Additional Proofs for: Crump, R., V. J. Hotz, G. Imbens and O. Mitnik, Nonparametric Tests for Treatment Effect Heterogeneity
|
|
- Ἡρακλῆς Βιτάλη
- 5 χρόνια πριν
- Προβολές:
Transcript
1 Aitional Proofs for: Crump, R., V. J. Hotz, G. Imbens an O. Mitnik, Nonparametric Tests for Treatment Effect Heterogeneity Final Draft: December 007
2 Aitional Proofs Proof of Lemma A. We will generalize the proof in Imbens, Newey an Rier 006. For i we will show [ E ˆΩ Ω ] C ζk K/N so that the result follows by Markov s inequality. [ E ˆΩ Ω ] [ E tr ˆΩ ] tr E [ˆΩ + tr Ω tr E + tr Ω ] tr ˆΩ Ω ] tr E [ˆΩ Ω Note first that E[ˆΩ ] Ω so that, [ˆΩ ] tr Ω. B. The first term of equation B. is ] tr E [ˆΩ N E w W i R kk X i R lk X i N w k l N i k l i j N E [ w W i w W j R kk X i R lk X i R kk X j R lk X j ]. We may partition this epression into terms with i j, N E [ w W i R kk X i R lk X i ] k l i B. an with terms i j, E [ w W i R kk X i R lk X i ] E [ w W j R kk X j R lk X j ]. k l i j B.3 For a ranom variable U with E U < an an event G with Pr G > 0, then E [U G] E [U G] Pr G. Using this we may rewrite equations B. an B.3 as, ] N tr E [ˆΩ π w k l E [ R kk X R lk X W w ] + π w N N N w trω. B.4 []
3 To eal with the first term of equation B.4 consier, E [ [ R kk X R lk X ] K W w E R kk X k l k l K ζk l K R lk X W w E [ R lk X W w ] ζk tr Ω λ ma Ω ζk K C ζk K. ] B.5 B.6 Equation B.5 follows by, K ζk sup R K sup RkK which then implies that RkK ζk. k k Equation B.6 follows since the maimum eigenvalue of Ω is O see below. Thus, the first term of equation B.4 is π w N N w k l E [ R kk X R lk X ] W w C ζk KN. B.7 To eal with secon term of equation B.4 an the secon term of equation B. we have, trω λ ma Ω K λma Ω K C K. Thus, πw N N trω N + o O K O KN. B.8 Combining the results from equations B.7 an B.8 yiels, [ E ˆΩ Ω ] O ζk KN + O KN O ζk KN. For ii, first note that for any two positive semi-efinite matrices A an B, an conformable vectors a an b, if A B in a positive semi-efinite sense, then for an λ min A min a a a Aa a Aa, λ min B min b b b Bb b Bb, λ ma A ma a a a Aa ā Aā, we have that, λ ma B ma b b b Bb b B b, λ min A λ min B B.9 []
4 an λ ma A λ ma B. B.0 Now, let f w f X W W w an efine q f 0 /f an note that by Assumptions.3 an 3. we have that 0 < q q q <. Thus we may efine q q + q so that, Ω 0,K E [R K R K W 0] R K R K f 0 R K R K qf R K R K q + q f q R K R K f + R K R K qf q Ω,K + R K R K qf q Ω,K + Q Q is a positive semi-efinite matri, which implies that Ω 0,K q Ω,K in a positive semi-efinite sense. Thus by equation B.9 λ min Ω 0,K q λ min Ω,K q an the minimum eigenvalue of Ω 0,K is boune away from zero. Net, observe that 0 q q q < an so by the above we have that Ω 0,K q Ω,K + R K R K qf q Ω,K + q q R K R K f q Ω,K, in a positive semi-efinite sense. Now by equation B.0 we have λ ma Ω 0,K λ ma q Ω,K q an the maimum eigenvalue of Ω 0,K is boune. Both the minimum an maimum eigenvalue of Ω,K are boune away from zero an boune, respectively, by construction. For iii consier the minimum [3]
5 eigenvalue of ˆΩ, λ min ˆΩ min min Ω + ˆΩ Ω ˆΩ min Ω + min λ min Ω + λ min ˆΩ Ω λ min Ω ˆΩ Ω ˆΩ Ω B. λ min Ω O p ζkk / N / B. Where B. follows since for a symmetric matri A A tr A λ min A, an since the norm is nonnegative an A λ min A A λ min A for all values of λ min A. Finally, B. follows by part i. Net, consier the maimum eigenvalue of ˆΩ. λ ma ˆΩ ma ma Ω + ˆΩ Ω ˆΩ ma Ω + ma λ ma Ω + λ ma ˆΩ Ω λ ma Ω + ˆΩ Ω ˆΩ Ω B.3 λ ma Ω + O p ζkk / N / B.4 Where B.3 follows by similar arguments as above an B.4 follows by part i. efine For iv let us first Σ R D R N w, D iag { w W i ε w,i; i,..., N }. Net recall that for matrices A an B we have that A + B A + B. [4]
6 Thus, E ˆΣ Σ E ˆΣ Σ + Σ Σ E ˆΣ Σ + E Σ Σ R ˆD D R E N w + E Σ Σ B.5 B.6 Before we eal with equations B.5 an B.6, we nee to establish conitions for consistency of the estimate errors. Note that, ˆε w,i ε w,i X i Y i ˆµ w X i Y i µ w X i µ w X i ˆµ w X i an so by Lemma A.6 v sup ˆε w,i ε w,i O p ζ K KN + O ζ K K s/. Moreover, ˆε w,i ε w,i ε w,i ˆε w,i ε w,i + O p ζ K KN + O ζ K K s/. an so, for M R ++ Pr ˆε w,i ε w,i > M E ˆε w,i ε w,i M E ε w,i ˆε w,i ε w,i + O p ζ K KN + O ζ K K s/ M O p ζ K KN + O ζ K K s/ an so E ε w,i ˆε w,i ε w,i M C sup + M ˆε w,i ε w,i E ε w,i M + O p ζ K KN + O ζ K K s/ ζ K K s/, O p ζ K KN + O ˆε w,i ε w,i O p ζ K KN + ζ K K s/. B.7 We begin with equation B.6 first. E Σ Σ E [tr Σ ] tr E [ Σ tr tr E ] First note that E [ Σ Σ an so, tr Σ Σ E + tr Σ ] [ Σ ] Σ + tr Σ [ Σ ] tr Σ. B.8 [5]
7 The first term of equation B.8 is ] tr E [ Σ N w + N w N N w + N k l i j k l i k l i j k l N N N w Equation B.9 may be rewritten as, N k l N N w N N w C C N E [ w W i w W j ε i ε jr kk X i R lk X i R kk X j R lk X j ] N E [ w W i ε 4 w,ir kk X i R lk X i ] E [ w W i ε w,ir kk X i R lk X i ] E [ {Wjw}ε w,jr kk X j R lk X j ] E [ w W ε 4 wr kk X R lk X ] k l E [ w W ε 4 wr kk X R lk X ] k l k l N k l N k l B.9 [ E w W ε wr kk XR lk X ]. B.0 E [ w W E [ ε 4 w X, w W ] R kk X R lk X ] E [ w W E [ ε 4 ] w X RkK X R lk X ] E [ w W R kk X R lk X ] E [ R kk X R lk X ] W w. Thus by equation B.7 we have that equation B.9 is, N E [ w W ε 4 wr kk X R lk X ] C ζk KN. k l B. Equation B.0 an the secon term of equation B.8 are, N N N w k l πw N N π w N w N N N w [ E w W ε wr kk XR lk X ] tr Σ k l [ E σ w X R kk XR lk X ] W w tr Σ tr Σ. B. The first factor is, πw N N N + o. [6]
