Probability maximization via Minkowski functionals: convex representations and tractable resolution

Abstract

In this paper, we consider the maximizing of the probability ${\mathbb {P}}\left\{ \, \zeta \, \mid \, \zeta \, \in \, {\mathbf {K}}({\mathbf {x}}) \, \right\}$ over a closed and convex set ${\mathcal {X}}$, a special case of the chance-constrained optimization problem. Suppose ${\mathbf {K}}({\mathbf {x}}) \, \triangleq \, \left\{ \, \zeta \, \in \, {\mathcal {K}}\, \mid \, c({\mathbf {x}},\zeta ) \, \ge \, 0 \right\}$, and $\zeta$ is uniformly distributed on a convex and compact set ${\mathcal {K}}$ and $c({\mathbf {x}},\zeta )$ is defined as either $c({\mathbf {x}},\zeta )\, \triangleq \, 1-\left| \zeta ^T{\mathbf {x}}\right| ^m$ where $m\ge 0$ (Setting A) or $c({\mathbf {x}},\zeta ) \, \triangleq \, T{\mathbf {x}}\, - \, \zeta$ (Setting B). We show that in either setting, by leveraging recent findings in the context of non-Gaussian integrals of positively homogenous functions, ${\mathbb {P}}\left\{ \,\zeta \, \mid \, \zeta \, \in \, {\mathbf {K}}({\mathbf {x}}) \, \right\}$ can be expressed as the expectation of a suitably defined continuous function $F(\bullet ,\xi )$ with respect to an appropriately defined Gaussian density (or its variant), i.e. ${\mathbb {E}}_{{{\tilde{p}}}} \left[ \, F({\mathbf {x}},\xi )\, \right]$. Aided by a recent observation in convex analysis, we then develop a convex representation of the original problem requiring the minimization of $g\left( {\mathbb {E}}\left[ \, F(\bullet ,\xi )\, \right] \right)$ over ${\mathcal {X}}$, where g is an appropriately defined smooth convex function. Traditional stochastic approximation schemes cannot contend with the minimization of $g\left( {\mathbb {E}}\left[ F(\bullet ,\xi )\right] \right)$ over $\mathcal X$, since conditionally unbiased sampled gradients are unavailable. We then develop a regularized variance-reduced stochastic approximation (r-VRSA) scheme that obviates the need for such unbiasedness by combining iterative regularization with variance-reduction. Notably, (r-VRSA) is characterized by almost-sure convergence guarantees, a convergence rate of $\mathcal {O}(1/k^{1/2-a})$ in expected sub-optimality where $a > 0$, and a sample complexity of $\mathcal {O}(1/\epsilon ^{6+\delta })$ where $\delta > 0$. To the best of our knowledge, this may be the first such scheme for probability maximization problems with convergence and rate guarantees. Preliminary numerics on a portfolio selection problem (Setting A) and a set-covering problem (Setting B) suggest that the scheme competes well with naive mini-batch SA schemes as well as integer programming approximation methods.

Appendix

Proof of Theorem 2:

• (a) When considering uniform distributions over a compact and convex set $\mathcal {K}$, the density is constant in this set and zero outside the set. It can then be concluded that $\zeta$ has a log-concave density. Furthermore, $\zeta$ has a symmetric density about the origin since $\mathcal {K}$ is a symmetric set about the origin. Hence by Lemma 6.2 in [16], h is convex where $h({\mathbf {x}})\triangleq 1/f({\mathbf {x}})$. $\square$

• (b) Since (11) is a convex program, any solution ${\mathbf {x}}^*$ satisfies ${ h({\mathbf {x}}^*) \le h({\mathbf {x}}), \ \forall \mathbf {x} \in \mathcal {X}.}$ From the positivity of f over ${\mathcal {X}}$, $\frac{1}{f({\mathbf {x}}^*)} \le \frac{1}{f({\mathbf {x}})}$ for every ${\mathbf {x}}\in \mathcal {X}$ implying that $f({\mathbf {x}}^*) \ge f({\mathbf {x}})$ for every ${\mathbf {x}}\in \mathcal {X}.$ Consequently, ${\mathbf {x}}^*$ is a global maximizer of (11). $\square$

Proof of Lemma 3:

