X k,it ] where we used X k,it is mean zero and independent across i, conditional on C, and our bounds on the moments of λ 0 ir and X k,it. We therefore have λ 0 Xk F = O p and the above inequality thus gives P λ 0 Xk = O p T = o p. i= # The proof for part b is similar. As above we first obtain X k P f 0 = P f 0 X k 6

27 O p T / f 0 X k F. Next, we have ] E C [ f 0 X k F = = R r= R r= N i= N [ R r= E C i= t,s= T ftr 0 X k,it t= T ftrf 0 sre 0 C Xk,it Xk,is max t f 0 tr ] N i= t,s= = O p T /4+ɛ O p = o p, T Cov C X k,it, X k,is where we used that uniformly bounded E f 0 t 4+ɛ implies max t f 0 tr = O p T /4+ɛ. We thus have f 0 X k F = o pt N and therefore X k P f 0 = o p. # Next, we show part c. First, we have [ λ 0 E {E C ex k ]} R F = E E C = E = { R R r= j= N r= i,j,l= t,s= N r= i,j= t= N N T λ 0 ire it X k,jt i= t= } T λ 0 irλ 0 lre C e it e ls X k,jt X k,js T E [ λ 0 ir E C e it Xk,jt] = ON T, where we used that E C e it e ls X k,jt X k,js is only non-zero if i = l because of cross-sectional independence conditional on C and t = s because regressors are pre-determined. We can thus conclude λ 0 ex k F = O p N T. Using this we find P λ 0eX k = λ 0 λ 0 λ 0 λ 0 ex k λ 0 λ 0 λ 0 λ 0 ex k λ 0 λ 0 λ 0 λ 0 ex k F = O p N / O p N T = O p. This is what we wanted to show. 7

28 # For part d, we first find f 0 eλ 0 F = O p, because E E C f 0 eλ 0 F = E E N T C i= t= { N N T T = E = = O, N i= i= T t= j= t= s= e it f 0 t λ 0 i E C e it e js f 0 t λ 0 i λ 0 j f 0 s E [ ] E C e it f 0 t λ 0 i λ 0 i ft 0 where we used e it is independent across i and over t, conditional on C. Thus we obtain P λ 0eP f 0 = λ 0 λ 0 λ 0 λ 0 ef 0 f 0 f 0 f 0 λ 0 λ 0 λ 0 λ 0 ef 0 f 0 f 0 f 0 where we used part i and ii of Lemma S.4.. O p N / O p N λ 0 ef 0 F O p T O p T / = O p, Lemma S.8.. Suppose A and B are T T and N N matrices that are independent of e, conditional on C, such that E C A F = Op and E C B F = Op, and let Assumption 5 be satisfied. Then there exists a finite non-random constant c 0 such that a E C {Tr [e e E C e e A]} c 0 N E C A F, b E C {Tr [ee E C ee B]} c 0 T E C B F. Proof. # Part a: Denote A ts to be the t, s th element of A. We have } Therefore, Tr {e e E C e e A} = = T T e e E C e e ts A ts t= T s= T N e it e is E C e it e is A ts. t= s= i= E C Tr {e e E C e e A} [ T T T T N N ] = E C e it e is E C e it e is e jp e jq E C e jp e jq E C A ts A pq. t= s= p= q= i= j= 8

29 Let Σ it = E C e it. Then we find { N N } E C e it e is E C e it e is e jp e jq E C e jp e jq i= i= = Therefore, j= N N = {E C e it e is e jp e jq E C e it e is E C e jp e jq } j= Σ it Σ is if t = p s = q and i = j Σ it Σ is if t = q s = p and i = j E C e 4 it Σ it if t = s = p = q and i = j 0 otherwise. E C Tr {e e E C e e A} T T N Σ it Σ is EC A ts + EC A ts A st + t= s= i= T N EC e 4 it Σ it EC A tt. t= i= Define Σ i = diag Σ i,..., Σ it. Then, we have Also, T t= T s= N N Σ it Σ is EC A ts = EC Tr A Σ i AΣ i i= i= N E C AΣ i N F Σ i E C A F i= i= N sup Σ it E C A F. it S.8. Finally, T t= T s= [ N N Σ it Σ is E C A ts A st = E C Tr Σ i AAΣ i] i= T t= i= i= N E C Σ i A N AΣ i F F Σ i E C A F i= i= N sup Σ it E C A F. S.8. it N EC e 4 it Σ it EC A tt N sup E C e 4 it E C A F, it 9 S.8.3

