High-Dimensional Mixed-Frequency IV Regression

Abstract

This article introduces a high-dimensional linear IV regression for the data sampled at mixed frequencies. We show that the high-dimensional slope parameter of a high-frequency covariate can be identified and accurately estimated leveraging on a low-frequency instrumental variable. The distinguishing feature of the model is that it allows handing high-dimensional datasets without imposing the approximate sparsity restrictions. We propose a Tikhonov-regularized estimator and study its large sample properties for time series data. The estimator has a closed-form expression that is easy to compute and demonstrates excellent performance in our Monte Carlo experiments. We also provide the confidence bands and incorporate the exogenous covariates via the double/debiased machine learning approach. In our empirical illustration, we estimate the real-time price elasticity of supply on the Australian electricity spot market. Our estimates suggest that the supply is relatively inelastic throughout the day.

Keywords:

• Confidence bands
• Double/debiased machine learning
• Electricity supply
• Real-time price elasticity
• Tikhonov regularization

Supplementary Materials

The Supplementary Material contains detailed proofs of Theorems 3.2 and 3.4.

Acknowledgments

This article is a substantially revised Chapter 2 of my Ph.D. thesis. I’m deeply indebted to my advisor Jean-Pierre Florens as well as Eric Gautier, Ingrid van Keilegom, and Timothy Christensen for helpful discussions and suggestions. I thank to participants at the Conference on Inverse Problems in Econometrics at Northwestern University, the ENTER seminar at Tilburg University, the 48èmes Journées de Statistique de la SFdS, the 3rd ISNPS Conference, and the Recent Advances in Econometrics Conference at TSE Toulouse as well as Aleksandra Babii, David Benatia, Bruno Biais, Pascal Lavergne, and Mario Rothfelder. I’m also grateful to anonymous referees whose comments helped me to improve significantly the article. All remaining errors are mine.

Appendix. Proofs

To prove Theorem 3.1, we need an additional lemma that bounds the expected norm of the sample mean of a covariance stationary zero-mean $L^2(S)$-valued stochastic process ${(X_t)}_{t\in \mathbf{Z}}$ by the norm of its auto-covariance function γ_h.

Lemma A.1.

1. Suppose that ${(X_t)}_{t\in \mathbf{Z}}$ is a zero-mean covariance stationary process in $L^2(S)$ with absolutely summable autocovariance function $\sum _{h\in \mathbf{Z}}\Vert {\gamma }_h{\Vert }_1<\infty$, where $\Vert {\gamma }_h{\Vert }_1={\int }_S|{\gamma }_h(s,s)|\mathrm{d}s$. Then$$E{\Vert \frac{1}{T}\sum _{t=1}^T{X}_t\Vert }^2\le \frac{1}{T}\sum _{h\in \mathbf{Z}}\Vert {\gamma }_h{\Vert }_1.$$

Proof.

We have$$\begin{array}{cc}E{\Vert \frac{1}{T}\sum _{t=1}^T{X}_t\Vert }^2\hfill & =\frac{1}{T^2}E〈\sum _{t=1}^T{X}_t,\sum _{k=1}^T{X}_k〉\hfill \\ \mathrm{}\hfill & =\frac{1}{T^2}\sum _{t,k=1}^T{\int }_{S}E[X_t(s)X_k(s)]\mathrm{d}s\hfill \\ \mathrm{}\hfill & =\frac{1}{T^2}\sum _{t,k=1}^T{\int }_S{\gamma }_{t-k}(s)\mathrm{d}s\hfill \\ \mathrm{}\hfill & =\frac{1}{T}\sum _{\left|h\right|<T}\frac{T-\left|h\right|}{T}{\int }_S{\gamma }_h(s,s)\mathrm{d}s\hfill \\ \mathrm{}\hfill & \le \frac{1}{T}\sum _{h\in \mathbf{Z}}{\int }_S\left|{\gamma }_h(s,s)\right|\mathrm{d}s,\hfill \end{array}$$where the second line follows by the bilinearity of the inner product and Fubini’s theorem and the third under the covariance stationarity. □

The following lemma allows controlling estimation errors appearing in the proof of Theorem 3.1 in terms of more primitive quantities.

Lemma A.1.