We prove this result by showing the unimodality of f on ${\mathbb {R}}_+$ where $f(u) = u^c e^{-u}$, implying that $f'(u) = cu^{c-1}e^{-u} - u^ce^{-u} = 0$ if $u = c.$ Furthermore, $f'(u) > 0$ when $u < c$ and $f'(u) < 0$ when $u > c$. Finally, $f(0) = 0$. It follows that $u^* = c$ is a maximizer of $u^ce^{-u}$ on $[0,\infty )$ where $f(c) = \frac{c^c}{e^{c}}$. $\square$

Proof of Proposition 2:

Recall the definition of $F({\mathbf {x}},\xi )$ from the statement of Lemma 4. We prove (a) by considering two cases. Case (i): $\xi \in \varXi _1({\mathbf {x}}) \cup \varXi _0({\mathbf {x}}).$ It follows that

$$\begin{aligned} \left| F({\mathbf {x}},\xi )\right| ^2 = \mathcal {C}^2_{{\mathcal {K}}} \left( (2\pi )^{{n}} e^{-2\mid \xi ^T{\mathbf {x}}\mid ^2+\Vert \xi \Vert _{{\mathcal {K}}}^2} \right)&\le \mathcal {C}^2_{{\mathcal {K}}} \left( (2\pi )^{{n}} e^{-2\mid \xi ^T{\mathbf {x}}\mid ^2+\mid \xi ^T{\mathbf {x}}\mid ^2} \right) \le \mathcal {C}^2_{{\mathcal {K}}} (2\pi )^n. \end{aligned}$$

Case (ii): $\xi \in \varXi _2({\mathbf {x}}).$ Proceeding similarly, we obtain that

$$\begin{aligned} \left| F({\mathbf {x}},\xi )\right| ^2&\le \mathcal {C}^2_{{\mathcal {K}}} \left( (2 \pi )^{{n}} e^{{-\Vert \xi \Vert _{{\mathcal {K}}}^2}} \right) \le \mathcal {C}^2_{{\mathcal {K}}} (2\pi )^n. \end{aligned}$$

Consequently, $\left| F({\mathbf {x}},\xi )\right| ^2 \le \mathcal {C}^2_{{\mathcal {K}}} (2\pi )^n$ for every $\xi \in {\mathbb {R}}^n$.

(b) We observe that $\partial F({\mathbf {x}},\xi )$ is defined as follows.

$$\begin{aligned} \partial F({\mathbf {x}},\xi ) = {\left\{ \begin{array}{ll} \left( \mathcal {C}_{{\mathcal {K}}}(2\pi )^{n/2} (-2\xi \xi ^T{\mathbf {x}}) e^{-|\xi ^{T} {\mathbf {x}}|^{2}+\frac{\Vert \xi \Vert _{{\mathcal {K}}}^2}{2}}\right) , &{} \xi \in \varXi _1({\mathbf {x}}) \triangleq \left\{ \xi \mid |\xi ^{T} {\mathbf {x}}|^{{2}} > \Vert \xi \Vert ^{2}_{\mathcal {K}}\right\} \\ \left( -\mathcal {C}_{{\mathcal {K}}} (2\pi )^{n/2} e^{-\max \{|\xi ^{T} {\mathbf {x}}|^{2}, {\Vert \xi \Vert ^{2}_{\mathcal {K}}}\}+\frac{\Vert \xi \Vert _{{\mathcal {K}}}^2}{2}}\right) \left[ 0,2\xi (\xi ^T{\mathbf {x}})\right] , &{} \xi \in \varXi _0({\mathbf {x}}) \triangleq \left\{ \xi \mid |\xi ^{T} {\mathbf {x}}|^{2} = \Vert \xi \Vert ^{2}_{\mathcal {K}}\right\} \\ \mathbf{0}. &{} \xi \in \varXi _2({\mathbf {x}}) \triangleq \left\{ \xi \mid |\xi ^{T} {\mathbf {x}}|^{2} < \Vert \xi \Vert ^{2}_{\mathcal {K}}\right\} \end{array}\right. } \end{aligned}$$

Consequently, it follows that ${\mathbb {E}}_{{\tilde{p}}} \left[ \, \Vert G({\mathbf {x}},\xi )\Vert ^2 \, \right]$ is bounded as follows.

