A Prediction–Correction ADMM for Multistage Stochastic Variational Inequalities

Abstract

The multistage stochastic variational inequality is reformulated into a variational inequality with separable structure through introducing a new variable. The prediction–correction ADMM which was originally proposed in He et al. (J Comput Math 24:693–710, 2006) for solving deterministic variational inequalities in finite-dimensional spaces is adapted to solve the multistage stochastic variational inequality. Weak convergence of the sequence generated by that algorithm is proved under the conditions of monotonicity and Lipschitz continuity. When the sample space is a finite set, the corresponding multistage stochastic variational inequality is actually defined on a finite-dimensional Hilbert space and the strong convergence of the algorithm naturally holds true. Some numerical examples are given to show the efficiency of the algorithm.

Appendix A: Proof of Lemma 2.5

In the appendix, we shall give the proof of Lemma 2.5. The main idea of the proof comes from [32, Lemma 4.6].

Proof

It is clear that (2.10) implies (2.9). Then we only need to prove that the converse is also true.

Define

$$\begin{aligned} A=\{\omega \in \Omega \mid \exists ~x\in C(\omega )~s.t.~\langle F(x^*)(\omega ), x-x^*(\omega )\rangle <0\}. \end{aligned}$$

To prove that  (2.10) holds, we only need to prove that  $P(A)=0$. Let

$$\begin{aligned} G=\{(\omega ,x)\in \Omega \times R^n \mid x\in C(\omega ),~\langle F(x^*)(\omega ), x-x^*(\omega )\rangle <0\}. \end{aligned}$$

Then G is ${\mathscr {F}}\otimes {\mathscr {B}}(R^n)$-measurable.By [4, Theorem III.23], A is ${\mathscr {F}}$-measurable.

Take $k,r=1,2,\cdots$, define

$$\begin{aligned} A_{k,r}=\left\{ \omega \in \Omega \mid \exists ~x\in C(\omega )\cap {\bar{B}}(0,r) ~s.t.~\langle F(x^*)(\omega ), x-x^*(\omega )\rangle \le -\frac{1}{k} \right\} \end{aligned}$$

and

$$\begin{aligned} \Phi _{k,r}(\omega )=\left\{ x\in C(\omega )\cap {\bar{B}}(0,r)\mid \langle F(x^*)(\omega ),x-x^*(\omega )\rangle \le -\frac{1}{k}\right\} . \end{aligned}$$

Here ${\bar{B}}(0,r)$ is the closed ball in $R^n$ of center 0 and radius r. Similarly, $A_{k,r}$ is ${\mathscr {F}}$-measurable. In addition, $\Phi _{k,r}$ is an ${\mathscr {F}}$-measurable set-valued map and

$$\begin{aligned} A=\bigcup _{r=1}^{\infty }\bigcup _{k=1}^{\infty }A_{k,r}. \end{aligned}$$

To prove $P(A)=0$, we only need to prove that for any k, r, $P(A_{k,r})=0$. Assume that there exist k, r such that $P(A_{k,r})>0$. By Lemma 2.2, there exists $\eta \in {\mathcal {L}}^2(\Omega ,{\mathscr {F}},P)$ such that

$$\begin{aligned} \eta (\omega )\in \Phi _{k,r}(\omega ),\quad \text{ a.s. }~\omega \in A_{k,r}. \end{aligned}$$

Define ${\tilde{\eta }}=\eta \chi _{A_{k,r}}+x^* \chi _{\Omega {\setminus } A_{k,r}}$, then ${\tilde{\eta }}\in {\mathcal {C}}$, and

$$\begin{aligned} \langle F(x^*),{\tilde{\eta }}-x^*\rangle _{{\mathcal {L}}^2}&=\int _{\Omega }\langle F(x^*)(\omega ),{\tilde{\eta }}(\omega )-x^*(\omega )\rangle P(d(\omega )) \\ {}&=\int _{A_{k,r}}\langle F(x^*)(\omega ),\eta (\omega )-x^*(\omega )\rangle P(d(\omega ))\\&\le -\frac{1}{k}P(A_{k,r}) <0, \end{aligned}$$

contradicting (2.9). Thus $P(A)=0$. $\square$