2. Suppose that $\widehat{k},k,\widehat{r},r,\beta$ are square-integrable. Then$$E{\Vert \widehat{K}-K\Vert }^2\le E{\Vert \widehat{k}-k\Vert }^2$$and$$E{\Vert \widehat{r}-\widehat{K}\beta \Vert }^2\le 2E{\Vert \widehat{r}-r\Vert }^2+2\Vert \beta {\Vert }^{2}E{\Vert \widehat{k}-k\Vert }^2.$$

Proof.

By the definition of the operator norm and the Cauchy–Schwartz inequality(A.1) $$\begin{array}{c}E{\Vert \widehat{K}-K\Vert }^2=E\left[\underset{\Vert \varphi \Vert \le 1}{\sup }{\Vert \widehat{K}\varphi -K\varphi \Vert }^2\right]\\ =E\left[\underset{\Vert \varphi \Vert \le 1}{\sup }\int {\left|\int \varphi (s)\left(\widehat{k}(s,u)-k(s,u)\right)\mathrm{d}s\right|}^2\mathrm{d}u\right]\\ \le E{\Vert \widehat{k}-k\Vert }^2.\end{array}$$(A.1)

For the second part, use $r=K\beta ,\Vert a+b{\Vert }^2\le 2\Vert a{\Vert }^2+2\Vert b{\Vert }^2,\Vert K\beta \Vert \le \Vert K\Vert \Vert \beta \Vert$, and the estimate in equation (A.1)(A.1) $$\begin{array}{c}E{\Vert \widehat{K}-K\Vert }^2=E\left[\underset{\Vert \varphi \Vert \le 1}{\sup }{\Vert \widehat{K}\varphi -K\varphi \Vert }^2\right]\\ =E\left[\underset{\Vert \varphi \Vert \le 1}{\sup }\int {\left|\int \varphi (s)\left(\widehat{k}(s,u)-k(s,u)\right)\mathrm{d}s\right|}^2\mathrm{d}u\right]\\ \le E{\Vert \widehat{k}-k\Vert }^2.\end{array}$$(A.1) $$\begin{array}{cc}E{\Vert \widehat{r}-\widehat{K}\beta \Vert }^2\hfill & \le 2E{\Vert \widehat{r}-r\Vert }^2+2E\Vert \widehat{K}\beta -K\beta {\Vert }^2\hfill \\ \hfill & \le 2E{\Vert \widehat{r}-r\Vert }^2+2\Vert \beta {\Vert }^{2}E{\Vert \widehat{k}-k\Vert }^2.\hfill \end{array}$$

Proof of Theorem 3.1.

The proof is based on the following decomposition:$$\widehat{\beta }-\beta ={(\alpha I+K^{*}K)}^{-1}{K}^{*}(\widehat{r}-\widehat{K}\beta )+\underset{\triangleq R_T}{\underbrace{R_1+R_2+R_3+R_4}}$$with$$\begin{array}{cc}{R}_1\hfill & =\left[{(\alpha I+{\widehat{K}}^{*}\widehat{K})}^{-1}{\widehat{K}}^{*}-{(\alpha I+K^{*}K)}^{-1}{K}^{*}\right](\widehat{r}-\widehat{K}\beta )\hfill \\ R_2\hfill & =\alpha {(\alpha I+{\widehat{K}}^{*}\widehat{K})}^{-1}{\widehat{K}}^{*}(\widehat{K}-K){(\alpha I+K^{*}K)}^{-1}\beta \hfill \\ R_3\hfill & =\alpha {(\alpha I+{\widehat{K}}^{*}\widehat{K})}^{-1}({\widehat{K}}^{*}-K^{*})K{(\alpha I+K^{*}K)}^{-1}\beta \hfill \\ R_4\hfill & ={(\alpha I+K^{*}K)}^{-1}{K}^{*}K\beta -\beta .\hfill \end{array}$$