$$\begin{aligned}&{\mathbb {E}}\left[ \, \Vert G({\mathbf {x}},\xi )\Vert ^2\, \right] = \int _{\varXi } \Vert G({\mathbf {x}},\xi )\Vert ^2 {\tilde{p}}(\xi ) d\xi \nonumber \\&= \int _{\varXi _1({\mathbf {x}})} \Vert G({\mathbf {x}},\xi )\Vert ^2 {\tilde{p}}(\xi ) d\xi + \int _{\varXi _2({\mathbf {x}})} \Vert \underbrace{G({\mathbf {x}},\xi )}_{ \ = \ \mathbf{0}}\Vert ^2 {\tilde{p}}(\xi ) d\xi \nonumber \\&\quad + \int _{\varXi _0({\mathbf {x}})} \Vert G({\mathbf {x}},\xi )\Vert ^2 {\tilde{p}}(\xi ) d\xi \nonumber \\&= \int _{\varXi _1({\mathbf {x}})} \Vert G({\mathbf {x}},\xi )\Vert ^2 {\tilde{p}}(\xi ) d\xi , \end{aligned}$$

(33)

where the last equality follows from observing that $G({\mathbf {x}},\xi ) = 0$ for $\xi \in \varXi _2({\mathbf {x}})$ and the integral in (33) is zero because $\varXi _0({\mathbf {x}})$ is a measure zero set. It follows that ${\mathbb {E}}\left[ \, \Vert G({\mathbf {x}},\xi )\Vert ^2\, \right]$ can be bounded as follows:

$$\begin{aligned}&{\mathbb {E}}\left[ \, \Vert G({\mathbf {x}},\xi )\Vert ^2\right] \nonumber \\&\quad = \int _{\varXi _1({\mathbf {x}})}\left( \mathcal {C}^2_{{\mathcal {K}}}(2\pi )^{n} (4\Vert \xi \Vert _2^2 (\xi ^T{\mathbf {x}})^2) e^{-2(\xi ^T{\mathbf {x}})^2 +{\Vert \xi \Vert ^2_{{\mathcal {K}}}}} \right) \frac{1}{D_{{\mathcal {K}}}}(2\pi )^{-\frac{n}{2}} e^{\frac{{\Vert \xi \Vert ^2_{{\mathcal {K}}}}}{2}} d \xi \end{aligned}$$

(34)

$$\begin{aligned}&\quad \le \int _{\varXi _1({\mathbf {x}})}\left( \mathcal {C}^2_{{\mathcal {K}}}(2\pi )^{n} (4\Vert \xi \Vert _2^2 (\xi ^T{\mathbf {x}})^2) e^{-(\xi ^T{\mathbf {x}})^2} \right) \frac{1}{D_{{\mathcal {K}}}}(2\pi )^{-\frac{n}{2}} e^{\frac{{\Vert \xi \Vert ^2_{{\mathcal {K}}}}}{2}} d\xi , \end{aligned}$$

(35)

where the inequality follows from $\xi \in \varXi _1({\mathbf {x}},u)$. Next, we consider the expression $(\xi ^T{\mathbf {x}})^2 e^{-(\xi ^T{\mathbf {x}})^2}$ or $ue^{-u}$. We note that by Lemma 3, $ue^{-u}$ is a unimodal function and $u^* = 1$ is a maximizer with value $e^{-1}$. Consequently, we have that

$$\begin{aligned} \max _{\{(\xi ^T{\mathbf {x}}) \mid \xi \in \varXi ({\mathbf {x}})\}} (\xi ^T{\mathbf {x}})^2 e^{-(\xi ^T{\mathbf {x}})^2} \le \max _{u \in {\mathbb {R}}_+} u e^{-u}\overset{\tiny \mathrm {Lemma}~3}{\le } \tfrac{1}{e}, \end{aligned}$$

