A direct approach to risk approximation for vast portfolios under gross-exposure constraint using high-frequency data

Abstract

It is well known that the traditional estimated risk for the Markowitz mean-variance optimization had been demonstrated to seriously depart from its theoretic optimal risk due to accumulation of input estimation errors. Fan et al. (in J. Am. Stat. Assoc. 107:592–606, 2012a) addressed the problem by introducing the gross-exposure constrained mean-variance portfolio selection. In this paper, we present a direct approach to estimate the risk for vast portfolios using asynchronous and noisy high-frequency data. This approach alleviates accumulation of the estimation error of tens of hundreds of integrated volatilities (or co-volatilities), and on the other hand it has the advantage of smoothing away the microstructure noise in the spatial direction. Based on the simple approach, together with the “pre-averaging” technique, we obtain a sharper bound of the risk approximation error than that in Fan et al. (in J. Am. Stat. Assoc. 107:412–428, 2012b). This bound is locally dependent on the allocation plan satisfying the gross-exposure constraint. The bound does not require exponential tail of the distribution of the microstructure noise. Finite fourth moment suffices. Our work also demonstrates that the mean squared error of the risk estimator can be decreased by choosing an optimal tuning parameter depending on the allocation plan. This is more pronounced for the moderately high-frequency data. Our theoretical results are further confirmed by simulations.

Appendix: Outlined proofs of main theorems

In this section, we provide the outline of the proof of main theorems. The detailed proof is available from the author upon request. In the sequel we assume that C is a generic constant which may take different values at different appearance. Recall that $X^{*}_{t_{j}}=\sum^{p}_{k=1}w_{k}X^{*k}_{j}$, $\epsilon^{*}_{t_{j}}=\sum^{p}_{k=1}w_{k}\epsilon^{*k}_{t_{j}}$, $X(t)=\sum^{p}_{k=1}w_{k}X^{k}(t)$, and $\epsilon_{t_{j}}=\sum^{p}_{k=1}w_{k}\epsilon^{k}_{t_{j}}$ for 0≤j≤n. Let $\overline{X}(i)=\sum^{L_{i}+1}_{j=L_{i-1}+1}g_{j}^{i}\varDelta ^{n}_{j}X$ and $\overline{\epsilon}(i)=\sum^{L_{i}+1}_{j=L_{i-1}+1}g_{j}^{i}\varDelta ^{n}_{j}\epsilon$, where $\varDelta ^{n}_{j}X=X(t_{j})-X(t_{j-1})$ and $\varDelta ^{n}_{j}\epsilon=\epsilon_{t_{j}}-\epsilon_{t_{j-1}}$. $\overline{X^{*}}(i)=\sum^{L_{i}+1}_{j=L_{i-1}+1}g^{i}_{j}\varDelta ^{n}_{j}X^{*}$ and $\overline{\epsilon^{*}}(i)=\sum^{L_{i}+1}_{j=L_{i-1}+1}g^{i}_{j}\varDelta ^{n}_{j}\epsilon^{*}$ where $\varDelta ^{n}_{j}X^{*}=X^{*}_{t_{j}}-X^{*}_{t_{j-1}}$ and $\varDelta ^{n}_{j}\epsilon ^{*}=\epsilon^{*}_{t_{j}}-\epsilon^{*}_{t_{j-1}}$. Then $\overline{Y}(i)=\overline{X}(i)+\overline{\epsilon}(i)$ and $\overline{Y*}(i)=\overline{X^{*}}(i)+\overline{\epsilon^{*}}(i)$. And our estimator of the integrated volatility of X is