To see that this decomposition holds, note that$$\begin{array}{c}{R}_2+R_3=\alpha {(\alpha I+{\widehat{K}}^{*}\widehat{K})}^{-1}\left[{\widehat{K}}^{*}\widehat{K}-K^{*}K\right]{(\alpha I+K^{*}K)}^{-1}\beta \\ =\alpha {(\alpha I+{\widehat{K}}^{*}\widehat{K})}^{-1}\left[(\alpha I+{\widehat{K}}^{*}\widehat{K})-(\alpha I+K^{*}K)\right]\\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{(\alpha I+K^{*}K)}^{-1}\beta \\ =\alpha {(\alpha I+K^{*}K)}^{-1}\beta -\alpha {(\alpha I+{\widehat{K}}^{*}\widehat{K})}^{-1}\beta \\ =\left[\mathrm{}I\mathrm{}-\mathrm{}\alpha {(\alpha I\mathrm{}+\mathrm{}{\widehat{K}}^{*}\widehat{K})}^{-1}\right]\beta \mathrm{}+\mathrm{}\left[\alpha {(\alpha I\mathrm{}+\mathrm{}{K}^{*}K)}^{-1}\mathrm{}-\mathrm{}I\right]\beta \\ ={(\alpha I+{\widehat{K}}^{*}\widehat{K})}^{-1}{\widehat{K}}^{*}\widehat{K}\beta -{(\alpha I+K^{*}K)}^{-1}{K}^{*}K\beta .\end{array}$$

Note also that$$\widehat{r}-\widehat{K}\beta =\frac{1}{T}\sum _{t=1}^T{U}_t\Psi (.,W_t).$$

Therefore, if we can show the desired order for $E\Vert R_T{\Vert }^2$, the conclusion of the theorem would follow. Under Assumption 3.1 by Lemma A.1.1$$\begin{array}{c}E{\Vert \widehat{r}-r\Vert }^2=E{\Vert \frac{1}{T}\sum _{t=1}^T\left\{Y_t\Psi (.,W_t)-E[Y_t\Psi (.,W_t)]\right\}\Vert }^2\\ \lesssim \frac{1}{T}\end{array}$$and$$\begin{array}{c}E{\Vert \widehat{k}-k\Vert }^2=E{\Vert \frac{1}{T}\sum _{t=1}^T\left\{Z_t(.)\Psi (.,W_t)-E[Z(.)\Psi (.,W)]\right\}\Vert }^2\\ \lesssim \frac{1}{T}.\end{array}$$

Since$$\Vert R_T\Vert \le \Vert R_1\Vert +\Vert R_2\Vert +\Vert R_3\Vert +\Vert R_4\Vert ,$$it is sufficient to control each of the four terms separately.

The fourth term is a regularization bias and its order follows directly from the Assumption 3.2 and the isometry of the functional calculus$$\begin{array}{cc}\Vert R_4\Vert \hfill & =\Vert \left[{(\alpha I+K^{*}K)}^{-1}{K}^{*}K-I\right]\beta \Vert \hfill \\ \hfill & \lesssim \Vert \left[I-{(\alpha I+K^{*}K)}^{-1}{K}^{*}K\right]{(K^{*}K)}^{\gamma }\Vert \hfill \\ \hfill & =\underset{\lambda \in \sigma (K^{*}K)}{\sup }\left|\left(1-\frac{\lambda }{\alpha +\lambda }\right){\lambda }^{\gamma }\right|\hfill \\ \hfill & =\underset{\lambda \in \sigma (K^{*}K)}{\sup }\left|\frac{{\lambda }^{\gamma }}{\alpha +\lambda }\right|\alpha .\hfill \end{array}$$

We can have two cases depending on the value of $\gamma >0$. For $\gamma \in (0,1)$, the function $\lambda \mapsto {\lambda }^{\gamma }/(\alpha +\lambda )$ admits maximum at $\lambda =\alpha \gamma /(1-\gamma )$. For $\gamma \ge 1$, the function $\lambda \mapsto {\lambda }^{\gamma }/(\alpha +\lambda )$ is strictly increasing on $[0,\infty )$, attaining maximum at the end of the spectrum $\lambda =\Vert K^{*}K\Vert$. Therefore, since ${\gamma }^{\gamma }{(1-\gamma )}^{1-\gamma }\le 1,\gamma \in (0,1)$, we have$$\underset{\lambda \in \sigma (K^{*}K)}{\sup }\frac{{\lambda }^{\gamma }}{\alpha +\lambda }\le \{\begin{array}{cc}\Vert K^{*}K{\Vert }^{\gamma -1},& \gamma \ge 1\\ {\alpha }^{\gamma -1},& \gamma \in (0,1).\end{array}$$

This gives $\Vert R_4\Vert \le {\alpha }^{\gamma }$.