30 and sup it E C e 4 it is assumed bounded by Assumption 5vi. # Part b: The proof is analogous to the proof of part a. Proof of Lemma B.. # For part a we have Tr P f 0 e P λ 0 Xk = Tr P f 0 e P λ 0P λ 0 Xk P f 0 R P λ 0 e P f 0 P λ 0 Xk Pf 0 = O p o p O p = o p, where the second-last equality follows by Lemma S.8. a and d. # To show statement b we define ζ k,ijt = e it Xk,jt. We then have Tr P λ 0 e X k = [ R λ 0 λ 0 ] r,q= N rq T N N λ 0 irλ 0 jqζ k,ijt. t= i,j= }{{} A k,rq We only have E C ζk,ijt ζ k,lms 0 if t = s because regressors are pre-determined and i = l and j = m because of cross-sectional independence. Therefore E { { } E C A T N } k,rq = E λ N 3 ir λ jq λ lr λ mq E C ζk,ijt ζ T k,lms = N 3 T T t,s= i,j,l,m= N t= i,j= E [ λ irλ jq E C ζ k,ijt] = O/N = op. We thus have A k,rq = o p and therefore also Tr P λ 0 e X k = o p. # The proof for statement c is similar to the proof of statement b. Define ξ k,its = e it Xk,is E C e it Xk,is. We then have { Tr P f 0 ]} [e Xk E C e Xk = [ R f ] f T r,q= rq N T T f tr f sq ξ k,its. i= t,s= }{{} B k,rq 30

31 Therefore E C B N T k,rq = T 3 N i,j= t,s,u,v= 4 max f t r t, r T 3 N 4 = max f t r t, r T 3 N = O p T 4/4+ɛ O p /T = o p, f tr f sq f ur f vq E C ξk,its ξ k,juv N T i,j= t,s,u,v= N T i= t,s,u,v= Cov C e it Xk,is, e ju Xk,jv Cov C e it Xk,is, e iu Xk,iv where we used uniformly bounded E f 0 t 4+ɛ implies max t f 0 tr = O p T /4+ɛ. # Part d and e: We have λ 0 λ 0 λ 0 f 0 f 0 f 0 = O p /, e = O p N /, X k = O p and P λ 0eP f 0 = O p, which was shown in Lemma S.8.. Therefore: Tr ep f 0 e M λ 0 X k f 0 f 0 f 0 λ 0 λ 0 λ 0 = Tr P λ 0eP f 0 e M λ 0 X k f 0 f 0 f 0 λ 0 λ 0 λ 0 R P λ 0eP f 0 e X k f 0 f 0 f 0 λ 0 λ 0 λ 0 = Op N / = o p. which shows statement d. The proof for part e is analogous. # To prove statement f we need to use in addition P λ 0 e X k = o pn 3/, which was also shown in Lemma S.8.. We find Tr e M λ 0 X k M f 0 e λ 0 λ 0 λ 0 f 0 f 0 f 0 = Tr e M λ 0 X k e P λ 0 λ 0 λ 0 λ 0 f 0 f 0 f 0 Tr e M λ 0 X k P f 0 e P λ 0 λ 0 λ 0 λ 0 f 0 f 0 f 0 R e P λ 0 e X k λ 0 λ 0 λ 0 f 0 f 0 f 0 = o p. R e X k P λ 0 e P f 0 λ 0 λ 0 λ 0 f 0 f 0 f 0 3

32 # Now we want to prove part g and h of the present lemma. For part g we have Tr { [ee E C ee ] M λ 0 X k f 0 f 0 f 0 λ 0 λ 0 λ 0 } = Tr { [ee E C ee ] M λ 0 X k f 0 f 0 f 0 λ 0 λ 0 λ 0 } + } Tr {[ee E C ee ] M λ 0 Xk P f 0 f 0 f 0 f 0 λ 0 λ 0 λ 0 = Tr { [ee E C ee ] M λ 0 X k f 0 f 0 f 0 λ 0 λ 0 λ 0 } + ee E C ee X k P f 0 f 0 f 0 f 0 λ 0 λ 0 λ 0 = Tr { [ee E C ee ] M λ 0 X k f 0 f 0 f 0 λ 0 λ 0 λ 0 } + o p. Thus, what is left to prove is Tr { [ee E C ee ] M λ 0 X k f 0 f 0 f 0 λ 0 λ 0 λ 0 } = o p. For this we define Using part i and ii of Lemma S.4. we find B k = M λ 0 X k f 0 f 0 f 0 λ 0 λ 0 λ 0. B k F R / B k R / X k f 0 f 0 f 0 λ 0 λ 0 λ 0 R / X k F f 0 f 0 f 0 λ 0 λ 0 λ 0. and therefore E C Bk F R f 0 f 0 f 0 λ 0 λ 0 λ 0 E C Xk F = O, where we used E C Xk F = O, which is true because we assumed uniformly bounded moments of X k,it. Applying Lemma S.8. we therefore find E C Tr {[ee E C ee T ] B k } c 0 E C Bk F = o, and thus Tr {[ee E C ee ] B k } = o p, 3