implying that

$$\begin{aligned}&{\mathbb {E}}[\Vert G({\mathbf {x}},\xi )\Vert ^2] \le \int _{\varXi _1({\mathbf {x}})} \left( \mathcal {C}_{{\mathcal {K}}}^2(2\pi )^{n} (\Vert \xi \Vert _2^2 (\xi ^T{\mathbf {x}})^2) e^{-(\xi ^T{\mathbf {x}})^2} \right) \tfrac{1}{D_{{\mathcal {K}}}}(2\pi )^{-\tfrac{n}{2}} e^{-\tfrac{{\Vert \xi \Vert ^2_{{\mathcal {K}}}}}{2}} d\xi \\&\quad \le e^{-1} \mathcal {C}^2_{{\mathcal {K}}}(2\pi )^{n}\int _{\varXi _1({\mathbf {x}})} \Vert \xi \Vert _2^2 \tfrac{1}{D_{{\mathcal {K}}}}(2\pi )^{-\tfrac{n}{2}} e^{-\tfrac{{\Vert \xi \Vert ^2_{{\mathcal {K}}}}}{2}} d\xi \\&\quad \le e^{-1} \mathcal {C}^2_{{\mathcal {K}}}(2\pi )^{n}\int _{{\mathbb {R}}^n} \Vert \xi \Vert _2^2 \tfrac{1}{D_{{\mathcal {K}}}}(2\pi )^{-\tfrac{n}{2}} e^{-\tfrac{{\Vert \xi \Vert ^2_{{\mathcal {K}}}}}{2}} d\xi = e^{-1} \mathcal {C}^2_{{\mathcal {K}}}(2\pi )^{n}{\mathbb {E}}_{{{\tilde{p}}}} \left[ \Vert \xi \Vert _2^2\right] . \end{aligned}$$

$\square$

Proof of Proposition 3:

(a) Since $\Vert \xi \Vert _{{\mathcal {K}}}^2 = \Vert \xi \Vert _p^2$, it follows from Theorem 1 that

$$\begin{aligned} f({\mathbf {x}})&= \int _{{\mathbb {R}}^n} \left( \mathcal {C}(2\pi \sigma ^2)^{\tfrac{n}{2}} e^{-\max \{ \mid \xi ^T {\mathbf {x}}\mid ^2, {\Vert \xi \Vert _{p}^2}\} } \right) d\xi \\&= \int _{{\mathbb {R}}^n} \underbrace{\left( \mathcal {C}(2\pi \sigma ^2)^{\tfrac{n}{2}} e^{-\max \{ \mid \xi ^T {\mathbf {x}}\mid ^2, {\Vert \xi \Vert _{p}^2} \}+\tfrac{{\Vert \xi \Vert _{2}^2}}{2\sigma ^2} } \right) }_{\triangleq F({\mathbf {x}},\xi )} \underbrace{(2\pi \sigma ^2)^{-\tfrac{n}{2}} e^{-\tfrac{{\Vert \xi \Vert _{2}^2}}{2\sigma ^2}}}_{\triangleq {\tilde{p}}(\xi )} d\xi . \end{aligned}$$

(b) Omitted (similar to proof of Proposition 2(a).

(c) Next, we derive a bound on the second moment of $\Vert G({\mathbf {x}},\xi )\Vert$ akin to Prop. 2(b). We observe that $\partial F({\mathbf {x}},\xi )$ is defined as

$$\begin{aligned} \partial F({\mathbf {x}},\xi ) = {\left\{ \begin{array}{ll} \left( \mathcal {C}(2\pi \sigma ^2)^{n/2} (-2\xi \xi ^T{\mathbf {x}}) e^{-|\xi ^{T} {\mathbf {x}}|^{{2}}+\frac{{\Vert \xi \Vert _{2}^2}}{2\sigma ^2}}\right) , &{} \xi \in \varXi _1({\mathbf {x}}) \triangleq \left\{ \xi \mid |\xi ^{T} {\mathbf {x}}|^{{2}} > \Vert \xi \Vert ^{2}_{p} \right\} \\ \left( -\mathcal {C} (2\pi \sigma ^2)^{n/2} e^{-\max \{|\xi ^{T} {\mathbf {x}}|^{{2}}, {\Vert \xi \Vert ^{{2}}_{p}}\}+\frac{{\Vert \xi \Vert _{2}^2}}{2\sigma ^2}}\right) \left[ 0,2\xi (\xi ^T{\mathbf {x}})\right] , &{} \xi \in \varXi _0({\mathbf {x}}) \triangleq \left\{ \xi \mid |\xi ^{T} {\mathbf {x}}|^{{2}} = \Vert \xi \Vert ^{2}_{p}\right\} \\ \mathbf{0}. &{} \xi \in \varXi _2({\mathbf {x}}) \triangleq \left\{ \xi \mid |\xi ^{T} {\mathbf {x}}|^{{2}} < \Vert \xi \Vert ^{2}_{p}\right\} \end{array}\right. } \end{aligned}$$