Next, note that $\widehat{K}$ is a finite-rank operator, and hence, compact. Therefore,$$\begin{array}{cc}\Vert R_2\Vert \hfill & \lesssim \Vert {(\alpha I+{\widehat{K}}^{*}\widehat{K})}^{-1}{\widehat{K}}^{*}\Vert \hfill \\ \hfill & \hspace{1em}\Vert \widehat{K}-K\Vert \Vert \alpha {(\alpha I+K^{*}K)}^{-1}{(K^{*}K)}^{\gamma }\Vert \hfill \\ \hfill & \le \underset{\lambda \in \sigma ({\widehat{K}}^{*}\widehat{K})}{\sup }\left|\frac{{\lambda }^{1/2}}{\alpha +\lambda }\right|\Vert \widehat{k}-k\Vert \hfill \\ \hfill & \hspace{1em}\underset{\lambda \in \sigma (K^{*}K)}{\sup }\left|\frac{{\lambda }^{\gamma }}{\alpha +\lambda }\right|\alpha {\lesssim }_P\frac{{\alpha }^{\gamma }}{\sqrt{\alpha T}},\hfill \end{array}$$where we use Lemma A.1.2. Similarly,$$\begin{array}{cc}\Vert R_3\Vert \hfill & \lesssim \Vert {(\alpha I+{\widehat{K}}^{*}\widehat{K})}^{-1}({\widehat{K}}^{*}-K^{*})\alpha K{(\alpha I+K^{*}K)}^{-1}\beta \Vert \hfill \\ \hfill & \le \Vert {(\alpha I+{\widehat{K}}^{*}\widehat{K})}^{-1}\Vert \hfill \\ \hfill & \hspace{1em}\Vert {\widehat{K}}^{*}-K^{*}\Vert \Vert K{(\alpha I+K^{*}K)}^{-1}{(K^{*}K)}^{\gamma }\Vert \alpha \hfill \\ \hfill & \lesssim \frac{1}{\alpha }\Vert \widehat{k}-k\Vert \underset{\lambda \in \sigma (K^{*}K)}{\sup }\left|\frac{{\lambda }^{\gamma +1/2}}{\alpha +\lambda }\right|\alpha {\lesssim }_P\frac{{\alpha }^{\gamma \wedge (1/2)}}{\sqrt{\alpha T}}.\hfill \end{array}$$

Lastly, similar computations yield$$\begin{array}{cc}\Vert R_1\Vert & \le \Vert {(\alpha I+{\widehat{K}}^{*}\widehat{K})}^{-1}{\widehat{K}}^{*}-{(\alpha I+K^{*}K)}^{-1}{K}^{*}\Vert \Vert \widehat{r}-\widehat{K}\beta \Vert \\ & =\Vert {(\alpha I+{\widehat{K}}^{*}\widehat{K})}^{-1}({\widehat{K}}^{*}-K^{*})\\ & \hspace{1em}+\left[{(\alpha I+{\widehat{K}}^{*}\widehat{K})}^{-1}-{(\alpha I+K^{*}K)}^{-1}\right]K^{*}\Vert \Vert \widehat{r}-\widehat{K}\beta \Vert \\ & {\lesssim }_P\Vert {(\alpha I+{\widehat{K}}^{*}{\widehat{K}}^{*})}^{-1}(K^{*}K-{\widehat{K}}^{*}\widehat{K}){(\alpha I+K^{*}K)}^{-1}{K}^{*}\Vert \\ & \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}\frac{1}{\sqrt{T}}+\frac{1}{\alpha T}\\ & \le \Vert {(\alpha I+{\widehat{K}}^{*}\widehat{K})}^{-1}{\widehat{K}}^{*}(\widehat{K}-K){(\alpha I+K^{*}K)}^{-1}{K}^{*}\Vert \frac{1}{\sqrt{T}}\\ & \hspace{1em}+\Vert {(\alpha I+{\widehat{K}}^{*}\widehat{K})}^{-1}({\widehat{K}}^{*}-K^{*})K{(\alpha I+K^{*}K)}^{-1}{K}^{*}\Vert \\ & \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}\frac{1}{\sqrt{T}}+\frac{1}{\alpha T}\\ & \le \frac{1}{\alpha \sqrt{T}}\Vert \widehat{K}-K\Vert +\frac{1}{\alpha \sqrt{T}}\Vert {\widehat{K}}^{*}-K^{*}\Vert +\frac{1}{\alpha T}{\lesssim }_P\frac{1}{\alpha T}.\end{array}$$