8 By Assumption 3. the secon factor is, tr Σ λma Σ K σ4 λ ma Ω K C K B.3 an so equation B. is, N N N w k l [ E w W ε wr kk XR lk X ] tr Σ O KN. Thus, by equations B. an B.3 we have, E Σ Σ O ζk KN + O KN O ζk KN. B.4 Now consier equation B.5. E R N w ˆD D N w N k l i j Then, by equation B.7 we have, N N k l i j R N E [ w W i w W j ˆε i ε i ˆε j ε jr kk X i R lk X i R kk X j R lk X j ]. E [ w W i w W j ˆε i ε i ˆε j ε jr kk X i R lk X i R kk X j R lk X j ] [O p ζ K KN + ζ K K s/] N k l i j [O p ζ K KN + O N E [ w W i w W j R kk X i R lk X i R kk X j R lk X j ] ζ K K s/] ] tr E [ˆΩ. From the proof of i, we have that ] tr E [ˆΩ O ζ K KN + O KN + O K O K, an so we have, E R D D N w R [O p ζ K KN + O ζ K K s/] O K. B.5 [7]
9 Combining equations B.4 an B.5 yiels, E ˆΣ Σ E Σ Σ + E R D D O ζk KN + [O p ζ K KN + O N w R ζ K K s/] O K O ζk KN + [O p ζ K 4 K N + O ζ K K s/ + O p ζ K 3 KK s/ N ] O K O ζk KN + O p ζ K 4 K 3 N + O ζ K KK s/ + O p ζ K 3 K K s/ N O p ζ K 4 K 3 N + O ζ K KK s/, an the result follows. For v note that, σ Ω Σ σ Ω in a positive semi-efinite sense. Thus, Similarly, λ min Σ λ min σ Ω σ λ min Ω σ min q,. λ ma Σ λ ma σ Ω σ λ ma Ω σ ma q,. For vi consier, λ min ˆΣ min min [ Σ + Σ ˆΣ ˆΣ min Σ + min λ min Σ + λ min Σ ˆΣ λ min Σ Σ ˆΣ ] Σ ˆΣ λ min Σ O p ζ K K 3/ N O p ζ K K / K s/. Net, λ ma ˆΣ ma ma [ Σ + Σ ˆΣ ˆΣ ma Σ + ma λ ma Σ + λ ma Σ ˆΣ λ ma Σ + Σ ˆΣ ] Σ ˆΣ λ ma Σ + O p ζ K K 3/ N + O p ζ K K / K s/. [8]
10 Proof of Lemma A. For this proof we nee two results. matri an B a conformable matri, then Let A be a symmetric positive efinite λ min B AB λ min A λ min B B, λ ma B AB λ ma A λ ma B B. Using the above result we have, λ min Ω Σ Ω λ min Σ λ min Ω σ min q, [λ ] ma Ω σ min q, [λ ma Ω ] σ min q, [ma q, ], B.6 an, λ ma Ω Σ Ω λ ma Σ λ ma Ω σ ma q, [λ ] min Ω σ ma q, [λ min Ω ] σ ma q, [min q, ]. B.7 Now consier, λ min N V min π 0 Ω 0,K π Σ 0,KΩ 0,K + Ω,K 0 π Σ,KΩ,K min Ω 0,K Σ 0,KΩ π 0 λ min Ω 0,K Σ 0,KΩ 0,K 0,K + min π Ω + λ min Ω,K π Σ,KΩ,K,K Σ,KΩ,K which is boune away from zero by equation B.6 an Assumption.3. λ ma N V ma Ω 0,K π Σ 0,KΩ 0,K + Ω,K 0 π Σ,KΩ,K ma π 0 Ω 0,K Σ 0,KΩ 0,K + ma π π 0 λ ma Ω 0,K Σ 0,KΩ 0,K,,K Σ,KΩ,K Ω + λ ma Ω,K π Σ,KΩ,K, Finally, consier [9]
11 which is boune by equation B.7 an Assumption.3. For ii we have, ˆΩ λ ˆΣ min ˆΩ λ min ˆΣ ˆΩ λ min [λ min Σ O p ζ K K 3/ N O p ζ K K / K s/] ˆΩ λ min [λ min Σ O p ζ K K 3/ N O p ζ K K / K s/] [λ min O p ζ K K / N /] Ω λ min Σ [λ min Ω ] + Op ζ K K 3/ N + O p ζ K K / K s/. In aition, ˆΩ λ ˆΣ ma ˆΩ λ ma ˆΣ ˆΩ λ ma [λ ma Σ + O p ζ K K 3/ N + O p ζ K K / K s/] ˆΩ λ ma [λ ma Σ + O p ζ K K 3/ N + O p ζ K K / K s/] [λ ma Ω + O p ζ K K / N /] ] λ ma Σ [λ ma Ω + Op ζ K K 3/ N + O p ζ K K / K s/. B.8 B.9 Thus, λ min N ˆV N min N ˆΩ ˆΣ 0,K ˆΩ 0,K 0,K + N 0 N ˆΩ ˆΣ,K ˆΩ,K,K N N 0 min ˆΩ ˆΣ 0,K ˆΩ 0,K 0,K N N 0 λ min ˆΩ 0,K ˆΣ 0,K ˆΩ 0,K + N N min ˆΩ ˆΣ,K ˆΩ,K,K + N N λ min ˆΩ,K ˆΣ,K ˆΩ,K is boune away from zero in probability by equation B.8, Assumption.3 an Assumptions 3. an 3.3. Finally, λ ma N ˆV N ma N ˆΩ ˆΣ 0,K ˆΩ 0,K 0,K + N 0 N ˆΩ ˆΣ,K ˆΩ,K,K N ma ˆΩ N ˆΣ 0 0,K ˆΩ 0,K 0,K + N ma ˆΩ N ˆΣ,K ˆΩ,K,K N N 0 λ ma ˆΩ 0,K ˆΣ 0,K ˆΩ 0,K + N N λ ma ˆΩ,K ˆΣ,K ˆΩ,K which is boune in probability by equation B.9, Assumption.3 an Assumptions 3. an 3.3. Proof of Lemma A.6 For i note that we woul like to provie regressors up to a certain power, say n, incluing cross prouct terms. However, we woul like to relate the number of covariates,, an the uppermost power esire, n, to the number of terms in the series, K. We may o so by, n i + n + K n +. i n i0 [0]
12 Then note that, n + n +! n n!! n + n + n j + j n +. Thus, we have that K C n +, or equivalently that n C K / an so n s C s K s/ for some C, C R ++. Finally, by Lorentz 986, Theorem 8 we have that sup µw R K γ 0 O n s O K s/, as esire. The proofs of ii an iv may be foun in Imbens, Newey an Rier 006. For iii consier, ˆγK γk 0 N ˆΩ R Y w w N ˆΩ R R γ 0 w N R Yw R γ 0 w λ ma R Yw R γ 0. By Lemma A., the first factor is, λ ma λ ma ˆΩ / Ω / N w ˆΩ / + O p ζ K K / N / O + O p ζ K K / N /. B.30 The secon factor may be broken up into, N ˆΩ / R Yw R γ 0 w N ˆΩ / R Yw µ w X + µ w X R γ 0 w N ˆΩ / R ε w B.3 w + R µw X R γ 0. B.3 N w ˆΩ / []
13 For equation B.3 we have, E N ˆΩ / R ε w w [ E tr ε wr ˆΩ [ E tr N w N w tr σ E N w σ R ε w ] R R R R ε w ε w ] [ ] E R R R R E [ε w ε w X] N w K C KN, [ ] tr R R R R an so N w ˆΩ / R ε w O p K / N / B.33 by Markov s inequality. For equation B.3 we have, R µw X R γ 0 N w ˆΩ / µw X R γ 0 R R N R R µw X R γ 0 w µw X R γ 0 µw X R γ N 0 w sup µw R K γk 0 O K s/, by i, an so N ˆΩ / R w µw X R γ 0 O K s/ B.34 Note that the thir line of the penultimate isplay follows since R R R R is a projection matri. Combining equations B.30, B.33 an B.34 yiels, [ ˆγK γk 0 O + O p ζ K K / N /] [ O p K / N / + O K s/] O p K / N / + O K s/ + O p ζ K KN + O p ζ K K / K s/ N / O K s/ + O p ζ K KN + O p ζ K K / K s/ N /. However, by Assumption 3.3, we have that, O p ζ K K / K s/ N / o p K s/ o p an so ˆγ K γk 0 Op ζ K KN + O K s/, []
14 as esire. sup Finally, for v we have, µ w ˆµ sup The first term is O K s/ by i. sup µ 0 w ˆµ sup RK γ 0 ˆγ sup R K γ 0 ˆγ [ ζ K O p ζ K KN + O O p ζ K KN + O µw µ 0 + sup For the secon term we have, K s/] ζ K K s/, µ 0 w ˆµ. where we use the efinition of ζ K an the result from iv. Thus, sup µ w ˆµ O K s/ + O p ζ K KN + O ζ K K s/ O p ζ K KN + O ζ K K s/, as esire. Proof of Lemma A.7 Aitional Details Equation A.4 is / [ˆΩ ˆΣ ] / ˆΩ ˆΩ ε [ˆΩ R ˆΣ ] / ˆΩ ˆΩ R ε w ] / [ˆΣ / ˆΩ ˆΩ / R ε ε w / The first factor is, / [ˆΣ ] ˆΩ ˆΩ / / ˆΣ ˆΣ / λ ˆΩ/ ma ˆΣ / ˆΩ/ λ ma λ ma K / [λ ma Σ / + O p ζ K K 3/ N + O p ζ K K / K s/] [λ ma Ω / + O p ζkk / N /] K / λ ma Σ / λ ma Ω / K / + O p ζ K K N + O ζ K KK s/ O K / + O p ζ K K N + O p ζ K KK s/. [3]