Consequently, ${\mathbb {E}}\left[ \Vert G({\mathbf {x}},\xi )\Vert ^2\right]$ can be bounded as follows.

$$\begin{aligned}&{\mathbb {E}}\left[ \Vert G({\mathbf {x}},\xi )\Vert ^2\right] = \int _{\varXi } \Vert G({\mathbf {x}},\xi )\Vert ^2 {\tilde{p}}(\xi ) d\xi \nonumber \\&= \int _{\varXi _1({\mathbf {x}})} \Vert G({\mathbf {x}},\xi )\Vert ^2 {\tilde{p}}(\xi ) d\xi + \int _{\varXi _2({\mathbf {x}})} \Vert \underbrace{G({\mathbf {x}},\xi )}_{ \ = \ \mathbf{0}}\Vert ^2 {\tilde{p}}(\xi ) d\xi \nonumber \\&\quad + \int _{\varXi _0({\mathbf {x}})} \Vert G({\mathbf {x}},\xi )\Vert ^2 {\tilde{p}}(\xi ) d\xi \nonumber \\&= \int _{\varXi _1({\mathbf {x}})} \Vert G({\mathbf {x}},\xi )\Vert ^2 {\tilde{p}}(\xi ) d\xi , \end{aligned}$$

(36)

where the last equality follows from observing that $G({\mathbf {x}},\xi ) = 0$ for $\xi \in \varXi _2({\mathbf {x}})$ and the integral in (36) is zero because $\varXi _0({\mathbf {x}})$ is a measure zero set. It follows that

$$\begin{aligned}&\ {\mathbb {E}}[\Vert G({\mathbf {x}},\xi )\Vert ^2] = \int _{\varXi _1({\mathbf {x}})}\left( \mathcal {C}^2(2\pi \sigma ^2)^{n} (4\Vert \xi \Vert _2^2 (\xi ^T{\mathbf {x}})^2) e^{-2(\xi ^T{\mathbf {x}})^2 +{\tfrac{\Vert \xi \Vert ^2_{2}}{\sigma ^2}}} \right) (2\pi \sigma ^2)^{-\tfrac{n}{2}} e^{\tfrac{{-\Vert \xi \Vert ^2_{2}}}{2\sigma ^2}} d\xi \nonumber \\&\le \int _{\varXi _1({\mathbf {x}})}\left( \mathcal {C}^2(2\pi \sigma ^2)^{n} (4\Vert \xi \Vert _2^2 (\xi ^T{\mathbf {x}})^2) e^{-2(\xi ^T{\mathbf {x}})^2 +{\Vert \xi \Vert ^2_{p}}} \right) (2\pi \sigma ^2)^{-\tfrac{n}{2}} e^{\tfrac{{-\Vert \xi \Vert ^2_{2}}}{2\sigma ^2}} d\xi \end{aligned}$$

(37)

$$\begin{aligned}&\le \int _{\varXi _1({\mathbf {x}})}\left( \mathcal {C}^2(2\pi \sigma ^2)^{n} (4\Vert \xi \Vert _2^2 (\xi ^T{\mathbf {x}})^2) e^{-(\xi ^T{\mathbf {x}})^2} \right) (2\pi \sigma ^2)^{-\tfrac{n}{2}} e^{\tfrac{{-\Vert \xi \Vert ^2_{2}}}{2\sigma ^2}} d\xi , \end{aligned}$$

(38)

where (38) follows from $\xi \in \varXi _1({\mathbf {x}})$ and (37) follows from

$$\begin{aligned} \frac{\Vert \xi \Vert _2^2}{\sigma ^2} \le \Vert \xi \Vert _p^2, \text{ where } \sigma ^2 = {\left\{ \begin{array}{ll} n^{1/2-1/p}, &{} p \ge 2 \\ 1. &{} 1 \le p < 2 \end{array}\right. } \end{aligned}$$

