Pure characteristics demand models and distributionally robust mathematical programs with stochastic complementarity constraints

Abstract

We formulate pure characteristics demand models under uncertainties of probability distributions as distributionally robust mathematical programs with stochastic complementarity constraints (DRMP-SCC). For any fixed first-stage variable and a random realization, the second-stage problem of DRMP-SCC is a monotone linear complementarity problem (LCP). To deal with ambiguity of probability distributions of the involved random variables in the stochastic LCP, we use the distributionally robust approach. Moreover, we propose an approximation problem with regularization and discretization to solve DRMP-SCC, which is a two-stage nonconvex-nonconcave minimax optimization problem. We prove the convergence of the approximation problem to DRMP-SCC regarding the optimal solution sets, optimal values and stationary points as the regularization parameter goes to zero and the sample size goes to infinity. Finally, preliminary numerical results for investigating distributional robustness of pure characteristics demand models are reported to illustrate the effectiveness and efficiency of our approaches.

Appendix

Proof

(The proof of Proposition 1) Denote $\bar{p}_i=P_0(\varXi _i)$ for $i=1,\cdots ,k$. We verify that $\bar{p}=(\bar{p}_1,\cdots ,\bar{p}_k)^\top \in \mathcal {P}_k$ for all sufficiently large k in the following. Since $\varPsi$ is continuous, we know from mean value theorem of integrals that

$$\begin{aligned} \mathbb {E}_{P_0}[\varPsi (\xi )]=\sum _{i=1}^k \int _{\varXi _i} \varPsi (\xi ) \,P_0(d\xi )=\sum _{i=1}^k \varPsi (\tilde{\xi }^i)P_0(\varXi _i) \end{aligned}$$

for some $\tilde{\xi }^i\in \varXi _i$, $i=1,\cdots ,k$. Then

$$\begin{aligned} \left\| \mathbb {E}_{P_0}[\varPsi (\xi )] - \sum _{i=1}^k \bar{p}_i \varPsi (\xi ^i) \right\| \le \sum _{i=1}^k \bar{p}_i \left\| \varPsi (\xi ^i) - \varPsi (\tilde{\xi }^i) \right\| . \end{aligned}$$

(45)

We first consider the case that $\varXi$ is bounded. For $\alpha >0$, there exists $\delta >0$ such that if $\max _{1\le i\le k}\mathrm {diam}(\varXi _i) <\delta$, then

$$\begin{aligned} \max _{1\le i\le k}\left\| \varPsi (\xi ^i) - \varPsi (\tilde{\xi }^i) \right\| \le \alpha . \end{aligned}$$

Since $\varXi$ is bounded, we can find a sequence $\{\xi ^k\}_{k=1}^\infty$ such that the corresponding Voronoi tessellation $\varXi _1,\cdots ,\varXi _k,\cdots$ satisfying

$$\begin{aligned} \lim _{k\rightarrow \infty }\max _{1\le i\le k}\mathrm {diam}(\varXi _i)=0. \end{aligned}$$

Hence there is $\bar{k}>0$ such that $\max _{1\le i\le k}\mathrm {diam}(\varXi _i)<\delta$ for any $k\ge \bar{k}$.

Then, it knows from (45) that

$$\begin{aligned} \left\| \mathbb {E}_{P_0}[\varPsi (\xi )] - \sum _{i=1}^k \bar{p}_i \varPsi (\xi ^i) \right\| \le \alpha . \end{aligned}$$

This, together with Assumption 1, indicates that $\sum _{i=1}^k \bar{p}_i \varPsi (\xi ^i) \in \varGamma ,$ which implies the nonemptiness of $\mathcal {P}_k$.

Now we consider the case that $\varXi$ is unbounded. Let $\varXi _b:=\{\xi \in \varXi :\left\| \xi \right\| \le b\}$ for $b>0$. Denote a probability distribution $\bar{P}_0$ supported on $\varXi _b$ by

$$\begin{aligned} \bar{P}_0(\varXi _a)=\frac{P_0(\varXi _a\cap \varXi _b)}{P_0(\varXi _b)} \end{aligned}$$

for any measurable $\varXi _a\subseteq \varXi$, where $P_0$ is defined in Assumption 1. Note that

$$\begin{aligned} \lim _{b\rightarrow \infty }\frac{1}{P_0(\varXi _b)}=1~\text {and}~\lim _{b\rightarrow \infty } \int _{\varXi _b} \varPsi (\xi ) P_0(d\xi ) = \int _{\varXi } \varPsi (\xi ) P_0(d\xi )=\mathbb {E}_{P_0}[\varPsi (\xi )]. \end{aligned}$$