33 which is what we wanted to show. The proof for part h is analogous. # Part i: Conditional on C the expression e itx it X it E C e it X it X it is mean zero, and it is also uncorrelated across i. This together with the bounded moments that we assume implies { N T [ ] } Var C e it X it X it E C e it X it X it = O p /N = o p, i= t= which shows the required result. # Part j: Define the K K matrix A = we have N i= T t= N i= e it X it X it X it X it = A + A. T t= e it X it + X it X it X it. Then Let B k be the N T matrix with elements B k,it = e it X k,it + X k,it. We have B k B k F = O p, because the moments of B k,it are uniformly bounded. The components of A can be written as A lk = Tr[B lx k X k ]. We therefore have We have X k X k = X k P f 0 + P λ 0 A lk R B l R B l A lk rankx k X k B l X k X k. Xk P f 0 Xk P f 0 + Xk M f 0. Therefore rankx k X k R and where we used Lemma S.8.. This shows the desired result. + P λ 0 Xk M f 0 P λ 0 Xk = R O p o p = o p, Proof of Lemma B.. Let c be a K-vector such that c =. The required result follows by the Cramer-Wold device, if we show N i= t= T e it X itc N 0, c Ωc. For this, define ξ it = e it X itc. Furthermore define ξ m = ξ M,m = ξ,it, with M = and m = T i + t {,..., M}. We then have the following: i Under Assumption 5i, ii, iii the sequence {ξ m, m =,..., M} is a martingale difference sequence under the filtration F m = C σ{ξ n : n < m}. 33

34 ii Eξ 4 it is uniformly bounded, because by Assumption 5vi E C e 8 it and E C X it 8+ɛ are uniformly bounded by a non-random constant applying Cauchy-Schwarz and the law of iterated expectations. iii M M m= ξ m = c Ωc + o p. ξ m E C ξ m ] } = { [ This is true, because firstly under our assumptions we have E M C M m= { E M C M m= ξ m E C ξ m } = O P /M = o P, implying we have M M m= ξ m = M M m= E Cξ m + o p. We furthermore have M M m= E Cξ m = Var C M / M m= ξ m, and using the result in equation 4 of the main text we find Var C M / M m= ξ m = Var C / N i= T t= ξ it = c Ωc + o p. These three properties of {ξ m, m =,..., M} allow us to apply Corollary 5.6 in White 00, which is based on Theorem.3 in Mcleish 974, to obtain M M m= ξ m d N 0, c Ωc. This concludes the proof, because M M m= ξ m = N i= T t= e itx itc. S.9 Expansions of Projectors and Residuals The incidental parameter estimators f and λ as well as the residuals ê enter into the asymptotic bias and variance estimators for the LS estimator β. To describe the properties of f, λ and ê, it is convenient to have asymptotic expansions of the projectors M λβ and M fβ that correspond to the minimizing parameters λβ and fβ in equation 4. Note the minimizing λβ and fβ can be defined for all values of β, not only for the optimal value β = β. The corresponding residuals are êβ = Y β X λβ f β. Theorem S.9.. Under Assumptions, 3, and 4i we have the following expansions K M λβ = M λ 0 + M λ,e + M λ,e k= K M fβ = M f 0 + M f,e + M f,e êβ = M λ 0 e M f 0 + ê e K k= k= βk β 0 k M βk β 0 k M λ,k + M rem λ f,k + M rem f βk β 0 k ê k + ê rem β, β, β, 34