$$\begin{aligned} \widetilde{\langle X, X \rangle}_t =& \Biggl(\sum^m_{i=1}\bigl( \overline{Y^{*}}(i)\bigr)^2-4\sum _{i=1}^m\sum^{L_i+1}_{j=L_{i-1}+1} \biggl(\frac{t_j-t_{j-1}}{\varDelta _m}\biggr)^2\hat{\sigma}_{\epsilon}^2 \Biggr)\big/\,\overline{G} \\ =& \Biggl(\sum^m_{i=1} \bigl( \overline{Y^{*}}^2(i)-\overline {Y}^2(i)+ \overline{X}^2(i)+\overline{\epsilon}^2(i)+2\overline {X}(i)\overline{\epsilon}(i) \bigr) \\ &{}-4\sum_{i=1}^m\sum ^{L_i+1}_{j=L_{i-1}+1}\biggl(\frac {t_j-t_{j-1}}{\varDelta _m} \biggr)^2 \bigl({\bf w}^T{\bf w}\bigr)\hat{ \sigma}^2_{\epsilon} \Biggr)\big/\,\overline {G}. \end{aligned}$$

(A.1)

We first give some lemmas related to the terms in (A.1), the proofs of which are omitted.

Lemma 1

Under Assumptions 1, 3 and 4,

$$ \Biggl \vert \frac{1}{\overline{G}}\sum^m_{i=1}E_{t_{L_{i-1}}} \overline {X}^2(i)-\int^t_0 \sigma_s^2\,ds\Biggr \vert \leq C\varDelta _m. $$

(A.2)

If m→∞,

$$ \frac{m}{\overline{G}^2}\sum^m_{i=1}E_{t_{L_{i-1}}} \bigl(\overline {X}^2(i)-E_{t_{L_{i-1}}}\overline{X}^2(i) \bigr)^2\rightarrow^p 2t\int^t_0 \sigma^4_s\,ds. $$

(A.3)

Lemma 2

Under Assumptions 2 and 3,

$$ \sum^m_{i=1}E \overline{\epsilon}^2(i)=4\sum_{i=1}^m \sum^{L_i+1}_{j=L_{i-1}+1}\biggl(\frac{t_j-t_{j-1}}{\varDelta _m} \biggr)^2\mu_2. $$

(A.4)

$$\begin{aligned} &\sum^m_{i=1}E \bigl(\overline{\epsilon}^2(i)-E\bigl(\overline{\epsilon }^2(i)\bigr) \bigr)^2 \\ &\quad {}=16\sum^m_{i=1}\sum ^{L_i+1}_{j=L_{i-1}+1}\biggl(\frac {t_j-t_{j-1}}{\varDelta _m} \biggr)^4\bigl(\mu_4-3\mu_2^2 \bigr) \\ &\qquad {}+32\sum^m_{i=1} \Biggl(\sum ^{L_i+1}_{j=L_{i-1}+1} \biggl(\frac{t_j-t_{j-1}}{\varDelta _m} \biggr)^2 \Biggr)^2\mu_2^2 \\ &\quad {}=32\sum^m_{i=1} \Biggl(\sum ^{L_i+1}_{j=L_{i-1}+1}\biggl(\frac {t_j-t_{j-1}}{\varDelta _m} \biggr)^2 \Biggr)^2\sigma_{\epsilon}^4 \bigl({\bf w}^T{\bf w}\bigr)^2\bigl[1+O( \varDelta _m)\bigr]. \end{aligned}$$

(A.5)

where $\mu_{2}=E\epsilon_{t_{1}}^{2}$ and $\mu_{4}=E\epsilon_{t_{1}}^{4}$.

Lemma 3

$$ \begin{aligned}[b] &\sum^m_{i=1}E_{t_{L_{i-1}}} \overline{X}^2(i)\overline{\epsilon }^2(i)\\ &\quad {}=4 \mu_2\overline{G}\bigl(1+O_P(\varDelta _m)\bigr) \sum^m_{i=1}\sigma _{t_{L_{i-1}}}^2 \varDelta _m \sum^{L_i+1}_{j=L_{i-1}+1}\biggl( \frac{t_j-t_{j-1}}{\varDelta _m}\biggr)^2. \end{aligned} $$

(A.6)

Lemma 4

$$ \hat{\sigma}_{\epsilon}^2- \sigma_{\epsilon}^2=O_p\biggl(\frac{1}{\sqrt {n^{*}}}\vee \frac{p}{n^{*}}\biggr), $$

(A.7)

where n ^∗ is the totality of sample sizes before data synchronization.