15 For the secon factor we have, E ˆΩ R ε ε w / [ ε ] E tr N ε w R ˆΩ R ε ε w w [ ε E ε w R R ] R R ε ε w [ ε ] E ε w ε ε w B.35 [ µw E X R γ µw X R γ ] N w sup µw R K γ N w sup µw R K γ 0 + RK γ 0 R K γ N w O K s + O ζkk K s B.36 O N O ζkk K s Thus, ˆΩ / R ε ε w / N / w O p ζkk / K s/ N / by Markov s inequality. Equation B.35 follows by the fact that R R R R is a projection matri an equation B.36 follows from Lemma A.6 i an iv. Then equation A.4 is / [ˆΩ ˆΣ ] / ˆΩ ˆΩ ε [ˆΩ R ˆΣ ] / ˆΩ ˆΩ R ε w [ O K / + O p ζ K K N + O p ζ K KK s/] [ O p ζkk / K s/ N /] O p ζkkk s/ N / + O p ζk 3 K 5/ K s/ N / + O p ζk K 3/ K s/ N / O p ζk K 3/ K s/ N /, by Assumption 3.3. Net, equation A.5 is / [ˆΩ ˆΣ ] / [ˆΩ ] / ˆΩ ˆΩ R ε w Σ Ω ˆΩ R ε w ] / [ ] / [ˆΣ ˆΩ Σ Ω N / w ˆΩ / R ε w The first factor is ] / [ ] / [ˆΣ ˆΩ Σ Ω O p ζ K K N +O p ζ K KN / +O p ζ K KK s/ [4]
16 by Lemma B.. The secon factor is, E ˆΩ R ε w / N w [ ] E tr ε N wr ˆΩ R ε w w [ E tr ε ] wr R R R ε w [ ] E tr R R R R ε w ε w [ ] tr E R R R R E [ε w ε X] σ w ] tr E [R R R R [ σ w ] E tr R R R R [ R σ w ] E tr R R R σ w tr I K σ w K. Thus, the the secon factor is O p K / by Markov s inequality. Putting this together yiels, / [ˆΩ ˆΣ ] / [ˆΩ ] / ˆΩ ˆΩ R ε w Σ Ω ˆΩ R ε w [O p ζ K K N + O p ζ K KN / + O p ζ K KK s/] O p K / O p ζ K K 3/ N / + O p ζ K K 3/ K s/. Finally, equation A.6 is / [ˆΩ [ Σ Ω Σ Ω ] / ˆΩ / ] / [ ] / ˆΩ R ε w Ω Σ Ω Ω R ε w Ω / / The first factor is, [ / Σ Ω] C I O K /. R ε w [5]
17 The secon factor is O p ζ K K / N / by Lemma A.. E R ε w / N w [ E tr ] ε N wr R ε w w [ E tr ] R N ε w ε wr w tr E [ R N E [ε w ε ] w X] R w σ w tr E [ R R ] /N w σ w tr Ω σ w K λ ma Ω C K For the thir factor consier, Thus, the thir factor is O p K / by Assumption 3., Lemma A. ii an Markov s inequality. Putting this together yiels, / [ˆΩ ] / [ ] / Σ Ω ˆΩ R ε w Ω Σ Ω Ω R ε w O p ζ K K 3/ N /. Before proving Theorem 3/3. we nee the following lemma. Lemma B. We may partition ˆV an V analogously to the partition of ˆV P an V P in Section 3.4. ˆV00 ˆV0 ˆV ˆV 0 ˆV an V00 V V 0 V 0 V where ˆV 00 an V 00 are scalars, ˆV 0 an V 0 are K vectors, ˆV 0 an V 0 are K vectors an ˆV an V are K K matrices. Then, λ min [N V ] λ min [N V ], λ ma [N V ] λ ma [N V ] an if O p ζ K K 3/ N + O p ζ K K / K s/ o p, [ λ min N ˆV ] [ λ min N ˆV ] [, λ ma N ˆV ] [ λ ma N ˆV ] with probability approaching one. Proof The proof follows by the interlacing theorem, see, for eample, Li an Mathias 00: If A is an n n positive semi-efinite Hermitian matri with eigenvalues λ... λ n, B is a k k principal submatri of A with eigenvalues λ... λ k, then λ i λ i λ i+n k, i,..., k. [6]
18 In our case, N ˆV an N V are positive semi-efinite, symmetric an thus positive semi-efinite, Hermitian. So then, by the interlacing theorem λ min N V λ min N V λ ma [N V ] λ ma [N V ] an λ ma N V λ ma N V λ min [N V ] λ min [N V ]. Moreover, by Lemma A., N ˆV is nonsingular with probability approaching one so that again by the interlacing theorem we obtain λ min N ˆV λ min N ˆV [ λ ma N ˆV ] [ λ ma N ˆV ] an λ ma N ˆV λ ma N ˆV [ λ min N ˆV ] [ λ min N ˆV ]. Proof of Theorem 3.3 When the conitional average treatment effect is constant we may choose the two approimating sequences, γ0,k 0 an γ0,k, to iffer only by way of the first element the coefficient of the constant term in the approimating sequence. To simplify notation, efine ˆδ K ˆγ,K ˆγ 0,K, δ K γ,k γ 0,K. We may again follow the logic of Lemmas A.3, A.4, an A.5 to conclue that / T ˆδK δk ˆV ˆδK δk K K N 0, We nee only show that T T o p to complete the proof. First, note that γ w,k γw,k 0 γ w,k,i γw,k,i 0 i i γ w,k,i γw,k,i 0 + γ w0,k γw0,k 0 γ γ 0 O KK s/, B.37 by Lemma A.6 ii an ˆγw,K γw,k 0 ˆγw,K,i γw,k,i 0 i i ˆγw,K,i γw,k,i 0 + ˆγw0,K γw0,k 0 ˆγ γ 0 [ O p ζ K KN + O K s/]. B.38 [7]
19 by Lemma A.6 iii. We may choose the last K elements of the approimating sequence to be equal, γ,k 0 γ0 0,K. This allows us to boun, ˆδ K ˆγ,K ˆγ 0,K ˆγ,K γ,k 0 + γ0,k 0 ˆγ 0,K ˆγ,K γ,k 0 + γ 0 0,K ˆγ 0,K O p ζ K KN + O K s/ B.39 by equation B.38. Also, δk γ,k γ0,k γ,k γ,k 0 + γ0,k 0 γ0,k γ,k γ,k 0 + γ 0 0,K γ0,k O K / K s/ B.40 by equation B.37. T T Net note that, ˆδK δk First consier, δk ˆV δ K tr δk ˆV K δ N δ [ tr K N ˆV ] δ K [ N λ ma N ˆV ] δk / ˆV ˆδK δk ˆδ K ˆV ˆδ K K δk ˆV δ K ˆδ K ˆV K/ δ K B.4 [ N λ ma N ˆV ] δk B.4 N [O + O p ζ K K 3/ N / + O p ζ K K 3/ K s/] O KK s/ O p ζ K K 5/ K s/ N /, B.43 where equation B.4 follows by Lemma B. an equation B.43 follows from Lemma B. an the proof [8]
20 of Theorem 3.. Now consier, ˆδ K ˆV δ K tr ˆδ K ˆV K δ [ N ˆδ tr K N ˆV ] δ K N λ ma [ N ˆV ] ˆδK δ K N λ ma [ N ˆV ] ˆδK δ K B.44 N [O + O p ζ K K 3/ N / + O p ζ K K 3/ K s/] [ O p ζ K KN + O K s/] [ O K / K s/] O ζ K K K 3s/ N B.45 where equation B.44 follows by Lemma B. an equation B.45 follows from Lemma B. an the proof of Theorem 3.. Thus, we have that equation B.4 is, T T O K / [ O p ζ K K 5/ K s/ N / ] + O p ζ K K K 3s/ N O p ζ K K K s/ N / + O p ζ K K 3/ K 3s/ N which is o p uner Assumptions 3. an 3.3. Proof of Theorem 3.4 First, note that we may partition R K as R R K. R K Net, consier ρ N sup sup µ µ 0 τ X sup X sup X sup X RK γ,k 0 µ + sup RK γ0,k 0 µ 0 X RK ˆγ 0,K R K γ0,k 0 RK ˆγ,K R K γ,k 0 + sup X + sup + sup X R K ˆγ,K R K ˆγ 0,K + R ˆγ 0,K R ˆγ 00,K τ X RK γ0,k 0 µ 0 + sup RK γ,k 0 µ X + sup R K ˆγ0,K γ0,k 0 + sup R K ˆγ,K γ 0,K X X + sup R K ˆγ,K ˆγ 0,K + R ˆγ 0,K R ˆγ 00,K τ X RK γ0,k 0 µ 0 + sup RK γ,k 0 µ X +ζk ˆγ0,K γ0,k 0 + ζk ˆγ,K γ,k 0 + ζk ˆγ,K ˆγ 0,K + R ˆγ 0,K R ˆγ 00,K τ. [9]