35 where the spectral norms of the remainders satisfy for any series η 0: M rem β λ sup {β: β β 0 η } β β 0 + / e β β 0 + 3/ e = O 3 p, M rem β f sup {β: β β 0 η } β β 0 + / e β β 0 + 3/ e = O 3 p, ê rem β sup {β: β β 0 η } / β β 0 + e β β 0 + e = O 3 p, and we have rankê rem β 7R, and the expansion coefficients are given by M λ,e = M λ 0 e f 0 f 0 f 0 λ 0 λ 0 λ 0 λ 0 λ 0 λ 0 f 0 f 0 f 0 e M λ 0, M λ,k = M λ 0 X k f 0 f 0 f 0 λ 0 λ 0 λ 0 λ 0 λ 0 λ 0 f 0 f 0 f 0 X k M λ 0, M λ,e = M λ 0 e f 0 f 0 f 0 λ 0 λ 0 λ 0 e f 0 f 0 f 0 λ 0 λ 0 λ 0 + λ 0 λ 0 λ 0 f 0 f 0 f 0 e λ 0 λ 0 λ 0 f 0 f 0 f 0 e M λ 0 M λ 0 e M f 0 e λ 0 λ 0 λ 0 f 0 f 0 λ 0 λ 0 λ 0 λ 0 λ 0 λ 0 f 0 f 0 λ 0 λ 0 λ 0 e M f 0 e M λ 0 M λ 0 e f 0 f 0 f 0 λ 0 λ 0 f 0 f 0 f 0 e M λ 0 + λ 0 λ 0 λ 0 f 0 f 0 f 0 e M λ 0 e f 0 f 0 f 0 λ 0 λ 0 λ 0, analogously M f,e = M f 0 e λ 0 λ 0 λ 0 f 0 f 0 f 0 f 0 f 0 f 0 λ 0 λ 0 λ 0 e M f 0, M f,k = M f 0 X k λ 0 λ 0 λ 0 f 0 f 0 f 0 f 0 f 0 f 0 λ 0 λ 0 λ 0 X k M f 0, M f,e = M f 0 e λ 0 λ 0 λ 0 f 0 f 0 f 0 e λ 0 λ 0 λ 0 f 0 f 0 f 0 + f 0 f 0 f 0 λ 0 λ 0 λ 0 e f 0 f 0 f 0 λ 0 λ 0 λ 0 e M f 0 M f 0 e M λ 0 e f 0 f 0 f 0 λ 0 λ 0 f 0 f 0 f 0 f 0 f 0 f 0 λ 0 λ 0 f 0 f 0 f 0 e M λ 0 e M f 0 M f 0 e λ 0 λ 0 λ 0 f 0 f 0 λ 0 λ 0 λ 0 e M f 0 + f 0 f 0 f 0 λ 0 λ 0 λ 0 e M f 0 e λ 0 λ 0 λ 0 f 0 f 0 f 0, 35

36 and finally ê k = M λ 0 X k M f 0, ê e = M λ 0 e M f 0 e λ 0 λ 0 λ 0 f 0 f 0 f 0 λ 0 λ 0 λ 0 f 0 f 0 f 0 e M λ 0 e M f 0 M λ 0 e f 0 f 0 f 0 λ 0 λ 0 λ 0 e M f 0. Proof. The general expansion of M λβ is given in Moon and Weidner 05, and in the theorem we just make this expansion explicit up to a particular order. The result for M fβ is just obtained by symmetry N T, λ f, e e, X k X k. For the residuals ê we have ê = M λ Y [ ] β k X k = M λ e β β 0 X + λ 0 f 0, k= and plugging in the expansion of M λ gives the expansion of ê. We have êβ = A 0 + λ 0 f 0 λβ f β, where A 0 = e k β k β 0 kx k. Therefore ê rem β = A + A + A 3 with A = A 0 M λ 0 A 0 M f 0, A = λ 0 f 0 λβ f β, and A 3 = ê e. We find ranka R, ranka R, ranka 3 3R, and thus rankê rem β 7R, as stated in the theorem. Having expansions for M λβ and M fβ, we also have expansions for P λβ = I N M λβ and P fβ = I T M fβ. The reason why we give expansions of the projectors and not expansions of λβ and fβ directly is for the latter we would need to specify a normalization, whereas the projectors are independent of any normalization choice. An expansion for λβ can, for example, be defined by λβ = P λβλ 0, in which case the normalization of λβ is implicitly defined by the normalization of λ 0. S.0 Consistency Proof for Bias and Variance Estimators Proof of Theorem 4.4 It is convenient to introduce some alternative notation for Definition in section 4.3 of the main text. 36