We have

$$\begin{aligned} \lim _{b\rightarrow \infty }\int _{\varXi _b} \varPsi (\xi ) \bar{P}_0(d\xi ) = \lim _{b\rightarrow \infty } \frac{1}{P_0(\varXi _b)}\int _{\varXi _b} \varPsi (\xi ) P_0(d\xi ) = \mathbb {E}_{P_0}[\varPsi (\xi )]. \end{aligned}$$

Therefore, there exists $b_0>0$ such that, for any $b\ge b_0$,

$$\begin{aligned} \left\| \mathbb {E}_{\bar{P}_0}[\varPsi (\xi )] - \mathbb {E}_{P_0}[\varPsi (\xi )]\right\| \le \frac{\alpha }{2}. \end{aligned}$$

From Assumption 1, we obtain

$$\begin{aligned} \mathbb {E}_{\bar{P}_0}[\varPsi (\xi )] + \frac{\alpha }{2}\mathbb {B}\subseteq \varGamma . \end{aligned}$$

(46)

Due to the boundedness of $\varXi _b$ and (46), by the same proof for the case that $\varXi$ is bounded, there exists a $\bar{k}>0$ such that $\mathcal {P}_k$ is nonempty for $k\ge \bar{k}$. $\square$

Proof

(The proof of Theorem 7) Since $(x^{k*},\mathbf {y}^{k*},p^{k*})$ is an accumulation point of $(x_\epsilon ^k,\mathbf {y}_\epsilon ^k, p_\epsilon ^k)$ as $\epsilon \downarrow 0$, there exists a sequence $\{\epsilon _j\}_{j= 1}^\infty$ with $\epsilon _j\downarrow 0$ as $j\rightarrow \infty$, such that $(x_{\epsilon _j}^k,\mathbf {y}_{\epsilon _j}^k,p_{\epsilon _j}^k)\rightarrow (x^{k*},\mathbf {y}^{k*},p^{k*})$ as $j\rightarrow \infty$. Based on (26), we have

$$\begin{aligned} {\left\{ \begin{array}{ll} 0\in \nabla _x G(x_{\epsilon _j}^k,\mathbf {y}_{\epsilon _j}^k,p_{\epsilon _j}^k) + \mathcal {N}_X(x_{\epsilon _j}^k),\\ \mathbf {y}_{\epsilon _j}^k=\hat{\mathbf {y}}_{\epsilon _j}(x_{\epsilon _j}^k),\\ 0\in -\nabla _{p}G(x_{\epsilon _j}^k,\mathbf {y}_{\epsilon _j}^k,p_{\epsilon _j}^k) + \mathcal {N}_{\mathcal {P}_k}(p_{\epsilon _j}^k). \end{array}\right. } \end{aligned}$$

We know from Definitions 3, 4 and Lemma 3 that

$$\begin{aligned} {\left\{ \begin{array}{ll} 0\in \nabla _x G(x_{\epsilon _j}^k,\mathbf {y}_{\epsilon _j}^k,p_{\epsilon _j}^k) + \mathcal {N}_X(x_{\epsilon _j}^k),\\ {\left\{ \begin{array}{ll} \nabla _\mathbf {y} G(x_{\epsilon _j}^k,\mathbf {y}_{\epsilon _j}^k,p_{\epsilon _j}^k) - \lambda ^{k,j} - (\mathbf {M}^\top +\epsilon _j I) \mu ^{k,j} = 0,\\ \lambda _i^{k,j} =0 ~\text {for}~ i\in \mathcal {I}_{+0}^{k,j}(\mathbf {y}_{\epsilon _j}^k),~ \mu _i^{k,j} =0 ~\text {for}~ i\in \mathcal {I}_{0+}^{k,j}(\mathbf {y}_{\epsilon _j}^k),\\ \lambda _i^{k,j}\ge 0,~\mu _i^{k,j}\ge 0~\text {for}~ i\in \mathcal {I}_{00}^{k,j}(\mathbf {y}_{\epsilon _j}^k), \end{array}\right. }\\ 0\in -\nabla _{p}G(x_{\epsilon _j}^k,\mathbf {y}_{\epsilon _j}^k,p_{\epsilon _j}^k) + \mathcal {N}_{\mathcal {P}_k}(p_{\epsilon _j}^k), \end{array}\right. } \end{aligned}$$

where $\{\lambda ^{k,j}\}_{j=1}^\infty$ and $\{\mu ^{k,j}\}_{j=1}^\infty$ are two sequences of multipliers, and