Lemma 5

1 and 2 in Theorem 1 hold.

Lemma 6

(A.8)

(A.9)

and

$$\begin{aligned} \sum^m_{i=1}E_{t_{L_{i-1}}}\bigl( \overline{X}(i)\bigr)^2\bigl(\overline {Y^{*}}(i)- \overline{Y}(i)\bigr)^2\leq c^2C\biggl(\frac{m}{n} \biggr)\frac{1}{n}. \end{aligned}$$

(A.10)

Proof of Theorem 1

Seen from Lemmas 1–4, we find that

$$\begin{aligned} &\widetilde{\langle X, X \rangle}_t-\int^t_0 \sigma^2_s\,ds \\ &\quad {}=\frac{1}{\overline{G}}\sum^m_{i=1}\bigl( \overline {X}^2(i)\,{-}\,E_{t_{L_{i-1}}}\overline{X}^2(i)\bigr) \,{+}\,\frac{1}{\overline{G}}\sum^m_{i=1}\bigl( \overline{\epsilon }^2(i)\,{-}\,E_{t_{L_{i-1}}}\overline{ \epsilon}^2(i)\bigr)\,{+}\,\frac{2}{\overline {G}}\sum ^m_{i=1}\overline{X}(i)\overline{\epsilon}(i) \\ &\qquad {}+o_P(1/\sqrt{m})+O_P \Biggl(\sum ^m_{i=1}\sum^{L_i+1}_{j=L_{i-1}+1} \biggl(\frac{t_j-t_{j-1}}{\varDelta _m}\biggr)^2{\bf w}^T{\bf w} \sigma^2_{\epsilon }\biggl(\frac{1}{\sqrt{n^{*}p}}\vee\frac{1}{n^{*}} \biggr) \Biggr) \\ &\qquad {}+\sum^m_{i=1} \bigl( \overline{Y^{*}}^2(i)-\overline{Y}^2(i) \bigr)/ \overline{G}, \end{aligned}$$

(A.11)

and 3–5 of Theorem 1 are proved. Now we go to the last term in (A.11) which can be decomposed as follows:

$$\begin{aligned} &\sum^m_{i=1} \bigl( \overline{Y^{*}}^2(i)-\overline{Y}^2(i) \bigr)/ \overline{G} \\ &\quad {}=\frac{1}{\overline{G}}\sum^m_{i=1}E_{t_{L_{i-1}}} \bigl(\overline {Y^{*}}^2(i)-\overline{Y}^2(i) \bigr) \\ &\qquad {}+\frac{1}{\overline{G}}\sum^m_{i=1} \bigl(\bigl(\overline{Y^{*}}(i)-\overline {Y}(i)\bigr)^2-E_{t_{L_{i-1}}} \bigl(\overline{Y^{*}}(i)-\overline {Y}(i)\bigr)^2 \bigr) \\ &\qquad {}+\frac{2}{\overline{G}}\sum^m_{i=1} \bigl( \overline {X}(i) \bigl(\overline{Y^{*}}(i)-\overline{Y}(i) \bigr)-E_{t_{L_{i-1}}}\overline {X}(i) \bigl(\overline{Y^{*}}(i)- \overline{Y}(i)\bigr) \bigr) \\ &\qquad {}+\frac{2}{\overline{G}}\sum^m_{i=1} \bigl( \overline{\epsilon }(i) \bigl(\overline{Y^{*}}(i)-\overline{Y}(i)\bigr) \bigr). \end{aligned}$$

(A.12)

By Lemma 6, the second and the third term are $O_{P}(\sqrt {(\frac{m}{n})^{3}}\frac{1}{\sqrt{n}})$, and $O_{P}(\sqrt{\frac {m}{n}}\frac{1}{\sqrt{n}})$. Therefore we proved the decomposition in Theorem 1. 1 and 2 of Theorem 1 are guaranteed by Lemma 5.