We may then conclude that

$$\begin{aligned}&\ {\mathbb {E}}[\Vert G({\mathbf {x}},\xi )\Vert ^2] \le \int _{\varXi _1({\mathbf {x}})} \left( \mathcal {C}^2(2\pi \sigma ^2)^{n} (\Vert \xi \Vert _2^2 (\xi ^T{\mathbf {x}})^2) e^{-(\xi ^T{\mathbf {x}})^2} \right) (2\pi \sigma ^2)^{-\tfrac{n}{2}} e^{-\frac{{\Vert \xi \Vert ^2_{2}}}{2\sigma ^2}} d\xi \nonumber \\&\le e^{-1} \mathcal {C}^2(2\pi \sigma ^2)^{n}\int _{\varXi _1({\mathbf {x}})} \left( \Vert \xi \Vert ^2 \right) (2\pi \sigma ^2)^{-\frac{n}{2}} e^{-\tfrac{{\Vert \xi \Vert ^2_{2}}}{2\sigma ^2}} d\xi \nonumber \\&\le e^{-1} \mathcal {C}^2(2\pi \sigma ^2)^{n}\int _{{\mathbb {R}}^n} \left( \Vert \xi \Vert _2^2\right) (2\pi \sigma ^2)^{-\frac{n}{2}} e^{-\tfrac{{\Vert \xi \Vert ^2_{2}}}{2\sigma ^2}} d\xi = e^{-1} \mathcal {C}^2(2\pi \sigma ^2)^{n}{\mathbb {E}}\left[ \Vert \xi \Vert ^2\right] . \end{aligned}$$

(39)

where (39) follows from Lemma 3. $\square$

Proof of Lemma 6:

Suppose $({\mathbf {x}},{\mathbf {y}})$ is feasible with respect to (PM$_{A,\mathrm{ext}}^{2}$). Then ${\mathbf {x}}\in \mathcal {X}$ and is therefore feasible with (PM$_{A}^{{\mathcal {E}}}$). In addition,

$$\begin{aligned} f({\mathbf {x}})&\triangleq {\mathbb {P}}\left\{ \zeta \in {\mathbb {R}}^n \, \left| \, \zeta \in {\mathcal {K}}_{{\mathcal {E}}}, \left| \zeta ^{T} {\mathbf {x}}\right| \ \le 1 \right. \, \right\} ={\mathbb {P}} \left\{ \zeta \, \left| \, \zeta ^{T} U^{T} {\varSigma ^{-1}} U \zeta \le 1, \left| \zeta ^{T} {\mathbf {x}}\right| \le 1 \right. \right\} \\&= {\mathbb {P}}\left\{ \zeta \in {\mathbb {R}}^n \, \left| \, \Vert {\varSigma ^{-1/2}} U\zeta \Vert ^2_2 \le 1, \left| \zeta ^{T} {\mathbf {x}}\right| \ \le 1 \right. \right\} \\&= {\mathbb {P}} \left\{ U^{T} {\varSigma ^{1/2}} {\eta } \in {\mathbb {R}}^n\, \left| \, \Vert {\eta }\Vert ^2_2 \le 1, \left| (U^{T} {\varSigma ^{1/2} \eta })^{T} {\mathbf {x}}\right| \ \le 1 \right. \right\} \\&= {\mathbb {P}} \left\{ U^{T} {\varSigma ^{1/2} \eta } \in {\mathbb {R}}^n\, \left| \, \Vert {\eta }\Vert ^2_2 \le 1, \left| {\eta ^{T} \varSigma ^{1/2}} U {\mathbf {x}}\right| \ \le 1 \right. \right\} \\&= {\mathbb {P}} \left\{ {\eta } \in {\mathbb {R}}^n \, \left| \, {\eta } \in {\mathcal {K}}_{2} \, \left| \, {\eta }^{T} \varSigma ^{1/2} U {\mathbf {x}}\right| \ \le 1 \right. \right\} \triangleq g({\mathbf {x}}). \end{aligned}$$

$\square$

Proof of Proposition 6:

(a) The result follows by a transformation argument. We define a new variable ${ {\tilde{\zeta }} \in {\tilde{{\mathcal {K}}}}}$ such that ${ {\tilde{\zeta }}\triangleq \zeta -\mu }$ where ${ {\tilde{{\mathcal {K}}}} \triangleq \{ {\tilde{\zeta }}: \Vert {\tilde{\zeta }} \Vert _p \le \alpha \}}$. The set ${\tilde{{\mathbf {K}}}}({\mathbf {x}})$ can be defined as the following

$$\begin{aligned} {\tilde{{\mathbf {K}}}}({\mathbf {x}}) =\left\{ {\tilde{\zeta }} \, \left| \, {\tilde{\zeta }} \in {\tilde{{\mathcal {K}}}} \right. \right\} \, \bigcap \, \left\{ {\tilde{\zeta }} \, \left| \, {\tilde{\zeta }} \le T{\mathbf {x}}-\mu \right. \right\} . \end{aligned}$$