$$\begin{aligned} \mathcal {I}_{+0}^{k,j}(\mathbf {y}_{\epsilon _j}^k)&=\{i:(\mathbf {y}_{\epsilon _j}^k)_i>0, ((\mathbf {M} + \epsilon _j I) \mathbf {y}_{\epsilon _j}^k + \mathbf {q}(x_{\epsilon _j}^k))_i=0\},\\ \mathcal {I}_{0+}^{k,j}(\mathbf {y}_{\epsilon _j}^k)&=\{i:(\mathbf {y}_{\epsilon _j}^k)_i=0, ((\mathbf {M} + \epsilon _j I) \mathbf {y}_{\epsilon _j}^k + \mathbf {q}(x_{\epsilon _j}^k))_i>0\},\\ \mathcal {I}_{00}^{k,j}(\mathbf {y}_{\epsilon _j}^k)&=\{i:(\mathbf {y}_{\epsilon _j}^k)_i=0, ((\mathbf {M} + \epsilon _j I) \mathbf {y}_{\epsilon _j}^k + \mathbf {q}(x_{\epsilon _j}^k))_i=0\}. \end{aligned}$$

Thus, for sufficiently large j, we have

$$\begin{aligned} {\left\{ \begin{array}{ll} 0\in \nabla _x G(x_{\epsilon _j}^k,\mathbf {y}_{\epsilon _j}^k,p_{\epsilon _j}^k) + \mathcal {N}_X(x_{\epsilon _j}^k),\\ {\left\{ \begin{array}{ll} \nabla _\mathbf {y} G(x_{\epsilon _j}^k,\mathbf {y}_{\epsilon _j}^k,p_{\epsilon _j}^k) - \lambda ^{k,j} - (\mathbf {M}^\top +\epsilon _j I) \mu ^{k,j} = 0,\\ \lambda _i^{k,j} =0 ~\text {for}~ i\in \mathcal {I}_{+0}^k(\mathbf {y}^{k*}),~ \mu _i^{k,j} =0 ~\text {for}~ i\in \mathcal {I}_{0+}^k(\mathbf {y}^{k*}), \end{array}\right. }\\ 0\in -\nabla _{p}G(x_{\epsilon _j}^k,\mathbf {y}_{\epsilon _j}^k,p_{\epsilon _j}^k) + \mathcal {N}_{\mathcal {P}_k}(p_{\epsilon _j}^k), \end{array}\right. } \end{aligned}$$

(47)

where

$$\begin{aligned} \mathcal {I}_{+0}(\mathbf {y}^{k*})&=\{i:\mathbf {y}^{k*}_i>0, (\mathbf {M} \mathbf {y}^{k*} + \mathbf {q}(x^{k*}))_i=0\},\\ \mathcal {I}_{0+}(\mathbf {y}^{k*})&=\{i:\mathbf {y}^{k*}_i=0, (\mathbf {M} \mathbf {y}^{k*} + \mathbf {q}(x^{k*}))_i>0\}. \end{aligned}$$

Next, we verify the boundedness of sequences $\{\lambda ^{k,j}\}_{j=1}^\infty$ and $\{\mu ^{k,j}\}_{j=1}^\infty$. Notice that if $\{\mu ^{k,j}\}_{j=1}^\infty$ is bounded, from the boundedness of $\{(x_{\epsilon _j}^k,\mathbf {y}_{\epsilon _j}^k,p_{\epsilon _j}^k)\}_{j=1}^\infty$ and continuous differentiability of G, $\{\lambda ^{k,j}\}_{j=1}^\infty$ is bounded. Now we assume that $\{\mu ^{k,j}\}_{j=1}^\infty$ is unbounded. Consider, by dividing $\left\| \mu ^{k,j}\right\|$, that

$$\begin{aligned} \frac{ \nabla _\mathbf {y} G(x_{\epsilon _j}^k,\mathbf {y}_{\epsilon _j}^k,p_{\epsilon _j}^k) }{\left\| \mu ^{k,j}\right\| } - \frac{\lambda ^j}{\left\| \mu ^{k,j}\right\| } - (\mathbf {M}^\top +\epsilon _j I) \frac{\mu ^{k,j}}{\left\| \mu ^{k,j}\right\| } = 0, \end{aligned}$$

which can deduce, according to $\left\| \mu ^{k,j}\right\| \rightarrow \infty$ as $j\rightarrow \infty$, that

$$\begin{aligned} \frac{\lambda ^{k,j}}{\left\| \mu ^{k,j}\right\| } + \mathbf {M}^\top \frac{\mu ^{k,j}}{\left\| \mu ^{k,j}\right\| } \rightarrow 0 \end{aligned}$$