Proof of Theorem 2

We have the following decomposition:

$$\begin{aligned} &\widetilde{\langle X, X \rangle}-\int^t_0 \sigma_s^2ds \\ &\quad {}=\frac{1}{\overline{G}}\sum^m_{i=1}\bigl( \overline {X}^2(i)-E_{t_{L_{i-1}}}\overline{X}^2(i)\bigr) +\frac{1}{\overline{G}}\sum^m_{i=1}\bigl( \overline{\epsilon }^2(i)-E\overline{\epsilon}^2(i)\bigr) \\ &\qquad {}+\frac{2}{\overline{G}}\sum^m_{i=1} \overline{\epsilon }(i)\overline{X}(i)+\sum^m_{i=1} \Biggl(E\overline{\epsilon }^2(i)-4\sum^{L_i+1}_{j=L_{i-1}+1} \biggl(\frac{t_j-t_{j-1}}{\varDelta _m}\biggr)^2\hat{\sigma}_{\epsilon}^2 \Biggr) \\ &\qquad {}+ \Biggl(\frac{1}{\overline{G}}\sum^m_{i=1}E_{t_{L_{i-1}}} \overline {X}^2(i)-\int^t_0 \sigma_s^2\,ds \Biggr)+\sum^m_{i=1}E_{t_{L_{i-1}}} \bigl(\overline{Y^{*}}^2(i)-\overline {Y}^2(i) \bigr) \\ &\qquad {}+\frac{1}{\overline{G}}\sum^m_{i=1} \bigl( \bigl(\overline {Y^{*}}(i)-\overline{Y}(i)\bigr)^2-E_{t_{L_{i-1}}} \bigl(\overline {Y^{*}}(i)-\overline{Y}(i)\bigr)^2 \bigr) \\ &\qquad {}+\frac{2}{\overline{G}}\sum^m_{i=1} \bigl( \overline {Y}(i) \bigl(\overline{Y^{*}}(i)-\overline{Y}(i) \bigr)-E_{t_{L_{i-1}}}\overline {Y}(i) \bigl(\overline{Y^{*}}(i)- \overline{Y}(i)\bigr) \bigr). \end{aligned}$$

(A.13)

By (A.2),

$$ \sup_{\|{\bf w}\|_1\leq c}\Biggl \vert \frac{1}{\overline{G}}\sum^m_{i=1}E_{t_{L_{i-1}}} \overline {X}^2(i)-\int^t_0 \sigma_s^2\,ds\Biggr \vert \leq C\varDelta _m. $$

(A.14)

By Lemma 5, we have

$$ \sup_{\|{\bf w}\|_1\leq c} \Biggl \vert \sum^m_{i=1}E_{t_{L_{i-1}}} \bigl(\overline{Y^{*}}^2(i)-\overline {Y}^2(i) \bigr)\Biggr \vert \leq c^2C\frac{m}{n}=O_P \biggl(\frac{1}{\sqrt{n}}\biggr). $$

(A.15)

Then Theorem 2 results from the following lemma.

Lemma 7

Under the conditions in Theorem 2, we have

$$ \sup_{\|{\bf w}\|_1\leq c}\Biggl \vert \frac{1}{\overline{G}}\sum ^m_{i=1}\bigl(\overline {X}^2(i)-E_{t_{L_{i-1}}}\overline{X}^2(i)\bigr)\Biggr \vert =O_p\biggl(\frac{1}{\sqrt{m}}\biggr); $$

(A.16)

$$ \sup_{\|{\bf w}\|_1\leq c}\Biggl \vert \frac{1}{\overline{G}}\sum ^m_{i=1}\bigl(\overline{\epsilon }^2(i)-E_{t_{L_{i-1}}}\overline{\epsilon}^2(i)\bigr) \Biggr \vert =O_p\biggl(\frac {1}{\sqrt{m}}\biggr); $$