37 Definition Let Γ : R R be the truncation kernel defined by Γx = for x, and Γx = 0 otherwise. Let M be a bandwidth parameter that depends on N and T. For an N N matrix A with elements A ij and a T T matrix B with elements B ts we define i the diagonal truncations A truncd = diag[a ii i=,...,n ] and B truncd = diag[b tt t=,...,t ]. ii the right-sided Kernel truncation of B, which is a T T matrix B truncr with elements Bts truncr = Γ s t Bts for t < s, and B truncr M ts = 0 otherwise. Here, we suppress the dependence of B truncr on the bandwidth parameter M. Using this notation we can represent the estimators for the bias in Definition as follows: B,k = N Tr [P f ê X k truncr], B,k = [ê T Tr ê truncd M λ X k f f f λ λ ] λ, B 3,k = [ê N Tr ê truncd M f X k λ λ λ f ] f f. Before proving Theorem 4.4 we establish some preliminary results. Corollary S.0.. Under the Assumptions of Theorem 4.3 we have β β 0 = O p. This corollary directly follows from Theorem 4.3. Corollary S.0.. Under the Assumptions of Theorem 4.4 we have P λ P λ 0 = M λ M λ 0 = Op N /, P f P f 0 = M f M f 0 = O p T /. Proof. Using e = O p N / and X k = O p N we find the expansion terms in Theorem S.9. satisfy M λ,e = O p N /, M λ,e = O p N, M λ,k = O p. Together with corollary S.0. the result for M λ M λ 0 immediately follows. In addition we have P λ P λ 0 = M λ + M λ 0. The proof for M f and P f is analogous. 37

38 Lemma S.0.3. Under the Assumptions of Theorem 4.4 we have A A N i= N i= T t= T t= e it X it X it X it X it = o p, e it ê it Xit X it = o p. Lemma S.0.4. Let f and f 0 be normalized as f f/t = IR and f 0 f 0 /T = I R. Then, under the assumptions of Theorem 4.4, there exists an R R matrix H = H such that f f 0 H = O p, λ λ 0 H = Op. Furthermore λ λ λ f f f λ 0 λ 0 λ 0 f 0 f 0 f 0 = Op N 3/. Here, the matrix H depends on N, T, but we write H instead of H simple. to keep notation Lemma S.0.5. Under the Assumptions of Theorem 4.4 we have i N EC e X k ê X k truncr = op, ii N EC e e ê ê truncd = op, iii T EC ee ê ê truncd = op. Lemma S.0.6. Under the Assumptions of Theorem 4.4 we have i N ê X k truncr = Op MT /8, ii N ê ê truncd = Op, iii T ê ê truncd = Op. The proof of the above lemmas is given section S. below. Using these lemmas we can now prove Theorem

39 Proof of Theorem 4.4, Part I: show Ŵ = W + o p. Using Tr C C rank C and corollary S.0. we find: Ŵk k W,k k [ = Tr M λ M ] ] λ 0 Xk M f X k + Tr [M λ 0 X k M f M f 0 X k R M λ M λ 0 Xk X k R M f M f 0 = R O pn O p + R O pt O p = o p. Thus we have Ŵ = W + o p = W + o p. X k X k Proof of Theorem 4.4, Part II: show Ω = Ω + o p. Let Ω N T i= t= e it X it X it. We have Ω = Ω + o P = Ω + A + A + o p = Ω + o P, where A and A are defined in Lemma S.0.3, and the lemma states A and A are o p. Proof of Theorem 4.4, Part III: show B = B + o p. Let B,k, = N Tr [P f 0 E C e X k ]. According to Assumption 6 we have B,k = B,k, +o p. What is left to show is B,k, = B,k + o p. Using Tr C C rank C we find B,k, B [ ] = E C N TrP f 0 e X k N [ Tr P f ê X k truncr] [P N Tr f 0 P f ê X k truncr] + { [ N Tr P f 0 E C e X k ê X k truncr]} R P f 0 P N f ê X k truncr + R N P f 0 EC e X k ê X k truncr. We have P f 0 =. We now apply Lemmas S.0.5, S.0. and S.0.6 to find B,k, B = N O p N / O p M /8 + o p N = o p. This is what we wanted to show. 39