21 Thus, ˆγ,K ˆγ 0,K ζk ρ N sup ζk sup RK γ0,k 0 µ 0 X X ζk sup RK γ,k 0 µ ˆγ0,K γ0,k 0 ˆγ,K γ,k 0 X ζk R ˆγ 0,K R ˆγ 00,K τ. We may follow the steps of the proof of Theorem 3. to obtain, for any M, Pr N / ζ K / K / ˆγ,K ˆγ 0,K > M. Net, we show that this implies that C ˆγ,K ˆγ 0,K V Pr ˆγ,K ˆγ 0,K K > M, K B.46 B.47 for an arbitrary constant C R ++. Denote λ min [N V ] an λ ma [N V ] by λ an λ, respectively an note that by Lemma A. an Lemma B. it follows that λ is boune away from zero an λ is boune. C ˆγ,K ˆγ 0,K V Pr ˆγ,K ˆγ 0,K K > M K C N ˆγ,K ˆγ 0,K [N V ] ˆγ,K ˆγ 0,K K Pr > M K Pr C N ˆγ,K ˆγ 0,K [N V ] ˆγ,K ˆγ 0,K > M K + K Pr λ C N ˆγ,K ˆγ 0,K ˆγ,K ˆγ 0,K > M K + K Pr Nζ K K ˆγ,K ˆγ 0,K ˆγ,K ˆγ 0,K > λ C [ ζ K K M ] K / + K / Pr N / ζ K / K / ˆγ,K ˆγ 0,K > λ C / ζ K M K K / + K / Since for any M, for large enough N, we have / λ C / ζ K M K K / + K / < λ C / it follows that this probability is for large N boune from below by the probability Pr N / ζ K / K / ˆγ,K ˆγ 0,K > λ C / which goes to one by B.46. To conclue we must show that this implies that PrT ˆγ,K ˆγ 0,K ˆV > M Pr ˆγ,K ˆγ 0,K K > M. K Let ˆλ λ min [N ˆV ] for simplicity of notation. Let A enote the event that ˆλ > λ / which satisfies Pr A as N by Lemmas A., A., an B. along with Assumptions 3. an 3.3. Also efine the event A, λ / N ˆγ,K ˆγ 0,K ˆγ,K ˆγ 0,K K K > M. [0]
22 Note that C ˆγ,K ˆγ 0,K V Pr ˆγ,K ˆγ 0,K K > M K C N ˆγ,K ˆγ 0,K [N V ] ˆγ,K ˆγ 0,K K Pr > M K λ C N ˆγ,K ˆγ 0,K ˆγ,K ˆγ 0,K K Pr > M K which goes to one as N by equation B.47. Since C was arbitrary we may choose C λ / λ an so Pr A as N. Thus, Pr A A as N. Finally, note that the event A A implies that T ˆγ,K ˆγ 0,K ˆV ˆγ,K ˆγ 0,K K K N ˆγ,K ˆγ 0,K [ N ˆV ] ˆγ,K ˆγ 0,K K K ˆλ N ˆγ,K ˆγ 0,K ˆγ,K ˆγ 0,K K K > λ / N ˆγ,K ˆγ 0,K ˆγ,K ˆγ 0,K K K > M. Hence PrT > M. Lemma B. Suppose Assumptions.-.3 an hol. Then i, ˆΣ ˆΩ Σ Ω O p ζ K K N + O p ζ K KN / + O p ζ K KK s/, an ii ˆΩ ˆΣ ˆΩ Ω Σ Ω O p ζ K K 3/ N / + O p ζ K K 3/ K s/. Proof For i note that, ˆΣ ˆΩ Σ Ω ˆΣ ˆΩ Σ ˆΩ + Σ ˆΣ Σ ˆΩ + Σ ˆΩ Ω ˆΩ Σ Ω B.48 B.49 First, consier equation B.48, ˆΣ Σ Op ζ K K 3/ N + O p ζ K K / K s/ by Lemma A.. Net, ˆΩ K / ˆΩ λ ma O K / + O p ζ K KN /. []
23 For equation B.49 we have that, Σ K / λ ma Σ O K /. Also, we have ˆΩ Ω O p ζ K K / N / B.50 by Lemma A.. Thus, ˆΣ ˆΩ Σ Ω [O p ζ K K 3/ N + O p ζ K K / K s/] [ O K / + O p ζ K KN /] [ + O K /] [ O p ζ K K / N /] O p ζ K K N + O p ζ K KN / + O p ζ K KK s/. For ii note that, ˆΩ ˆΩ ˆΩ + Ω ˆΣ ˆΩ Ω Σ Ω ˆΣ ˆΩ Ω ˆΣ ˆΩ + Ω ˆΣ ˆΩ Ω Σ Ω Ω ˆΣ ˆΩ ˆΣ ˆΩ Σ Ω B.5 B.5 For equation B.5 the first factor is O p ζ K K / N / by equation B.50 an the secon factor is ˆΣ ˆΩ λ ma ˆΣ ˆΩ λ ma ˆΣ λ ma ˆΩ K / λ ma Σ λ ma Ω K / + O p ζ K K N + O p ζ K KK s/ O K / + O p ζ K K N + O p ζ K KK s/. Thus, equation B.5 is ˆΩ Ω ˆΣ ˆΩ For equation B.5 the first factor is, Ω λ ma Ω I O K /. O p ζ K KN / + O p ζ K K 3/ K s/ N /. The secon factor is ˆΣ ˆΩ Σ Ω O p ζ K K N + O p ζ K KN / + O p ζ K KK s/ by i. Thus, equation B.5 is Ω ˆΣ ˆΩ Σ Ω O p ζ K K 3/ N / + O p ζ K K 3/ K s/ []
24 Finally, ˆΩ ˆΣ ˆΩ Ω Σ Ω O p ζ K KN / + O p ζ K K 3/ K s/ N / + O p ζ K K 3/ N / + O p ζ K K 3/ K s/ O p ζ K K 3/ N / + O p ζ K K 3/ K s/. [3]
25 Aitional References Imbens, Guio, Whitney Newey an Geert Rier, Mean-square-error Calculations for Average Treatment Effects, unpublishe manuscript, Department of Economics, Harvar University 006. Li, Chi-Kwong, an Roy Mathias, Interlacing Inequalities for Totally Nonnegative Matrices, Linear Algebra an its Applications 34 January 00, Lorentz, George, Approimation of Functions New York: Chelsea Publishing Company, 986. [4]
A Two-Sided Laplace Inversion Algorithm with Computable Error Bounds and Its Applications in Financial Engineering
Electronic Companion A Two-Sie Laplace Inversion Algorithm with Computable Error Bouns an Its Applications in Financial Engineering Ning Cai, S. G. Kou, Zongjian Liu HKUST an Columbia University Appenix
Διαβάστε περισσότεραw o = R 1 p. (1) R = p =. = 1
Πανεπιστήµιο Κρήτης - Τµήµα Επιστήµης Υπολογιστών ΗΥ-570: Στατιστική Επεξεργασία Σήµατος 205 ιδάσκων : Α. Μουχτάρης Τριτη Σειρά Ασκήσεων Λύσεις Ασκηση 3. 5.2 (a) From the Wiener-Hopf equation we have:
Διαβάστε περισσότερα2 Composition. Invertible Mappings
Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan Composition. Invertible Mappings In this section we discuss two procedures for creating new mappings from old ones, namely,
Διαβάστε περισσότεραSrednicki Chapter 55
Srednicki Chapter 55 QFT Problems & Solutions A. George August 3, 03 Srednicki 55.. Use equations 55.3-55.0 and A i, A j ] = Π i, Π j ] = 0 (at equal times) to verify equations 55.-55.3. This is our third
Διαβάστε περισσότεραST5224: Advanced Statistical Theory II
ST5224: Advanced Statistical Theory II 2014/2015: Semester II Tutorial 7 1. Let X be a sample from a population P and consider testing hypotheses H 0 : P = P 0 versus H 1 : P = P 1, where P j is a known
Διαβάστε περισσότεραSection 8.3 Trigonometric Equations
99 Section 8. Trigonometric Equations Objective 1: Solve Equations Involving One Trigonometric Function. In this section and the next, we will exple how to solving equations involving trigonometric functions.