We first show that $\zeta \in {\mathbf {K}}({\mathbf {x}})$ if and only if ${\tilde{\zeta }} \in {\tilde{{\mathbf {K}}}}({\mathbf {x}})$. Suppose $\zeta \in {\mathbf {K}}({\mathbf {x}})$. Then $\zeta \in \mathcal {K}$ and $c({\mathbf {x}},\zeta ) = T{\mathbf {x}}-\zeta \ge 0$. If $\zeta \in \mathcal {K}$, then $\Vert \zeta -\mu \Vert _p \le \alpha$ or $\Vert {\tilde{\zeta }}\Vert _p \le \alpha$ where ${\tilde{\zeta }} = \zeta - \mu$. Furthermore, $T{\mathbf {x}}\ge \zeta$ can be rewritten as $T{\mathbf {x}}- \mu \ge \zeta - \mu$ or $T{\mathbf {x}}- \mu \ge {\tilde{\zeta }}$. It follows that

$$\begin{aligned} {\tilde{\zeta }} \in {\tilde{{\mathbf {K}}}}({\mathbf {x}}) \, = \, \left\{ {\tilde{\zeta }} \, \left| \, {\tilde{\zeta }} \in {\tilde{{\mathcal {K}}}} \right. \right\} \, \bigcap \, \left\{ {\tilde{\zeta }} \, \left| \, T{\mathbf {x}}- \mu \ge {\tilde{\zeta }}\right. \right\} . \end{aligned}$$

The reverse direction follows similarly. Consequently, ${\mathbb {P}}\left\{ \zeta \, \left| \, \zeta \in {\mathbf {K}}({\mathbf {x}}) \right. \right\} = {\mathbb {P}}\left\{ {\tilde{\zeta }} \, \left| \, {\tilde{\zeta }} \in {\tilde{{\mathbf {K}}}}({\mathbf {x}}) \right. \right\} .$ We now analyze the latter probability. It may be observed that the Minkowski functional associated with ${\tilde{{\mathcal {K}}}}$ is given by $\Vert {\tilde{\zeta }}\Vert _{{\tilde{{\mathcal {K}}}}} = \tfrac{1}{\alpha }\Vert {\tilde{\zeta }}\Vert _p$. Since ${T_{i,\bullet } {\mathbf {x}}-\mu _i \ge \delta >0 }$ for $i=1,\ldots ,d$, it follows that

$$\begin{aligned} {\tilde{{\mathbf {K}}}}({\mathbf {x}})&= \Bigg \{{\tilde{\zeta }} \, \left| \, \tfrac{1}{\alpha } \Vert {\tilde{\zeta }}\Vert _p \le 1 \right. \Bigg \} \, \bigcap \, \left\{ {\tilde{\zeta }} \, \left| \, \bigcap _{i=1}^d \tfrac{\max \{{\tilde{\zeta }}_i,0\}}{T_{i,\bullet } {\mathbf {x}}-\mu _i}\le 1 \right. \right\} \\&= \Bigg \{{\tilde{\zeta }} \, \left| \, \tfrac{1}{\alpha ^2}\Vert {\tilde{\zeta }} \Vert ^{2}_p \le 1 \right. \Bigg \} \, \bigcap \, \left\{ {\tilde{\zeta }} \, \left| \, \bigcap _{i=1}^d \left( \tfrac{\max \{{\tilde{\zeta }}_i,0\}}{T_{i,\bullet } {\mathbf {x}}-\mu _i}\right) ^2 \le 1 \right. \right\} \\&= \left\{ {\tilde{\zeta }} \, \left| \, \max \left\{ \tfrac{1}{\alpha ^2}\Vert {\tilde{\zeta }}\Vert _{p}^2,\left( \tfrac{\max \{{\tilde{\zeta }}_1,0\}}{T_{1,\bullet }{\mathbf {x}}-\mu _1}\right) ^2,\cdots , \left( \tfrac{\max \{{\tilde{\zeta }}_d,0\}}{T_{d,\bullet }{\mathbf {x}}-\mu _d}\right) ^2 \right\} \le 1 \right. \right\} . \end{aligned}$$

