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- ≤ Ks. (31) Set Fn i = F(i−1)∆n . By (4.10) in Jacod and Rosenbaum (2013) we have, E αn i q Fn i ≤ Kq∆q n for all q ≥ 0 and E kn−1 X j=0 αn i+j q Fn i ≤ Kq∆q nkq/2 n whenever q ≥ 2. (32) Combining (40), (38), (39) with Z = c and the H older inequality yields for q ≥ 2, E βn i q Fn i ≤ Kq∆q/4 , and E γn i q Fn i
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- ≤ Ks. (39) Set Fn i = F(i−1)∆n . By (4.10) in Jacod and Rosenbaum (2013) we have, E αn i q Fn i ≤ Kq∆q n for all q ≥ 0 and E kn−1 X j=0 αn i+j q Fn i ≤ Kq∆q nkq/2 n whenever q ≥ 2. (40) Combining (40), (38), (39) with Z = c and the H older inequality yields for q ≥ 2, E βn i q Fn i ≤ Kq∆q/4 , and E γn i q Fn i ≤ Kq∆q/4 . (41) For any c` adl` ag bounded process Z, we set ηt,s(Z) = s E sup 0<u≤s kZt+u − Ztk2|Fn i , ηn i,j(Z) = s E sup 0≤u≤j∆n kZ(i−1)∆n+u − Z(i−1)∆n k2|Fn i . Lemma 6. For any c` adl` ag bounded process Z, for all t, s > 0, j, k ≥ 0, set ηt,s = ηt,s(Z), then we have: ∆nE [t/∆n] X i=1 ηi,kn −→ 0, ∆nE [t/∆n] X i=1 ηi,2kn −→ 0, E ηi+j,k|Fn i ≤ ηi,j+k and ∆nE [t/∆n] X i=1 ηi,4kn −→ 0.
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- EXC Figure 2: Monthly contribution of the idiosyncratic volatility to the total volatility (1- b R2 Y j) over the period 2003:2012. The dotted blue line plots this measure calculated in CAPM model. The solid red line plots the same measure obtained in the FF3 model. We use the ticker of the stocks to label the graphs.
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- i+j|Fn i−1) = 0, where the first inequality is a consequence of E(kξ 0 n i+jkq |Fn i−1) ≤ E(kξn i+jkq |Fn i−1) ≤ LqLq which can be proved using the Jensen inequality and the law of iterated expectation. Hence applying Lemma B.2 of A ıt-Sahalia and Jacod (2014) we have E(k 2kn−1 X j=1 ξ 00 n i+jkq |Fn i−1) ≤ KqLq Lqkq/2 n . To see the latter, we first prove that the required condition E(kξn i kq |Fn i−1) ≤ LqLq ) in the Lemma B.2 of A
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- IBM Figure 1: Monthly contribution of the idiosyncratic volatility to the total volatility (1- b R2 Y j) over the period 2003:2012. The dotted blue line plots this measure calculated in CAPM model. The solid red line plots the same measure obtained in the FF3 model. We use the ticker of the stocks to label the graphs. 2003 2005 2007 2010 2012 0 0.5 1
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- In this section, we set for convenience, cn i−1 = c(i−1)∆n and Fi = F(i−1)∆n . Given the polynomial growth assumption satisfied by H and G and the fact that kn = θ(∆n)−1/2 , by Theorem 2.2 in Jacod and Rosenbaum (2012) we have 1 √ ∆n \ [H(c), G(c)] (A2) T − 3 θ2 d X g,h,a,b=1 Z T 0 ∂ghH∂abG (ct)(cga t chb t + cgb i cha t )dt ! = Op(1), which yields 1 ∆ 1/4 n \ [H(c), G(c)] (A2) T − 3 θ2 d X g,h,a,b=1 Z T 0 ∂ghH∂abG (ct)(cga t chb t + cgb i cha t )dt ! P −→ 0.
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- It is easy to see that the ξ(H, gh, u; G, ab, v)n i are martingale increments, relative to the discrete filtration (Fn i ). Therefore, by Theorem 2.2.15 of Jacod and Protter (2012), to obtain the joint asymptotic distribution of 1 ∆ 1/4 n Z(H, gh, u; G, ab, v)n T , it is enough to prove the following three properties, for all t > 0, all (H, gh, u; G, ab, v), (H0 , g0 h0 , u0 ; G0 , a0 b0 , v0 ) and all martingales N which are either bounded and orthogonal to W, or equal to one component Wj , A (H, gh, u; G, ab, v), (H0 , g0 h0 , u0 ; G0 , a0 b0 , v0 ) n t := [t/∆n] X i=2kn E(ξ(H, gh, u; G, ab, v)n i ξ(H0 , g0 h0 , u0 ; G0 , a0 b0 , v0 )n i |Fn i−1) P =⇒ A (H, gh, u; G, ab, v), (H0 , g0 h0 , u0 ; G0 , a0 b0 , v0 ) t [t/∆n] X i=2kn E(|ξ(H, gh, u; G, ab, v)n i |4 |Fn i−1) P =⇒ 0 B(N; H, gh, u; G, ab, v)n t := [t/∆n] X i=2kn E(ξ(H, gh, u; G, ab, v)n i ∆n i N|Fn i−1) P =⇒ 0.
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- We introduce some new notations. Following Jacod and Rosenbaum (2012), we define αn i = (∆n i Y 0 )(∆n i Y 0 )> − cn i ∆n, βn i = b c 0 n i − cn i , and γn i = b c 0 n i+kn − b c 0 n i , (37) which satisfy βn i = 1 kn∆n kn−1 X j=0 (αn i+j + (cn i+j − cn i )∆n) and γn i = βi+kn − βn i + ∆n(cn i+kn − cn i ). (38) We recall some well-known results. For any continuous Itˆ o process Zt, we have E sup w∈[0,s] Zt+w − Zt q Ft ≤ Kqsq/2 , and E Zt+s − Zt Ft
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- We make use of lemma B.8 in A ıt-Sahalia and Jacod (2014) to prove that b E0 (1) P ⇒ 0. To this end, we write b E0 (1) = 1 ∆ 1/4 n " [T/∆n]−2kn+1 X i=1 Θ(u, v) (c),i−1+2kn,n 0 V(i−1)∆n # .
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