Διαβάστε περισσότεραMatrices and Determinants
Matrices and Determinants SUBJECTIVE PROBLEMS: Q 1. For what value of k do the following system of equations possess a non-trivial (i.e., not all zero) solution over the set of rationals Q? x + ky + 3z
Διαβάστε περισσότεραEE512: Error Control Coding
EE512: Error Control Coding Solution for Assignment on Finite Fields February 16, 2007 1. (a) Addition and Multiplication tables for GF (5) and GF (7) are shown in Tables 1 and 2. + 0 1 2 3 4 0 0 1 2 3
Διαβάστε περισσότεραLecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3
Lecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3 1 State vector space and the dual space Space of wavefunctions The space of wavefunctions is the set of all
Διαβάστε περισσότεραHomework 3 Solutions
Homework 3 Solutions Igor Yanovsky (Math 151A TA) Problem 1: Compute the absolute error and relative error in approximations of p by p. (Use calculator!) a) p π, p 22/7; b) p π, p 3.141. Solution: For
Διαβάστε περισσότεραC.S. 430 Assignment 6, Sample Solutions
C.S. 430 Assignment 6, Sample Solutions Paul Liu November 15, 2007 Note that these are sample solutions only; in many cases there were many acceptable answers. 1 Reynolds Problem 10.1 1.1 Normal-order
Διαβάστε περισσότεραEvery set of first-order formulas is equivalent to an independent set
Every set of first-order formulas is equivalent to an independent set May 6, 2008 Abstract A set of first-order formulas, whatever the cardinality of the set of symbols, is equivalent to an independent
Διαβάστε περισσότεραMATH423 String Theory Solutions 4. = 0 τ = f(s). (1) dτ ds = dxµ dτ f (s) (2) dτ 2 [f (s)] 2 + dxµ. dτ f (s) (3)
1. MATH43 String Theory Solutions 4 x = 0 τ = fs). 1) = = f s) ) x = x [f s)] + f s) 3) equation of motion is x = 0 if an only if f s) = 0 i.e. fs) = As + B with A, B constants. i.e. allowe reparametrisations
Διαβάστε περισσότεραSpace-Time Symmetries
Chapter Space-Time Symmetries In classical fiel theory any continuous symmetry of the action generates a conserve current by Noether's proceure. If the Lagrangian is not invariant but only shifts by a
Διαβάστε περισσότεραFinite Field Problems: Solutions
Finite Field Problems: Solutions 1. Let f = x 2 +1 Z 11 [x] and let F = Z 11 [x]/(f), a field. Let Solution: F =11 2 = 121, so F = 121 1 = 120. The possible orders are the divisors of 120. Solution: The
Διαβάστε περισσότεραStatistical Inference I Locally most powerful tests
Statistical Inference I Locally most powerful tests Shirsendu Mukherjee Department of Statistics, Asutosh College, Kolkata, India. shirsendu st@yahoo.co.in So far we have treated the testing of one-sided
Διαβάστε περισσότεραCHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS
CHAPTER 5 SOLVING EQUATIONS BY ITERATIVE METHODS EXERCISE 104 Page 8 1. Find the positive root of the equation x + 3x 5 = 0, correct to 3 significant figures, using the method of bisection. Let f(x) =
Διαβάστε περισσότερα4.6 Autoregressive Moving Average Model ARMA(1,1)
84 CHAPTER 4. STATIONARY TS MODELS 4.6 Autoregressive Moving Average Model ARMA(,) This section is an introduction to a wide class of models ARMA(p,q) which we will consider in more detail later in this
Διαβάστε περισσότερα3.4 SUM AND DIFFERENCE FORMULAS. NOTE: cos(α+β) cos α + cos β cos(α-β) cos α -cos β
3.4 SUM AND DIFFERENCE FORMULAS Page Theorem cos(αβ cos α cos β -sin α cos(α-β cos α cos β sin α NOTE: cos(αβ cos α cos β cos(α-β cos α -cos β Proof of cos(α-β cos α cos β sin α Let s use a unit circle
Διαβάστε περισσότεραExample Sheet 3 Solutions
Example Sheet 3 Solutions. i Regular Sturm-Liouville. ii Singular Sturm-Liouville mixed boundary conditions. iii Not Sturm-Liouville ODE is not in Sturm-Liouville form. iv Regular Sturm-Liouville note
Διαβάστε περισσότεραOrdinal Arithmetic: Addition, Multiplication, Exponentiation and Limit
Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit Ting Zhang Stanford May 11, 2001 Stanford, 5/11/2001 1 Outline Ordinal Classification Ordinal Addition Ordinal Multiplication Ordinal
Διαβάστε περισσότερα6.3 Forecasting ARMA processes
122 CHAPTER 6. ARMA MODELS 6.3 Forecasting ARMA processes The purpose of forecasting is to predict future values of a TS based on the data collected to the present. In this section we will discuss a linear
Διαβάστε περισσότεραA Note on Intuitionistic Fuzzy. Equivalence Relation
International Mathematical Forum, 5, 2010, no. 67, 3301-3307 A Note on Intuitionistic Fuzzy Equivalence Relation D. K. Basnet Dept. of Mathematics, Assam University Silchar-788011, Assam, India dkbasnet@rediffmail.com
Διαβάστε περισσότεραNowhere-zero flows Let be a digraph, Abelian group. A Γ-circulation in is a mapping : such that, where, and : tail in X, head in
Nowhere-zero flows Let be a digraph, Abelian group. A Γ-circulation in is a mapping : such that, where, and : tail in X, head in : tail in X, head in A nowhere-zero Γ-flow is a Γ-circulation such that
Διαβάστε περισσότεραTridiagonal matrices. Gérard MEURANT. October, 2008
Tridiagonal matrices Gérard MEURANT October, 2008 1 Similarity 2 Cholesy factorizations 3 Eigenvalues 4 Inverse Similarity Let α 1 ω 1 β 1 α 2 ω 2 T =......... β 2 α 1 ω 1 β 1 α and β i ω i, i = 1,...,
Διαβάστε περισσότεραMath221: HW# 1 solutions
Math: HW# solutions Andy Royston October, 5 7.5.7, 3 rd Ed. We have a n = b n = a = fxdx = xdx =, x cos nxdx = x sin nx n sin nxdx n = cos nx n = n n, x sin nxdx = x cos nx n + cos nxdx n cos n = + sin
Διαβάστε περισσότεραSection 7.6 Double and Half Angle Formulas
09 Section 7. Double and Half Angle Fmulas To derive the double-angles fmulas, we will use the sum of two angles fmulas that we developed in the last section. We will let α θ and β θ: cos(θ) cos(θ + θ)
Διαβάστε περισσότερα6.1. Dirac Equation. Hamiltonian. Dirac Eq.