Combining all the estimates, we obtain $\Vert R_T\Vert \lesssim \frac{1}{\alpha T}+\frac{{\alpha }^{\gamma \wedge (1/2)}}{\sqrt{\alpha T}}+{\alpha }^{\gamma }$. □

Proof of Theorem 3.3.

Decompose ${\widehat{\beta }}_m-\beta ={\widehat{\beta }}_m-\widehat{\beta }+\widehat{\beta }-\beta .$ By Theorem 3.1, we know that $\Vert \widehat{\beta }-\beta \Vert {\lesssim }_P\frac{1}{\alpha T}+\frac{{\alpha }^{\gamma \wedge (1/2)}}{\sqrt{\alpha T}}+{\alpha }^{\gamma }$. Consequently, it remains to control the discretization error $\Vert {\widehat{\beta }}_m-\widehat{\beta }\Vert$. To that end, note that if ${\widehat{\psi }}_m$ solves $(\alpha I+{\widehat{K}}_m{\widehat{K}}^{*}){\widehat{\psi }}_m=\widehat{r}$, then ${\widehat{\beta }}_m={\widehat{K}}^{*}{\widehat{\psi }}_m$. Therefore, ${\widehat{\beta }}_m={\widehat{K}}^{*}{(\alpha I+{\widehat{K}}_m{\widehat{K}}^{*})}^{-1}\widehat{r}$. Next, decompose(A.2) $$\begin{array}{cc}{\widehat{\beta }}_m\mathrm{}-\mathrm{}\widehat{\beta }& ={\widehat{K}}^{*}{(\alpha I+{\widehat{K}}_m{\widehat{K}}^{*})}^{-1}\widehat{r}-{\widehat{K}}^{*}{(\alpha I+\widehat{K}{\widehat{K}}^{*})}^{-1}\widehat{r}\\ & ={\widehat{K}}^{*}\left[{(\alpha I+{\widehat{K}}_m{\widehat{K}}^{*})}^{-1}-{(\alpha I+\widehat{K}{\widehat{K}}^{*})}^{-1}\right]\widehat{r}\\ & ={\widehat{K}}^{*}{(\alpha I+\widehat{K}{\widehat{K}}^{*})}^{-1}\left[\widehat{K}{\widehat{K}}^{*}-{\widehat{K}}_m{\widehat{K}}^{*}\right]{(\alpha I+{\widehat{K}}_m{\widehat{K}}^{*})}^{-1}\widehat{r}\\ & ={\widehat{K}}^{*}{(\alpha I+\widehat{K}{\widehat{K}}^{*})}^{-1}(\widehat{K}-{\widehat{K}}_m){\widehat{K}}^{*}{(\alpha I+{\widehat{K}}_m{\widehat{K}}^{*})}^{-1}\widehat{r}.\end{array}$$(A.2)

Then$$\begin{array}{cc}\Vert {\widehat{\beta }}_m-\widehat{\beta }\Vert & \le \Vert {\widehat{K}}^{*}{(\alpha I+\widehat{K}{\widehat{K}}^{*})}^{-1}\Vert \Vert (\widehat{K}-{\widehat{K}}_m){\widehat{K}}^{*}\Vert \\ & \hspace{1em}\Vert {(\alpha I+{\widehat{K}}_m{\widehat{K}}^{*})}^{-1}\Vert \Vert \widehat{r}\Vert \\ & {\lesssim }_P\frac{1}{{\alpha }^{3/2}}\Vert (\widehat{K}-{\widehat{K}}_m){\widehat{K}}^{*}\Vert .\end{array}$$

