Ambiguous risk constraints with moment and unimodality information

Abstract

Optimization problems face random constraint violations when uncertainty arises in constraint parameters. Effective ways of controlling such violations include risk constraints, e.g., chance constraints and conditional Value-at-Risk constraints. This paper studies these two types of risk constraints when the probability distribution of the uncertain parameters is ambiguous. In particular, we assume that the distributional information consists of the first two moments of the uncertainty and a generalized notion of unimodality. We find that the ambiguous risk constraints in this setting can be recast as a set of second-order cone (SOC) constraints. In order to facilitate the algorithmic implementation, we also derive efficient ways of finding violated SOC constraints. Finally, we demonstrate the theoretical results via computational case studies on power system operations.

Appendices

Appendix A: proof of Observation 2

Proof

As $f^k_+(z) = \bigl (\frac{\alpha }{\alpha + 1}\bigr )(1 - k^{-\alpha -1})z - (1 - k^{-\alpha }) \beta$, we have

$$\begin{aligned} \Bigl [f^k_+(z)\Bigr ]_+ = \left\{ \begin{array}{ll} 0, &{}\quad {\text {if }} z < z_0(k) := \bigl (\frac{\alpha +1}{\alpha }\bigr ) \bigl (\frac{1 - k^{-\alpha }}{1 - k^{-\alpha -1}}\bigr ) \beta \\ \bigl (\frac{\alpha }{\alpha + 1}\bigr )(1 - k^{-\alpha -1})z - (1 - k^{-\alpha }) \beta , &{}\quad {\text {if }} z \ge z_0(k) \end{array}\right. \end{aligned}$$

for all $k \ge 1$. As $[f^{k+1}_+(z)]_+ \ge 0$ for all $z \in \mathbb {R}$, to show that $[f^{k+1}_+(z)]_+ \ge [f^k_+(z)]_+$, it suffices to prove that $[f^{k+1}_+(z)]_+ \ge f^k_+(z)$ for all $z \in \mathbb {R}$. First, as $\beta \le 0$ and $\frac{1 - k^{-\alpha }}{1 - k^{-\alpha -1}}$ increases in k, we have $\bigl (\frac{\alpha +1}{\alpha }\bigr ) \bigl (\frac{1 - k^{-\alpha }}{1 - k^{-\alpha -1}}\bigr ) \beta \ge \bigl (\frac{\alpha +1}{\alpha }\bigr ) \bigl (\frac{1 - (k+1)^{-\alpha }}{1 - (k+1)^{-\alpha -1}}\bigr ) \beta$, i.e., $z_0(k) \ge z_0(k+1)$. It follows that, when $z < z_0(k)$, $f^k_+(z) \le 0$ and hence $[f^{k+1}_+(z)]_+ \ge f^k_+(z)$. Second, when $z \ge z_0(k)$, $f^{k+1}_+(z) \ge 0$ because $z \ge z_0(k) \ge z_0(k+1)$ and $f^{k+1}_+(z)$ increases in z. As both $f^k_+(z)$ and $f^{k+1}_+(z)$ are affine functions of z, we have $f^{k+1}_+(z) = f^{k+1}_+(z_0(k)) + (\frac{\alpha }{\alpha + 1})(1 - (k+1)^{-\alpha -1})(z - z_0(k))$ and $f^k_+(z) = (\frac{\alpha }{\alpha + 1})(1 - k^{-\alpha -1})(z - z_0(k))$ for $z \ge z_0(k)$. It follows that $f^{k+1}_+(z) - f^k_+(z) = f^{k+1}_+(z_0(k)) + (\frac{\alpha }{\alpha + 1})[k^{-\alpha -1} - (k+1)^{-\alpha -1}](z - z_0(k)) \ge 0$. Hence, $[f^{k+1}_+(z)]_+ \ge f^k_+(z)$ when $z \ge z_0(k)$ and the proof is complete. $\square$

Appendix B

For random variable Z and constant $\beta \in \mathbb {R}$, we make the following observation on the worst-case expectation $\sup _{{\mathbb P}_Z \in \mathcal {D}(\mu _0, \Sigma _0)} \mathbb {E}_{{\mathbb P}_Z}[Z-\beta ]_+$. Note that this observation can be made following the derivations in [30], and we present a proof below for completeness.