(A.17)

$$ \sup_{\|{\bf w}\|_1\leq c}\Biggl \vert \sum ^m_{i=1} \Biggl(E\overline{\epsilon}^2(i)-4 \sum^{L_i+1}_{j=L_{i-1}+1}\biggl(\frac{t_j-t_{j-1}}{\varDelta _m} \biggr)^2\hat{\sigma }_{\epsilon}^2 \Biggr)\Biggr \vert =O_p\biggl(\frac{1}{\sqrt{m}}\biggr); $$

(A.18)

$$ \sup_{\|{\bf w}\|_1\leq c}\Biggl \vert \frac{2}{\overline{G}}\sum ^m_{i=1}\overline{\epsilon }(i) \overline{X}(i)\Biggr \vert =O_p\biggl(\frac{1}{\sqrt{m}}\biggr); $$

(A.19)

$$ \sup_{\|{\bf w}\|_1\leq c}\Biggl \vert \frac{1}{\overline{G}}\sum ^m_{i=1} \bigl(\bigl(\overline {Y^{*}}(i)-\overline{Y}(i)\bigr)^2-E_{t_{L_{i-1}}}\bigl( \overline {Y^{*}}(i)-\overline{Y}(i)\bigr)^2 \bigr) \Biggr \vert =O_P\biggl(\frac{1}{\sqrt{m}}\biggr); $$

(A.20)

$$ \sup_{\|{\bf w}\|_1\leq c}\Biggl \vert \frac{2}{\overline{G}}\sum ^m_{i=1} \bigl(\overline {Y}(i) \bigl( \overline{Y^{*}}(i)-\overline{Y}(i)\bigr)-E_{t_{L_{i-1}}}\overline {Y}(i) \bigl(\overline{Y^{*}}(i)-\overline{Y}(i)\bigr) \bigr) \Biggr \vert =O_P\biggl(\frac{1}{\sqrt{m}}\biggr). $$

(A.21)

Proof of Proposition 1

To save space, we only prove the result when k=l. For the result when $k\not=l$, the proof is similar. As in (A.13), we have

$$\begin{aligned} &n^{1/4} \biggl(\widehat{\bigl\langle X^k, X^k \bigr\rangle}-\int^t_0\bigl( \sigma^{k}_s\bigr)^2ds \biggr) \\ &\quad {}=n^{1/4} \Biggl(\frac{1}{\overline{G}}\sum^m_{i=1} \bigl(\overline {X^k}^2(t_i)-E_{t_{L_{i-1}}} \overline{X^k}^2(t_i)\bigr) \Biggr) \\ &\qquad {}+n^{1/4} \Biggl(\frac{1}{\overline{G}}\sum^m_{i=1} \bigl(\overline {\epsilon^k}^2(t_i)-E_{t_{L_{i-1}}} \overline{\epsilon ^k}^2(t_i)\bigr) \Biggr) \\ &\qquad {}+n^{1/4}\frac{1}{\overline{G}}\sum^m_{i=1} \overline{\epsilon ^k}(t_i)\overline{X^k}(t_i) \\ &\qquad {}+n^{1/4} \sum^m_{i=1} \Biggl(E\overline { \epsilon^k}^2(t_i)-4\sum ^{L_i+1}_{j=L_{i-1}+1}\biggl(\frac {t_j-t_{j-1}}{\varDelta _m} \biggr)^2\hat{\sigma}_{\epsilon}^2 \Biggr) \\ &\qquad {}+n^{1/4} \Biggl(\frac{1}{\overline{G}}\sum^m_{i=1}E_{t_{L_{i-1}}} \overline{X^k}^2(t_i)-\int ^t_0\bigl(\sigma_s^{k} \bigr)^ 2\,ds \Biggr) \\ &\qquad {}+n^{1/4}\frac{1}{\overline{G}}\sum ^m_{i=1}E_{t_{L_{i-1}}} \bigl(\overline{Z^{*k}}^2(i)- \overline {Z^{k}}^2(i) \bigr) \\ &\qquad {}+n^{1/4} \Biggl(\frac{1}{\overline{G}}\sum^m_{i=1} \bigl(\bigl(\overline {Z^{*k}}(i)-\overline{Z^{k}}(i) \bigr)^2-E_{t_{L_{i-1}}}\bigl(\overline {Z^{*k}}(i)- \overline{Z^{k}}(i)\bigr)^2 \bigr) \Biggr) \\ &\qquad {}+n^{1/4} \Biggl(\frac{2}{\overline{G}}\sum^m_{i=1} \bigl(\overline {Z^k}(i) \bigl(\overline{Z^{*k}}(i)- \overline {Z^{k}}(i)\bigr)-E_{t_{L_{i-1}}}\overline{Z^k}(i) \bigl(\overline {Z^{*k}}(i)-\overline{Z^{k}}(i)\bigr) \bigr) \Biggr). \end{aligned}$$