6.1. Dirac Equation Ref: M.Kaku, Quantum Field Theory, Oxford Univ Press (1993) η μν = η μν = diag(1, -1, -1, -1) p 0 = p 0 p = p i = -p i p μ p μ = p 0 p 0 + p i p i = E c 2 - p 2 = (m c) 2 H = c p 2
Διαβάστε περισσότεραSCHOOL OF MATHEMATICAL SCIENCES G11LMA Linear Mathematics Examination Solutions
SCHOOL OF MATHEMATICAL SCIENCES GLMA Linear Mathematics 00- Examination Solutions. (a) i. ( + 5i)( i) = (6 + 5) + (5 )i = + i. Real part is, imaginary part is. (b) ii. + 5i i ( + 5i)( + i) = ( i)( + i)
Διαβάστε περισσότεραk A = [k, k]( )[a 1, a 2 ] = [ka 1,ka 2 ] 4For the division of two intervals of confidence in R +
Chapter 3. Fuzzy Arithmetic 3- Fuzzy arithmetic: ~Addition(+) and subtraction (-): Let A = [a and B = [b, b in R If x [a and y [b, b than x+y [a +b +b Symbolically,we write A(+)B = [a (+)[b, b = [a +b
Διαβάστε περισσότεραUniform Convergence of Fourier Series Michael Taylor
Uniform Convergence of Fourier Series Michael Taylor Given f L 1 T 1 ), we consider the partial sums of the Fourier series of f: N 1) S N fθ) = ˆfk)e ikθ. k= N A calculation gives the Dirichlet formula
Διαβάστε περισσότεραF19MC2 Solutions 9 Complex Analysis
F9MC Solutions 9 Complex Analysis. (i) Let f(z) = eaz +z. Then f is ifferentiable except at z = ±i an so by Cauchy s Resiue Theorem e az z = πi[res(f,i)+res(f, i)]. +z C(,) Since + has zeros of orer at
Διαβάστε περισσότεραPhys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required)
Phys460.nb 81 ψ n (t) is still the (same) eigenstate of H But for tdependent H. The answer is NO. 5.5.5. Solution for the tdependent Schrodinger s equation If we assume that at time t 0, the electron starts
Διαβάστε περισσότεραReminders: linear functions
Reminders: linear functions Let U and V be vector spaces over the same field F. Definition A function f : U V is linear if for every u 1, u 2 U, f (u 1 + u 2 ) = f (u 1 ) + f (u 2 ), and for every u U
Διαβάστε περισσότεραChapter 6: Systems of Linear Differential. be continuous functions on the interval
Chapter 6: Systems of Linear Differential Equations Let a (t), a 2 (t),..., a nn (t), b (t), b 2 (t),..., b n (t) be continuous functions on the interval I. The system of n first-order differential equations
Διαβάστε περισσότεραderivation of the Laplacian from rectangular to spherical coordinates
derivation of the Laplacian from rectangular to spherical coordinates swapnizzle 03-03- :5:43 We begin by recognizing the familiar conversion from rectangular to spherical coordinates (note that φ is used
Διαβάστε περισσότεραANSWERSHEET (TOPIC = DIFFERENTIAL CALCULUS) COLLECTION #2. h 0 h h 0 h h 0 ( ) g k = g 0 + g 1 + g g 2009 =?
Teko Classes IITJEE/AIEEE Maths by SUHAAG SIR, Bhopal, Ph (0755) 3 00 000 www.tekoclasses.com ANSWERSHEET (TOPIC DIFFERENTIAL CALCULUS) COLLECTION # Question Type A.Single Correct Type Q. (A) Sol least
Διαβάστε περισσότεραLecture 21: Properties and robustness of LSE
Lecture 21: Properties and robustness of LSE BLUE: Robustness of LSE against normality We now study properties of l τ β and σ 2 under assumption A2, i.e., without the normality assumption on ε. From Theorem
Διαβάστε περισσότεραSolutions to Exercise Sheet 5
Solutions to Eercise Sheet 5 jacques@ucsd.edu. Let X and Y be random variables with joint pdf f(, y) = 3y( + y) where and y. Determine each of the following probabilities. Solutions. a. P (X ). b. P (X
Διαβάστε περισσότεραMock Exam 7. 1 Hong Kong Educational Publishing Company. Section A 1. Reference: HKDSE Math M Q2 (a) (1 + kx) n 1M + 1A = (1) =
Mock Eam 7 Mock Eam 7 Section A. Reference: HKDSE Math M 0 Q (a) ( + k) n nn ( )( k) + nk ( ) + + nn ( ) k + nk + + + A nk... () nn ( ) k... () From (), k...() n Substituting () into (), nn ( ) n 76n 76n
Διαβάστε περισσότεραPractice Exam 2. Conceptual Questions. 1. State a Basic identity and then verify it. (a) Identity: Solution: One identity is csc(θ) = 1
Conceptual Questions. State a Basic identity and then verify it. a) Identity: Solution: One identity is cscθ) = sinθ) Practice Exam b) Verification: Solution: Given the point of intersection x, y) of the
Διαβάστε περισσότεραCoefficient Inequalities for a New Subclass of K-uniformly Convex Functions
International Journal of Computational Science and Mathematics. ISSN 0974-89 Volume, Number (00), pp. 67--75 International Research Publication House http://www.irphouse.com Coefficient Inequalities for
Διαβάστε περισσότεραProblem Set 3: Solutions
CMPSCI 69GG Applied Information Theory Fall 006 Problem Set 3: Solutions. [Cover and Thomas 7.] a Define the following notation, C I p xx; Y max X; Y C I p xx; Ỹ max I X; Ỹ We would like to show that C
Διαβάστε περισσότεραSCITECH Volume 13, Issue 2 RESEARCH ORGANISATION Published online: March 29, 2018
Journal of rogressive Research in Mathematics(JRM) ISSN: 2395-028 SCITECH Volume 3, Issue 2 RESEARCH ORGANISATION ublished online: March 29, 208 Journal of rogressive Research in Mathematics www.scitecresearch.com/journals
Διαβάστε περισσότεραPg The perimeter is P = 3x The area of a triangle is. where b is the base, h is the height. In our case b = x, then the area is
Pg. 9. The perimeter is P = The area of a triangle is A = bh where b is the base, h is the height 0 h= btan 60 = b = b In our case b =, then the area is A = = 0. By Pythagorean theorem a + a = d a a =
Διαβάστε περισσότεραCongruence Classes of Invertible Matrices of Order 3 over F 2
International Journal of Algebra, Vol. 8, 24, no. 5, 239-246 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/.2988/ija.24.422 Congruence Classes of Invertible Matrices of Order 3 over F 2 Ligong An and
Διαβάστε περισσότεραMINIMAL CLOSED SETS AND MAXIMAL CLOSED SETS
MINIMAL CLOSED SETS AND MAXIMAL CLOSED SETS FUMIE NAKAOKA AND NOBUYUKI ODA Received 20 December 2005; Revised 28 May 2006; Accepted 6 August 2006 Some properties of minimal closed sets and maximal closed
Διαβάστε περισσότεραORDINAL ARITHMETIC JULIAN J. SCHLÖDER
ORDINAL ARITHMETIC JULIAN J. SCHLÖDER Abstract. We define ordinal arithmetic and show laws of Left- Monotonicity, Associativity, Distributivity, some minor related properties and the Cantor Normal Form.
Διαβάστε περισσότεραHOMEWORK 4 = G. In order to plot the stress versus the stretch we define a normalized stretch:
HOMEWORK 4 Problem a For the fast loading case, we want to derive the relationship between P zz and λ z. We know that the nominal stress is expressed as: P zz = ψ λ z where λ z = λ λ z. Therefore, applying
Διαβάστε περισσότεραExercises 10. Find a fundamental matrix of the given system of equations. Also find the fundamental matrix Φ(t) satisfying Φ(0) = I. 1.