Observation 3

Given $\beta \in {\mathbb R}$, we have

$$\begin{aligned} \sup _{{\mathbb P}_Z \in \mathcal {D}(\mu _0, \Sigma _0)} \mathbb {E}_{{\mathbb P}_Z}[Z-\beta ]_+ \ = \ \frac{1}{2}\left[ \sqrt{(\beta - \mu _0)^2 + (\Sigma _0 - \mu _0^2)} - \beta + \mu _0 \right] . \end{aligned}$$

Proof

We represent $\sup _{{\mathbb P}_Z \in \mathcal {D}(\mu _0, \Sigma _0)} \mathbb {E}_{{\mathbb P}_Z}[Z-\beta ]_+$ as the following optimization problem

$$\begin{aligned} v_P \ = \ \max _{{\mathbb P}_Z} \ \mathbb {E}_{{\mathbb P}_Z}[Z-\beta ]_+ \\ ({\text {P}}) \quad \quad \quad {\text {s.t.}}\, \ \mathbb {E}_{{\mathbb P}_Z}[Z] = \mu _0, \\ \ \mathbb {E}_{{\mathbb P}_Z}[Z^2] = \Sigma _0, \\ \ \mathbb {E}_{{\mathbb P}_Z}[1] = 1, \\ {\text {whose dual is}} \quad v_D \ = \ \min _{q, p, r} \ \mu _0 p + \Sigma _0 q + r\\ {\text {(D)}} \quad \quad \quad {\text {s.t.}} \ \ qz^2 + pz + r \ge [z-\beta ]_+, \ \ \forall z \in {\mathbb R}. \end{aligned}$$

The weak duality between (P) and (D), i.e., $v_D \le v_P$, holds because $\mu _0 p + \Sigma _0 q + r = \mathbb {E}_{{\mathbb P}_Z}[qZ^2 + pZ + r] \le \mathbb {E}_{{\mathbb P}_Z}[Z-\beta ]_+$ for any feasible solution (q, p, r) to (D) and feasible solution ${\mathbb P}_Z$ to (P). Now we prove the strong duality by constructing two feasible solutions to (P) and (D), respectively, that have the same objective value. On the one hand, the primal solution $\hat{{\mathbb P}}_Z$ is supported on two points $z_1$ and $z_2$ with probability masses $p_1$ and $p_2$, respectively, where $\Delta = \sqrt{(\beta - \mu _0)^2 + (\Sigma _0 - \mu _0^2)}$ and

$$\begin{aligned} p_1 = \frac{\beta - \mu _0 + \Delta }{2\Delta }, \quad p_2 = \frac{\mu _0 - \beta + \Delta }{2\Delta }, \quad z_1 = \beta - \Delta , \quad {\text {and}} \quad z_2 = \beta + \Delta . \end{aligned}$$

We have $p_1, p_2 \ge 0$ because $\Delta \ge |\beta - \mu _0|$. Meanwhile, we have

$$\begin{aligned} p_1z_1 + p_2 z_2 \ = \ \frac{(\beta - \mu _0 + \Delta )(\beta - \Delta )}{2\Delta } + \frac{(\mu _0 - \beta + \Delta )(\beta + \Delta )}{2\Delta } \ = \ \mu _0, \end{aligned}$$

and

$$\begin{aligned} p_1z_1^2 + p_2 z_2^2 \ =&\ \frac{(\beta - \mu _0 + \Delta )(\beta - \Delta )^2}{2\Delta } + \frac{(\mu _0 - \beta + \Delta )(\beta + \Delta )^2}{2\Delta } \\ =&\ \frac{(\beta - \mu _0)\left[ (\beta - \Delta )^2 - (\beta + \Delta )^2 \right] + \Delta \left[ (\beta - \Delta )^2 + (\beta + \Delta )^2 \right] }{2\Delta } \\ =&\ -\beta ^2 + 2\mu _0\beta + \Delta ^2 = -\beta ^2 + 2\mu _0\beta + (\beta - \mu _0)^2 + (\Sigma _0 - \mu _0^2) = \Sigma _0. \end{aligned}$$