40 Proof of Theorem 4.4, final part: show B = B + o p and B 3 = B 3 + o p. Define B,k, = T Tr [ E C ee M λ 0 X k f 0 f 0 f 0 λ 0 λ 0 λ 0 ]. According to Assumption 6 we have B,k = B,k, + o p. What is left to show is B,k, = B,k + o p. We have B,k B,k = T Tr [ E C ee M λ 0 X k f 0 f 0 f 0 λ 0 λ 0 λ 0 ] T Tr [ ê ê truncd M λ X k f f f λ λ λ ] = T Tr [ ê ê truncd M λ X k f 0 f 0 f 0 λ 0 λ 0 λ 0 f f f λ λ λ ] + T Tr [ê ê truncd M λ 0 M λ Xk f 0 f 0 f 0 λ 0 λ 0 λ 0 ] + T Tr {[ E C ee ê ê truncd] M λ 0 X k f 0 f 0 f 0 λ 0 λ 0 λ 0 } Using Tr C C rank C which is true for every square matrix C we find B,k B R,k ê ê truncd Xk f 0 f 0 f 0 λ 0 λ 0 λ 0 T f f f λ λ λ + R ê ê truncd Mλ 0 M λ Xk f 0 f 0 f 0 λ 0 λ 0 λ 0 T + R E C ee ê ê truncd Xk f 0 f 0 f 0 λ 0 λ 0 λ 0. T Here we used M f 0 = =. Using X k = O p, and applying Lemmas S.0., M f S.0.4, S.0.5 and S.0.6, we now find B,k B [,k = T O p T O p / O p N 3/ + O p T O p N / O p / O p / ] + o p T O p / O p / = o p. This is what we wanted to show. The proof of B 3 = B 3 + o p is analogous.. S. Proof of Intermediate Lemma Here we provide the proof of some intermediate lemmas that were stated and used in section S.0. The following lemma gives a useful bound on the maximum of correlated random variables 40

41 Lemma S... Let Z i, i =,,..., n, be n real valued random variables, and let γ and B > 0 be finite constants independent of n. Assume max i E C Z i γ B, i.e., the γ th moment of the Z i are finite and uniformly bounded. For n we then have max i Z i = O p n /γ. S.. Proof. Using Jensen s inequality one obtains E C max i Z i E C max i Z i γ /γ E C n i= Z i γ /γ n max i E C Z i γ /γ n /γ B /γ. Markov s inequality then gives equation S... Lemma S... Let Z k,tτ = N / Z t = N / Z 3 i = T / N [e it X k,iτ E C e it X k,iτ ], i= N [ ] e it E C e it, i= T [ ] e it E C e it. t= Under assumption 5 we have E Z C k,tτ 4 B, E Z C tτ 4 B, E Z3 C i 4 B, for some B > 0, i.e., the conditional expectations Z k,tτ, Z tτ, and Z 3 i are uniformly bounded over t, τ, or i, respectively. Proof. # We start with the proof for Z k,tτ. Define Z k,tτ,i = e itx k,iτ E C e it X k,iτ. By assumption we have finite 8th moments for e it and X k,iτ uniformly across k, i, t, τ, and thus using Cauchy Schwarz inequality we have finite 4th moment of Z k,tτ,i uniformly across k, i, t, τ. For ease of notation we now fix k, t, τ and write Z i = Z k,tτ,i. We have E CZ i = 0 and E C Z i Z j Z k Z l = 0 if i / {j, k, l} and the same holds for permutations of i, j, k, l. Using 4

42 this we compute N 4 N E C Z i = E C Z i Z j Z k Z l i= i,j,k,l= = 3 E C Z i Zj + E C Z 4 i i j i N N [ ] } = 3 E C Z i EC Z j + {E C Z 4i 3 EC Z i i,j= i=, Because we argued E C Z 4 i is bounded uniformly, the last equation shows is bounded uniformly across k, t, τ. This is what we wanted to show. # The proofs for Z t and Z 3 i are analogous. Z k,tτ = N / N i= Z k,tτ,i Lemma S..3. For a T T matrix A we have A truncr M A truncr max M max max A tτ, t t<τ t+m Here, for the bounds on τ we could write max, t M instead of t M, and mint, t + M instead of t + M, to guarantee τ T. Since this would complicate notation, we prefer the convention A tτ = 0 for t < or τ < of t > T or τ > T. Proof. For the -norm of A truncr we find A truncr = max t=...t M t+m τ=t+ A tτ max A tτ = M A truncr t<τ t+m max, and analogously we find the same bound for the -norm A truncr. Applying part vii of Lemma S.4. we therefore also get this bound for the operator norm A truncr. Proof of Lemma S.0.3. # We first show A N T i= t= e it X it X it X it X it = o p. Let B,it = X it X it, B,it = e itx it, and B 3,it = e it X it. Note B, B, and B 3 can either be viewed as K-vectors for each pair i, t, or equivalently as N T matrices B,k, B,k, and B 3,k for each k =,..., K. We have A = i t B,it B,it + B 3,it B,it, or equivalently A,k k = Tr B,k B 3,k + B,k B,k. 4