A branch-and-bound algorithm for instrumental variable quantile regression

Abstract

This paper studies a statistical problem called instrumental variable quantile regression (IVQR). We model IVQR as a convex quadratic program with complementarity constraints and—although this type of program is generally NP-hard—we develop a branch-and-bound algorithm to solve it globally. We also derive bounds on key variables in the problem, which are valid asymptotically for increasing sample size. We compare our method with two well known global solvers, one of which requires the computed bounds. On random instances, our algorithm performs well in terms of both speed and robustness.

Appendices

Appendix A Proof of Lemma 1

Proof

Since $Ax^* + \epsilon = b$, we have from the triangle inequality that $\Vert \epsilon \Vert _\infty \le \Vert b\Vert _\infty + \Vert Ax^* \Vert _\infty$ and $\Vert Ax^*\Vert _\infty \le \Vert b\Vert _\infty + \Vert \epsilon \Vert _\infty$. So the event $\Vert \epsilon \Vert _\infty > C \Vert b\Vert _\infty$ implies

$$\begin{aligned} \Vert \epsilon \Vert _\infty > C \Vert b\Vert _\infty \ge C (\Vert \epsilon \Vert _\infty - \Vert Ax^*\Vert _\infty ) = C \Vert \epsilon \Vert _\infty - C\Vert Ax^*\Vert _\infty \end{aligned}$$

and

$$\begin{aligned} \Vert \epsilon \Vert _\infty > C \Vert b\Vert _\infty \ge C (\Vert Ax^*\Vert _\infty - \Vert \epsilon \Vert _\infty ) = C \Vert Ax^*\Vert _\infty - C\Vert \epsilon \Vert _\infty , \end{aligned}$$

which together imply

$$\begin{aligned} \frac{C}{C+1}< \frac{\Vert \epsilon \Vert _\infty }{\Vert Ax^*\Vert _\infty } < \frac{C}{C-1}. \end{aligned}$$

This proves the first bound. To prove the second, we note that, by definition,

$$\begin{aligned} \Vert \epsilon \Vert _\infty> C \Vert b\Vert _\infty \Longleftrightarrow \max _{i=1}^m |\epsilon _i| > C \max _{k=1}^m \left| a_k^{\text {T}} x^* + \epsilon _k\right| , \end{aligned}$$

and so $\Vert \epsilon \Vert _\infty > C \Vert b\Vert _\infty$ implies that $|\epsilon _i| > C \max _{k=1}^m |a_k^{\text {T}} x^* + \epsilon _k|$ holds for at least one specific i. Hence,

$$\begin{aligned} P\left( \Vert \epsilon \Vert _\infty> C \Vert b\Vert _\infty \right)&\le \sum _{i=1}^m P\left( |\epsilon _i|> C\max _{k=1}^m \left| a_k^{\text {T}}x^* + \epsilon _k\right| \right) \nonumber \\&\le m \max _{i=1}^m P\left( |\epsilon _i| > C\max _{k} \left| a_k^{\text {T}}x^* + \epsilon _k\right| \right) . \end{aligned}$$

(14)

Next, we claim $|a_k^{\text {T}} x^* + \epsilon _k| \ge |\epsilon _k| \cdot \mathbbm {1}\{\epsilon _k a_k^{\text {T}} x^* \ge 0 \}$ for each k. If $\epsilon _k a_k^{\text {T}} x^* < 0$, this is certainly true. Otherwise, if $\epsilon _ka_k^{\text {T}}x^* \ge 0$, then $\epsilon _k$ and $a_k^{\text {T}}x^*$ have the same (possibly zero) signs and hence $|a_k^{\text {T}}x^* + \epsilon _k| \ge |\epsilon _k|$. Thus,

$$\begin{aligned} |\epsilon _i|> C\max _{k=1}^m \left| a_k^{\text {T}}x^* + \epsilon _k\right| \Longrightarrow |\epsilon _i| > C \max _{k=1}^m \left\{ |\epsilon _k| \cdot \mathbbm {1}\left\{ \epsilon _k a_k^{\text {T}} x^* \ge 0\right\} \right\} \end{aligned}$$

and so

$$\begin{aligned} P\left( |\epsilon _i|> C\max _{k=1}^m \left| a_k^{\text {T}}x^* + \epsilon _k\right| \right) \le P\left( |\epsilon _i| > C \max _{k=1}^m \left\{ |\epsilon _k| \cdot \mathbbm {1}\left\{ \epsilon _k a_k^{\text {T}} x^* \ge 0\right\} \right\} \right) . \end{aligned}$$

(15)

Combining inequalities (14) and (15), we achieve the second bound. $\square$

Appendix B Proof of Lemma 2

Proof

The first two-sided inequality is a standard fact about the normal distribution. For the last inequality, we have

$$\begin{aligned} P \left( \max _{1 \le p \le q} Z_p \le \theta \sigma \right) = \Pi _{p=1}^q P(Z_p \le \theta \sigma ) = (1 - P(Z > \theta \sigma ))^q. \end{aligned}$$

By the first part of the lemma and the standard fact that $0< x < 1$ and $a > 0$ imply $(1-x)^a \le \exp (-ax)$,

$$\begin{aligned} (1 - P(Z > \theta \sigma ))^q&\le \left( 1 - \frac{1}{2\theta } \cdot \epsilon (\theta ) \right) ^q \\&\le \exp \left( -\frac{q}{2\theta } \cdot \epsilon (\theta ) \right) . \end{aligned}$$

Now substituting the definition of $\epsilon (\theta )$ and $\theta = \sqrt{\log (q)}$, we see

$$\begin{aligned} \exp \left( -\frac{q}{2\theta } \cdot \epsilon (\theta ) \right)&= \exp \left( -\frac{q}{2\theta } \cdot \frac{1}{\sqrt{2\pi }} \exp (-\log (q)/2) \right) \\&= \exp \left( -\frac{q}{2\theta } \cdot \frac{1}{\sqrt{2\pi }} \, q^{-1/2} \right) \\&= \exp \left( -\frac{ \sqrt{q}}{2\sqrt{\log (q)}} \cdot \frac{1}{\sqrt{2\pi }} \right) \\&\le \exp \left( -q^{1/4}\right) , \end{aligned}$$

where the last inequality follows from the assumption that $q/(8\pi \log (q)) \ge \sqrt{q}$. This proves the result. $\square$