Hence, $\hat{{\mathbb P}}_Z$ is feasible to (P). On the other hand, the dual solution $(\hat{q}, \hat{p}, \hat{r})$ is such that

$$\begin{aligned} \hat{q} = \frac{1}{4\Delta }, \quad \hat{p} = \frac{\Delta - \beta }{2\Delta }, \quad {\text {and}} \quad \hat{r} = \frac{(\Delta - \beta )^2}{4\Delta }. \end{aligned}$$

Hence, $\hat{q}z^2 + \hat{p}z + \hat{r} = \frac{1}{4\Delta }(z + \Delta - \beta )^2$. It follows that $\hat{q}z^2 + \hat{p}z + \hat{r} \ge 0$ for all $z \in {\mathbb R}$. Meanwhile, $(\hat{q}z^2 + \hat{p}z + \hat{r}) - (z - \beta ) = \frac{1}{4\Delta }(z - \beta - \Delta )^2 \ge 0$, i.e., $\hat{q}z^2 + \hat{p}z + \hat{r} \ge z - \beta$. Thus, $\hat{q}z^2 + \hat{p}z + \hat{r} \ge [z-\beta ]_+$ and so $(\hat{q}, \hat{p}, \hat{r})$ is feasible to (D).

Finally, the primal objective value associated with $\hat{{\mathbb P}}_Z$ is $p_2(z_2 - \beta ) = \frac{(\mu _0 - \beta + \Delta )\Delta }{2\Delta } = \frac{1}{2}(\Delta - \beta + \mu _0)$. Meanwhile, the dual objective value associated with $(\hat{q}, \hat{p}, \hat{r})$ is

$$\begin{aligned}&\mu _0 \left( \frac{\Delta - \beta }{2\Delta }\right) + \Sigma _0 \left( \frac{1}{4\Delta }\right) + \frac{(\Delta - \beta )^2}{4\Delta } \\&\quad = \ \frac{\Delta ^2 + (\beta ^2 - 2\mu _0\beta + \mu _0^2) + (\Sigma _0 - \mu _0^2) + 2\mu _0 \Delta - 2 \Delta \beta }{4\Delta } \\&\quad = \ \frac{2\Delta ^2 + 2\mu _0 \Delta - 2 \Delta \beta }{4\Delta } \ = \ \frac{1}{2}(\Delta - \beta + \mu _0), \end{aligned}$$

which coincides with the primal objective value associated with $\hat{{\mathbb P}}_Z$. $\square$

Appendix C: corrected ACC representation of Example 3.4.4 in [17]

The ACC representation (3.50) in Example 3.4.4 in [17] has typos and is corrected as follows:

$$\begin{aligned}&\beta - (\mu - m)^{\top } \gamma - \langle \Sigma + (\mu - m)(\mu - m)^{\top }, \ \Gamma \rangle \ \ge \ {(1 - \epsilon )\tau }, \\&\quad \begin{bmatrix} \tau - \beta&\frac{1}{2} \frac{\alpha }{\alpha +1}\gamma ^{\top } \\ \frac{1}{2} \frac{\alpha }{\alpha +1}\gamma&\frac{\alpha }{\alpha +2}\Gamma \end{bmatrix} \ \succeq \ 0, \\&\quad \begin{bmatrix} {\left[ \alpha ^{\frac{1}{\alpha +1}} + \left( \frac{1}{\alpha }\right) ^{\frac{\alpha }{\alpha +1}}\right] } \tau ^{\frac{1}{\alpha +1}} \left( b(x) - m^{\top }a(x)\right) ^{\frac{\alpha }{\alpha +1}} - \beta&\frac{1}{2} \left( \frac{\alpha }{\alpha +1}\gamma - a(x)\right) ^{\top } \\ \frac{1}{2} \left( \frac{\alpha }{\alpha +1}\gamma - a(x)\right)&\frac{\alpha }{\alpha +2}\Gamma \end{bmatrix} \ \succeq \ 0. \end{aligned}$$