Exercises 0 More exercises are available in Elementary Differential Equations. If you have a problem to solve any of them, feel free to come to office hour. Problem Find a fundamental matrix of the given
Διαβάστε περισσότεραSection 9.2 Polar Equations and Graphs
180 Section 9. Polar Equations and Graphs In this section, we will be graphing polar equations on a polar grid. In the first few examples, we will write the polar equation in rectangular form to help identify
Διαβάστε περισσότεραAreas and Lengths in Polar Coordinates
Kiryl Tsishchanka Areas and Lengths in Polar Coordinates In this section we develop the formula for the area of a region whose boundary is given by a polar equation. We need to use the formula for the
Διαβάστε περισσότεραLecture 2. Soundness and completeness of propositional logic
Lecture 2 Soundness and completeness of propositional logic February 9, 2004 1 Overview Review of natural deduction. Soundness and completeness. Semantics of propositional formulas. Soundness proof. Completeness
Διαβάστε περισσότερα5. Choice under Uncertainty
5. Choice under Uncertainty Daisuke Oyama Microeconomics I May 23, 2018 Formulations von Neumann-Morgenstern (1944/1947) X: Set of prizes Π: Set of probability distributions on X : Preference relation
Διαβάστε περισσότεραforms This gives Remark 1. How to remember the above formulas: Substituting these into the equation we obtain with
Week 03: C lassification of S econd- Order L inear Equations In last week s lectures we have illustrated how to obtain the general solutions of first order PDEs using the method of characteristics. We
Διαβάστε περισσότεραAreas and Lengths in Polar Coordinates
Kiryl Tsishchanka Areas and Lengths in Polar Coordinates In this section we develop the formula for the area of a region whose boundary is given by a polar equation. We need to use the formula for the
Διαβάστε περισσότεραMath 446 Homework 3 Solutions. (1). (i): Reverse triangle inequality for metrics: Let (X, d) be a metric space and let x, y, z X.
Math 446 Homework 3 Solutions. (1). (i): Reverse triangle inequalit for metrics: Let (X, d) be a metric space and let x,, z X. Prove that d(x, z) d(, z) d(x, ). (ii): Reverse triangle inequalit for norms:
Διαβάστε περισσότεραSupplementary Material For Testing Homogeneity of. High-dimensional Covariance Matrices
Supplementary Material For Testing Homogeneity of High-dimensional Covariance Matrices Shurong Zheng, Ruitao Lin, Jianhua Guo, and Guosheng Yin 3 School of Mathematics & Statistics and KLAS, Northeast
Διαβάστε περισσότεραInverse trigonometric functions & General Solution of Trigonometric Equations. ------------------ ----------------------------- -----------------
Inverse trigonometric functions & General Solution of Trigonometric Equations. 1. Sin ( ) = a) b) c) d) Ans b. Solution : Method 1. Ans a: 17 > 1 a) is rejected. w.k.t Sin ( sin ) = d is rejected. If sin
Διαβάστε περισσότεραOther Test Constructions: Likelihood Ratio & Bayes Tests
Other Test Constructions: Likelihood Ratio & Bayes Tests Side-Note: So far we have seen a few approaches for creating tests such as Neyman-Pearson Lemma ( most powerful tests of H 0 : θ = θ 0 vs H 1 :
Διαβάστε περισσότεραPartial Differential Equations in Biology The boundary element method. March 26, 2013
The boundary element method March 26, 203 Introduction and notation The problem: u = f in D R d u = ϕ in Γ D u n = g on Γ N, where D = Γ D Γ N, Γ D Γ N = (possibly, Γ D = [Neumann problem] or Γ N = [Dirichlet
Διαβάστε περισσότερα557: MATHEMATICAL STATISTICS II RESULTS FROM CLASSICAL HYPOTHESIS TESTING
Most Powerful Tests 557: MATHEMATICAL STATISTICS II RESULTS FROM CLASSICAL HYPOTHESIS TESTING To construct and assess the quality of a statistical test, we consider the power function β(θ). Consider a
Διαβάστε περισσότεραCRASH COURSE IN PRECALCULUS
CRASH COURSE IN PRECALCULUS Shiah-Sen Wang The graphs are prepared by Chien-Lun Lai Based on : Precalculus: Mathematics for Calculus by J. Stuwart, L. Redin & S. Watson, 6th edition, 01, Brooks/Cole Chapter
Διαβάστε περισσότεραAppendix to On the stability of a compressible axisymmetric rotating flow in a pipe. By Z. Rusak & J. H. Lee
Appendi to On the stability of a compressible aisymmetric rotating flow in a pipe By Z. Rusak & J. H. Lee Journal of Fluid Mechanics, vol. 5 4, pp. 5 4 This material has not been copy-edited or typeset
Διαβάστε περισσότεραNumerical Analysis FMN011
Numerical Analysis FMN011 Carmen Arévalo Lund University carmen@maths.lth.se Lecture 12 Periodic data A function g has period P if g(x + P ) = g(x) Model: Trigonometric polynomial of order M T M (x) =
Διαβάστε περισσότεραΑπόκριση σε Μοναδιαία Ωστική Δύναμη (Unit Impulse) Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο. Απόστολος Σ.
Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο The time integral of a force is referred to as impulse, is determined by and is obtained from: Newton s 2 nd Law of motion states that the action
Διαβάστε περισσότεραSOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM
SOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM Solutions to Question 1 a) The cumulative distribution function of T conditional on N n is Pr T t N n) Pr max X 1,..., X N ) t N n) Pr max
Διαβάστε περισσότεραb. Use the parametrization from (a) to compute the area of S a as S a ds. Be sure to substitute for ds!
MTH U341 urface Integrals, tokes theorem, the divergence theorem To be turned in Wed., Dec. 1. 1. Let be the sphere of radius a, x 2 + y 2 + z 2 a 2. a. Use spherical coordinates (with ρ a) to parametrize.
Διαβάστε περισσότεραAppendix to Technology Transfer and Spillovers in International Joint Ventures
Appendix to Technology Transfer and Spillovers in International Joint Ventures Thomas Müller Monika Schnitzer Published in Journal of International Economics, 2006, Volume 68, 456-468 Bundestanstalt für
Διαβάστε περισσότεραMath 6 SL Probability Distributions Practice Test Mark Scheme
Math 6 SL Probability Distributions Practice Test Mark Scheme. (a) Note: Award A for vertical line to right of mean, A for shading to right of their vertical line. AA N (b) evidence of recognizing symmetry
Διαβάστε περισσότεραEcon 2110: Fall 2008 Suggested Solutions to Problem Set 8 questions or comments to Dan Fetter 1
Eon : Fall 8 Suggested Solutions to Problem Set 8 Email questions or omments to Dan Fetter Problem. Let X be a salar with density f(x, θ) (θx + θ) [ x ] with θ. (a) Find the most powerful level α test
Διαβάστε περισσότεραCHAPTER 48 APPLICATIONS OF MATRICES AND DETERMINANTS
CHAPTER 48 APPLICATIONS OF MATRICES AND DETERMINANTS EXERCISE 01 Page 545 1. Use matrices to solve: 3x + 4y x + 5y + 7 3x + 4y x + 5y 7 Hence, 3 4 x 0 5 y 7 The inverse of 3 4 5 is: 1 5 4 1 5 4 15 8 3
Διαβάστε περισσότεραFourier Series. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics
Fourier Series MATH 211, Calculus II J. Robert Buchanan Department of Mathematics Spring 2018 Introduction Not all functions can be represented by Taylor series. f (k) (c) A Taylor series f (x) = (x c)
Διαβάστε περισσότεραRisk! " #$%&'() *!'+,'''## -. / # $
Risk! " #$%&'(!'+,'''## -. / 0! " # $ +/ #%&''&(+(( &'',$ #-&''&$ #(./0&'',$( ( (! #( &''/$ #$ 3 #4&'',$ #- &'',$ #5&''6(&''&7&'',$ / ( /8 9 :&' " 4; < # $ 3 " ( #$ = = #$ #$ ( 3 - > # $ 3 = = " 3 3, 6?3
Διαβάστε περισσότεραOnline Appendix I. 1 1+r ]}, Bψ = {ψ : Y E A S S}, B W = +(1 s)[1 m (1,0) (b, e, a, ψ (0,a ) (e, a, s); q, ψ, W )]}, (29) exp( U(d,a ) (i, x; q)
Online Appendix I Appendix D Additional Existence Proofs Denote B q = {q : A E A S [0, +r ]}, Bψ = {ψ : Y E A S S}, B W = {W : I E A S R}. I slightly abuse the notation by defining B q (L q ) the subset
Διαβάστε περισσότεραA General Note on δ-quasi Monotone and Increasing Sequence
International Mathematical Forum, 4, 2009, no. 3, 143-149 A General Note on δ-quasi Monotone and Increasing Sequence Santosh Kr. Saxena H. N. 419, Jawaharpuri, Badaun, U.P., India Presently working in
Διαβάστε περισσότεραPARTIAL NOTES for 6.1 Trigonometric Identities
PARTIAL NOTES for 6.1 Trigonometric Identities tanθ = sinθ cosθ cotθ = cosθ sinθ BASIC IDENTITIES cscθ = 1 sinθ secθ = 1 cosθ cotθ = 1 tanθ PYTHAGOREAN IDENTITIES sin θ + cos θ =1 tan θ +1= sec θ 1 + cot
Διαβάστε περισσότεραOptimal Parameter in Hermitian and Skew-Hermitian Splitting Method for Certain Two-by-Two Block Matrices
Optimal Parameter in Hermitian and Skew-Hermitian Splitting Method for Certain Two-by-Two Block Matrices Chi-Kwong Li Department of Mathematics The College of William and Mary Williamsburg, Virginia 23187-8795
Διαβάστε περισσότεραSecond Order Partial Differential Equations
Chapter 7 Second Order Partial Differential Equations 7.1 Introduction A second order linear PDE in two independent variables (x, y Ω can be written as A(x, y u x + B(x, y u xy + C(x, y u u u + D(x, y
Διαβάστε περισσότεραSolution Series 9. i=1 x i and i=1 x i.