The expression inside of the operator norm is the integral operator such that for every $\psi \in L^2$ $$\begin{array}{c}(\widehat{K}-{\widehat{K}}_m){\widehat{K}}^{*}\psi \\ =\int \mathrm{}\psi (u)\mathrm{}\left(\mathrm{}\int \widehat{k}(s,v)\widehat{k}(s,u)\mathrm{d}s-\sum _{j=1}^m\widehat{k}(s_j,v)\widehat{k}(s_j,u){\delta }_j\mathrm{}\right)\mathrm{d}u.\end{array}$$

Therefore, by the same computations as in Equation (A.1)(A.1) $$\begin{array}{c}E{\Vert \widehat{K}-K\Vert }^2=E\left[\underset{\Vert \varphi \Vert \le 1}{\sup }{\Vert \widehat{K}\varphi -K\varphi \Vert }^2\right]\\ =E\left[\underset{\Vert \varphi \Vert \le 1}{\sup }\int {\left|\int \varphi (s)\left(\widehat{k}(s,u)-k(s,u)\right)\mathrm{d}s\right|}^2\mathrm{d}u\right]\\ \le E{\Vert \widehat{k}-k\Vert }^2.\end{array}$$(A.1) and the triangle inequality$$\begin{array}{c}\Vert (\widehat{K}-{\widehat{K}}_m){\widehat{K}}^{*}\Vert \\ \hspace{1em}\le \Vert \int \widehat{k}(s,.)\widehat{k}(s,.)\mathrm{d}s-\sum _{j=1}^m\widehat{k}(s_j,.)\widehat{k}(s_j,.){\delta }_j\Vert \\ \hspace{1em}=\Vert \frac{1}{T^2}\sum _{t=1}^T\sum _{k=1}^T\Psi (.,W_t)\Psi (.,W_k)\{\int Z_t(s)Z_k(s)\mathrm{d}s\\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}-\sum _{j=1}^m{Z}_t(s_j)Z_k(s_j){\delta }_j\}\Vert \\ \hspace{1em}\le \frac{1}{T^2}\sum _{t=1}^T\sum _{k=1}^T\Vert \Psi (.,W_t)\Vert \Vert \Psi (.,W_k)\Vert \\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}\left|\int Z_t(s)Z_k(s)\mathrm{d}s-\sum _{j=1}^m{Z}_t(s_j)Z_k(s_j){\delta }_j\right|\\ \hspace{1em}\le {\max }_{1\le t\le T}\Vert \Psi (.,W_t){\Vert }^2{\max }_{1\le t,k\le T}\\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}\left|\sum _{j=1}^m{Z}_t(s_j)Z_k(s_j){\delta }_j-\int Z_t(s)Z_k(s)\mathrm{d}s\right|.\end{array}$$

Under Assumption 3.4 (i)$$\begin{array}{c}\left|Z_t(s_j)Z_k(s_j)-Z_t(s)Z_k(s)\right|\\ \hspace{1em}\le |Z_t(s_j)-Z_t(s)||Z_k(s_j)|+|Z_t(s)||Z_k(s_j)-Z_k(s)|\\ \hspace{1em}\le 2L^2|s_j-s{|}^{\kappa },\end{array}$$and whence$$\begin{array}{c}\left|\sum _{j=1}^m{Z}_t(s_j)Z_k(s_j){\delta }_j-\int Z_t(s)Z_k(s)\mathrm{d}s\right|\\ \hspace{1em}=\left|\sum _{j=1}^m{\int }_{s_{j-1}}^{s_j}\left\{Z_t(s_j)Z_k(s_j)-Z_t(s)Z_k(s)\right\}\mathrm{d}s\right|\\ \hspace{1em}\le \sum _{j=1}^m{\int }_{s_{j-1}}^{s_j}\left|Z_t(s_j)Z_k(s_j)-Z_t(s)Z_k(s)\right|\mathrm{d}s\\ \hspace{1em}\le 2L^2\sum _{j=1}^m{\int }_{s_{j-1}}^{s_j}|s_j-s{|}^{\kappa }\mathrm{d}s\le 2L^2{\max }_{1\le j\le m}{\delta }_j^{\kappa }.\end{array}$$

This shows that $\Vert {\widehat{\beta }}_m-\widehat{\beta }\Vert {\lesssim }_P{\Delta }_m^{\kappa }/{\alpha }^{3/2}$. □