(A.22)

Similarly to (A.2), one has

$$ n^{1/4}\Biggl \vert \frac{1}{\overline{G}}\sum ^m_{i=1}E_{t_{iL}}\overline {X^k}^2(t_i)-\int^t_0 \bigl(\sigma_s^{k}\bigr)^ 2\,ds\Biggr \vert \leq\frac{C}{n^{1/4}}. $$

(A.23)

Similarly to (A.15), we have

$$ n^{1/4}\Biggl \vert \sum^m_{i=1}E_{t_{L_{i-1}}} \bigl(\overline {Z^{*k}}^2(i)-\overline{Z^{k}}^2(i) \bigr)\Biggr \vert \leq\frac{C}{n^{1/4}}. $$

(A.24)

Therefore the fifth and sixth term in (A.22) is negligible. Now it suffices to prove Lemma 8, the proof of which is omitted.

Lemma 8

For x∈(0,c ₅ n ^1/4], we have

$$ P \Biggl(n^{1/4}\Biggl \vert \frac{1}{\overline{G}}\sum ^m_{i=1}\bigl(\overline {X^k}^2(t_i)-E_{t_{L_{i-1}}} \overline{X^k}^2(t_i)\bigr)\Biggr \vert >x \Biggr)\leq \frac{1}{C}\exp\bigl(-Cx^2\bigr). $$

(A.25)

$$ P \Biggl(n^{1/4}\Biggl \vert \frac{1}{\overline{G}}\sum ^m_{i=1}\bigl(\overline { \epsilon^k}^2(t_i)-E_{t_{L_{i-1}}}\overline{ \epsilon ^k}^2(t_i)\bigr)\Biggr \vert >x \Biggr)\leq \frac{1}{C}\exp{\bigl(-Cx^2\bigr)}. $$

(A.26)

$$ P \Biggl(n^{1/4}\Biggl \vert \frac{1}{\overline{G}}\sum ^m_{i=1}\overline {\epsilon^k}(t_i) \overline{X^k}(t_i)\Biggr \vert >x \Biggr)\leq \frac{1}{C}\,\mathrm{exp}\bigl(-Cx^2\bigr). $$

(A.27)

$$ P \Biggl(n^{1/4}\Biggl \vert \sum ^m_{i=1} \Biggl(E\overline{\epsilon ^k}^2(t_i)-4\sum ^{L_i+1}_{j=L_{i-1}+1}\biggl(\frac{t_j-t_{j-1}}{\varDelta _m} \biggr)^2\hat{\sigma}_{\epsilon^k}^2 \Biggr)\Biggr \vert >x \Biggr)\leq \frac{1}{C}\,\mathrm{exp}\bigl(-Cx^2\bigr). $$

(A.28)

$$\begin{aligned} &P \Biggl(n^{1/4}\Biggl \vert \frac{1}{\overline{G}}\sum ^m_{i=1} \bigl(\bigl(\overline{Z^{*k}}(i)- \overline {Z^{k}}(i)\bigr)^2-E_{t_{L_{i-1}}}\bigl( \overline{Z^{*k}}(i)-\overline {Z^{k}}(i)\bigr)^2 \bigr)\Biggr \vert >x \Biggr) \\ &\quad {}\leq \frac{1}{C}\,\mathrm{exp}\bigl(-Cx^2\bigr). \end{aligned}$$