Lecturer: Prof. Dr. Mete SONER Coordinator: Yilin WANG Solution Series 9 Q1. Let α, β >, the p.d.f. of a beta distribution with parameters α and β is { Γ(α+β) Γ(α)Γ(β) f(x α, β) xα 1 (1 x) β 1 for < x
Διαβάστε περισσότεραBayesian statistics. DS GA 1002 Probability and Statistics for Data Science.
Bayesian statistics DS GA 1002 Probability and Statistics for Data Science http://www.cims.nyu.edu/~cfgranda/pages/dsga1002_fall17 Carlos Fernandez-Granda Frequentist vs Bayesian statistics In frequentist
Διαβάστε περισσότεραFractional Colorings and Zykov Products of graphs
Fractional Colorings and Zykov Products of graphs Who? Nichole Schimanski When? July 27, 2011 Graphs A graph, G, consists of a vertex set, V (G), and an edge set, E(G). V (G) is any finite set E(G) is
Διαβάστε περισσότεραSOME PROPERTIES OF FUZZY REAL NUMBERS
Sahand Communications in Mathematical Analysis (SCMA) Vol. 3 No. 1 (2016), 21-27 http://scma.maragheh.ac.ir SOME PROPERTIES OF FUZZY REAL NUMBERS BAYAZ DARABY 1 AND JAVAD JAFARI 2 Abstract. In the mathematical
Διαβάστε περισσότεραA Bonus-Malus System as a Markov Set-Chain. Małgorzata Niemiec Warsaw School of Economics Institute of Econometrics
A Bonus-Malus System as a Markov Set-Chain Małgorzata Niemiec Warsaw School of Economics Institute of Econometrics Contents 1. Markov set-chain 2. Model of bonus-malus system 3. Example 4. Conclusions
Διαβάστε περισσότεραChapter 6: Systems of Linear Differential. be continuous functions on the interval
Chapter 6: Systems of Linear Differential Equations Let a (t), a 2 (t),..., a nn (t), b (t), b 2 (t),..., b n (t) be continuous functions on the interval I. The system of n first-order differential equations
Διαβάστε περισσότεραPartial Trace and Partial Transpose
Partial Trace and Partial Transpose by José Luis Gómez-Muñoz http://homepage.cem.itesm.mx/lgomez/quantum/ jose.luis.gomez@itesm.mx This document is based on suggestions by Anirban Das Introduction This
Διαβάστε περισσότεραJesse Maassen and Mark Lundstrom Purdue University November 25, 2013
Notes on Average Scattering imes and Hall Factors Jesse Maassen and Mar Lundstrom Purdue University November 5, 13 I. Introduction 1 II. Solution of the BE 1 III. Exercises: Woring out average scattering
Διαβάστε περισσότεραDerivation of Optical-Bloch Equations
Appendix C Derivation of Optical-Bloch Equations In this appendix the optical-bloch equations that give the populations and coherences for an idealized three-level Λ system, Fig. 3. on page 47, will be
Διαβάστε περισσότεραHomework 8 Model Solution Section
MATH 004 Homework Solution Homework 8 Model Solution Section 14.5 14.6. 14.5. Use the Chain Rule to find dz where z cosx + 4y), x 5t 4, y 1 t. dz dx + dy y sinx + 4y)0t + 4) sinx + 4y) 1t ) 0t + 4t ) sinx
Διαβάστε περισσότεραDESIGN OF MACHINERY SOLUTION MANUAL h in h 4 0.
DESIGN OF MACHINERY SOLUTION MANUAL -7-1! PROBLEM -7 Statement: Design a double-dwell cam to move a follower from to 25 6, dwell for 12, fall 25 and dwell for the remader The total cycle must take 4 sec
Διαβάστε περισσότεραMath 248 Homework 1. Edward Burkard. Exercise 1. Prove the following Fourier Transforms where a > 0 and c R: f (x) = b. f(x c) = e.
Math 48 Homework Ewar Burkar Exercise. Prove the following Fourier Transforms where a > an c : a. f(x) f(ξ) b. f(x c) e πicξ f(ξ) c. eπixc f(x) f(ξ c). f(ax) f(ξ) a e. f (x) πiξ f(ξ) f. xf(x) f(ξ) πi ξ
Διαβάστε περισσότεραJordan Journal of Mathematics and Statistics (JJMS) 4(2), 2011, pp
Jordan Journal of Mathematics and Statistics (JJMS) 4(2), 2011, pp.115-126. α, β, γ ORTHOGONALITY ABDALLA TALLAFHA Abstract. Orthogonality in inner product spaces can be expresed using the notion of norms.
Διαβάστε περισσότεραLecture 13 - Root Space Decomposition II
Lecture 13 - Root Space Decomposition II October 18, 2012 1 Review First let us recall the situation. Let g be a simple algebra, with maximal toral subalgebra h (which we are calling a CSA, or Cartan Subalgebra).
Διαβάστε περισσότεραProblem Set 9 Solutions. θ + 1. θ 2 + cotθ ( ) sinθ e iφ is an eigenfunction of the ˆ L 2 operator. / θ 2. φ 2. sin 2 θ φ 2. ( ) = e iφ. = e iφ cosθ.
Chemistry 362 Dr Jean M Standard Problem Set 9 Solutions The ˆ L 2 operator is defined as Verify that the angular wavefunction Y θ,φ) Also verify that the eigenvalue is given by 2! 2 & L ˆ 2! 2 2 θ 2 +
Διαβάστε περισσότεραConcrete Mathematics Exercises from 30 September 2016
Concrete Mathematics Exercises from 30 September 2016 Silvio Capobianco Exercise 1.7 Let H(n) = J(n + 1) J(n). Equation (1.8) tells us that H(2n) = 2, and H(2n+1) = J(2n+2) J(2n+1) = (2J(n+1) 1) (2J(n)+1)
Διαβάστε περισσότεραCHAPTER 101 FOURIER SERIES FOR PERIODIC FUNCTIONS OF PERIOD
CHAPTER FOURIER SERIES FOR PERIODIC FUNCTIONS OF PERIOD EXERCISE 36 Page 66. Determine the Fourier series for the periodic function: f(x), when x +, when x which is periodic outside this rge of period.
Διαβάστε περισσότερα557: MATHEMATICAL STATISTICS II HYPOTHESIS TESTING
557: MATHEMATICAL STATISTICS II HYPOTHESIS TESTING A statistical hypothesis test is a decision rule that takes as an input observed sample data and returns an action relating to two mutually exclusive
Διαβάστε περισσότερα( ) 2 and compare to M.
Problems and Solutions for Section 4.2 4.9 through 4.33) 4.9 Calculate the square root of the matrix 3!0 M!0 8 Hint: Let M / 2 a!b ; calculate M / 2!b c ) 2 and compare to M. Solution: Given: 3!0 M!0 8
Διαβάστε περισσότερα2. THEORY OF EQUATIONS. PREVIOUS EAMCET Bits.
EAMCET-. THEORY OF EQUATIONS PREVIOUS EAMCET Bits. Each of the roots of the equation x 6x + 6x 5= are increased by k so that the new transformed equation does not contain term. Then k =... - 4. - Sol.
Διαβάστε περισσότεραExample of the Baum-Welch Algorithm
Example of the Baum-Welch Algorithm Larry Moss Q520, Spring 2008 1 Our corpus c We start with a very simple corpus. We take the set Y of unanalyzed words to be {ABBA, BAB}, and c to be given by c(abba)
Διαβάστε περισσότερα