(48)

as $j\rightarrow \infty$. Since $\lambda _i^{k,j} =0$ for $i\in \mathcal {I}_{+0}^k(\mathbf {y}^{k*})$ and $\mu _i^{k,j} =0$ for $i\in \mathcal {I}_{0+}^k(\mathbf {y}^{k*})$, we rewrite (48) as

$$\begin{aligned} \sum _{i\in \mathcal {I}_{0+}^k(\mathbf {y}^{k*})\cup \mathcal {I}_{00}^k(\mathbf {y}^{k*})}\frac{\lambda _i^{k,j}}{\left\| \mu ^{k,j}\right\| } \mathbf {e}_i + \sum _{i\in \mathcal {I}_{+0}^k(\mathbf {y}^{k*})\cup \mathcal {I}_{00}^k(\mathbf {y}^{k*})} \frac{\mu _i^{k,j}}{\left\| \mu ^{k,j}\right\| } \mathbf {M}^\top _i \rightarrow 0, \end{aligned}$$

where $\mathcal {I}_{00}^k(\mathbf {y}^{k*})=\{i:\mathbf {y}^{k*}_i=0, (\mathbf {M} \mathbf {y}^{k*} + \mathbf {q}(x^{k*}))_i=0\}.$

Then, by MPEC-LICQ at $\mathbf {y}^{k*}$ for problem (23) with $(\bar{x},\bar{p})=(x^{k*},p^{k*})$ and $\epsilon =0$, we obtain

$$\begin{aligned} \frac{\lambda _i^{k,j}}{\left\| \mu ^{k,j}\right\| }\rightarrow 0, i\in \mathcal {I}_{0+}^k(\mathbf {y}^{k*})\cup \mathcal {I}_{00}^k(\mathbf {y}^{k*}) ~\text {and}~ \frac{\mu _i^{k,j}}{\left\| \mu ^{k,j}\right\| }\rightarrow 0, i\in \mathcal {I}_{+0}^k(\mathbf {y}^{k*})\cup \mathcal {I}_{00}^k(\mathbf {y}^{k*}) \end{aligned}$$

as $k\rightarrow \infty$, which contradicts $\frac{\mu _i^{k,j}}{\left\| \mu ^{k,j}\right\| }\nrightarrow 0$ for some $i\in \mathcal {I}_{+0}^k(\mathbf {y}^{k*})\cup \mathcal {I}_{00}^k(\mathbf {y}^{k*})$. Therefore, both $\{\lambda ^{k,j}\}_{j=1}^\infty$ and $\{\mu ^{k,j}\}_{j=1}^\infty$ are bounded. Without loss of generality, we assume that $\lambda ^{k,j}\rightarrow \lambda ^{k*}$ and $\mu ^{k,j}\rightarrow \mu ^{k*}$ as $j\rightarrow \infty$. Therefore, by letting $j\rightarrow \infty$, we have from (47) that

$$\begin{aligned} {\left\{ \begin{array}{ll} 0\in \nabla _x G(x^{k*},\mathbf {y}^{k*},p^{k*}) + \mathcal {N}_X(x^{k*}),\\ {\left\{ \begin{array}{ll} \nabla _\mathbf {y} G(x^{k*},\mathbf {y}^{k*},p^{k*}) - \lambda ^{k*} - \mathbf {M}^\top \mu ^{k*} = 0,\\ \lambda _i^{k*} =0 ~\text {for}~ i\in \mathcal {I}_{+0}^k(\mathbf {y}^{k*}),~ \mu _i^{k*} =0 ~\text {for}~ i\in \mathcal {I}_{0+}^k(\mathbf {y}^{k*}), \end{array}\right. }\\ 0\in -\nabla _{p}G(x^{k*},\mathbf {y}^{k*},p^{k*}) + \mathcal {N}_{\mathcal {P}_k}(p^{k*}). \end{array}\right. } \end{aligned}$$

(49)

Moreover, for $\lambda _i^{k,j}\mu _i^{k,j}\ge 0$ for $i=1,\ldots , mk$, we have $\lambda _i^{k*}\mu _i^{k*}\ge 0$ for $i\in \mathcal {I}_{00}^k(\mathbf {y}^{k*})$. This together with (49) means that $(x^{k*},\mathbf {y}^{k*},p^{k*})$ is a block coordinatewise C-stationary point of problem ($\mathrm {P}_k$). $\square$