(A.29)

$$\begin{aligned} &P \Biggl(n^{1/4}\Biggl \vert \frac{2}{\overline{G}}\sum ^m_{i=1} \bigl(\overline{Z^k}(i) \bigl(\overline{Z^{*k}}(i)-\overline {Z^{k}}(i) \bigr)-E_{t_{L_{i-1}}}\overline{Z^k}(i) \bigl(\overline {Z^{*k}}(i)-\overline{Z^{k}}(i)\bigr) \bigr)\Biggr \vert >x \Biggr) \\ &\quad {}\leq \frac{1}{C}\,\mathrm{exp}\bigl(-Cx^2\bigr). \end{aligned}$$

(A.30)

Proof of Theorem 3

It suffices to prove that for any B _n →∞ satisfying B _n ≤logn,

$$ (\tilde{{\bf w}}_{\mathrm{opt}}-{\bf w}_{\mathrm{opt}})^T( \tilde{{\bf w}}_{\mathrm{opt}}-{\bf w}_{\mathrm{opt}})=O_P\biggl( \frac{B_n}{n^{1/4}}\biggr). $$

(A.31)

By the expansion, we have on $\{(\tilde{{\bf w}}_{\mathrm{opt}}-{\bf w}_{\mathrm{opt}})^{T}(\tilde{{\bf w}}_{\mathrm{opt}}-{\bf w}_{\mathrm{opt}})\geq\frac {MB_{n}}{n^{1/4}}\}$ for any M>0,

$$\begin{aligned} &R(\tilde{{\bf w}}_{\mathrm{opt}})-R({\bf w}_{\mathrm{opt}})=( \tilde{{\bf w}}_{\mathrm{opt}}-{\bf w}_{\mathrm{opt}})^T \boldsymbol {\Sigma }_{t, \tau} (\tilde{{\bf w}}_{\mathrm{opt}}-{\bf w}_{\mathrm{opt}}) \\ &\quad {}\geq \lambda_{\mathrm{min}} (\tilde{{\bf w}}_{\mathrm{opt}}-{\bf w}_{\mathrm{opt}})^T(\tilde {{\bf w}}_{\mathrm{opt}}-{\bf w}_{\mathrm{opt}})\geq\frac{\lambda_{\mathrm{min}}MB_n}{n^{1/4}}. \end{aligned}$$

(A.32)

On the other hand, by Theorem 2 and Corollary 1, on $\{(\tilde{{\bf w}}_{\mathrm{opt}}-{\bf w}_{\mathrm{opt}})^{T}(\tilde{{\bf w}}_{\mathrm{opt}}-{\bf w}_{\mathrm{opt}})\geq\frac{MB_{n}}{n^{1/4}}\}$,

$$\begin{aligned} &\tilde{R}(\tilde{{\bf w}}_{\mathrm{opt}})-\tilde{R}({\bf w}_{\mathrm{opt}}) \\ &\quad {}= \tilde{R}(\tilde{\bf w}_{\mathrm{opt}})-R(\tilde{{\bf w}}_{\mathrm{opt}})+R( \tilde {{\bf w}}_{\mathrm{opt}})-R({\bf w}_{\mathrm{opt}})+R({\bf w}_{\mathrm{opt}})-\tilde{R}({\bf w}_{\mathrm{opt}}) \\ &\quad {}\geq \frac{\lambda_{\mathrm{min}}MB_n}{n^{1/4}}+O_P\bigl(1/n^{1/4}\bigr)>0, \end{aligned}$$

(A.33)

with probability arbitrarily close to 1 for large enough M and n. This is a contradiction with the definition of $\tilde{{\bf w}}_{\mathrm{opt}}$. □