Since $g_i({\mathbf {x}},{\tilde{\zeta }}) \ \triangleq \ \left( \frac{\max \{{\tilde{\zeta }}_i,0\}}{T_{i,\bullet } {\mathbf {x}}-\mu _i}\right) ^2$ for $i=1,\ldots , d$ and $g_{d+1}({\mathbf {x}}, {\tilde{\zeta }}) \triangleq \tfrac{1}{\alpha ^2}\Vert {\tilde{\zeta }}\Vert ^{2}_{p}$ are PHFs with degree 2, then $g({{\mathbf {x}}}, {\tilde{\zeta }}) \triangleq \max \{ g_1({\mathbf {x}}, {\tilde{\zeta }}),\ldots ,g_{d+1}({\mathbf {x}},{\tilde{\zeta }}) \}$ is positively homogeneous with degree 2. By selecting $h(\zeta ) = 1$ and ${\varLambda } = {\tilde{{\mathbf {K}}}}({\mathbf {x}})$, we may invoke Lemma 2, leading to the following equality.

$$\begin{aligned} f({\mathbf {x}}) \, = \, \int _{{\tilde{{\mathbf {K}}}}({\mathbf {x}})} 1 \ d {\tilde{\zeta }} = \frac{1}{\mathrm {Vol}({\mathcal {K}})} \frac{1}{\varGamma (1+d/2)} \int _{{\mathbb {R}}^d} e^{-g({\mathbf {x}},\xi )} \ d\xi . \end{aligned}$$

(40)

The Eq. (40) can be rewritten as

$$\begin{aligned} f({\mathbf {x}})&= \int _{{\mathbb {R}}^d} \underbrace{\left( \mathcal {C} (2\pi \sigma ^2)^{d/2} e^{-g({\mathbf {x}},\xi ) +\frac{\Vert \xi \Vert ^2_{2}}{2\sigma ^2}}\right) }_{\triangleq F({\mathbf {x}},\xi )}\underbrace{\left( \tfrac{1}{(2\pi \sigma ^2)^{d/2}} {e^{-\tfrac{\Vert \xi \Vert ^2_{2}}{2\sigma ^2}}} \right) }_{\triangleq p(\xi )} \ d\xi \\&= \int _{{\mathbb {R}}^{d}} F({\mathbf {x}},\xi ) \ {{\tilde{p}}}(\xi ) \ d\xi = \mathcal {C} \ {\mathbb {E}}_{{{\tilde{p}}}(\xi )}[F({\mathbf {x}},\xi )], \text{ where } \mathcal {C} \triangleq \tfrac{1}{\text {Vol}({\mathcal {K}})} \ {\tfrac{1}{\varGamma (1+d/2)}}, \end{aligned}$$

(b) Omitted (similar to proof of Lemma 8 (a)).

(c) When ${\mathcal {K}}$ satisfies Assumption 2, the proof of Lemma 8(b) requires slight modification. Suppose $F({\mathbf {x}},\xi )$ and $p(\xi )$ are defined as in (a). Then we may define $\partial F({\mathbf {x}},\xi )$ as

$$\begin{aligned} \partial F({\mathbf {x}},\xi ) = {\left\{ \begin{array}{ll} \left( {\mathcal {C}}(2\pi \sigma ^2)^{{d}/2} \frac{2(\max \{\xi _i,0\})^2 T_{i,\bullet }^T}{(T_{i,\bullet } {\mathbf {x}}-\mu _i)^3} e^{-g_i({\mathbf {x}},\xi )+\frac{{\Vert \xi \Vert _{2}^2}}{2\sigma ^2}}\right) , &{} \xi \in \varXi _i({\mathbf {x}}), i = 1, \cdots , d \\ \left( -{\mathcal {C}}(2\pi \sigma ^2)^{{d}/2} e^{-g({\mathbf {x}},\xi ) +\frac{{\Vert \xi \Vert _{2}^2}}{2\sigma ^2}}\right) H({\mathbf {x}},\xi ), &{} \xi \in \varXi _0({\mathbf {x}})\\ \mathbf{0}. &{} \xi \in \varXi _{d+1}({\mathbf {x}}), \end{array}\right. } \end{aligned}$$

where $H({\mathbf {x}},\xi )$ denotes the Clarke generalized gradient of $g({\mathbf {x}},\xi )$, defined as in (17). Consequently, it follows that ${\mathbb {E}} \left[ \Vert G({\mathbf {x}},\xi )\Vert ^2\right]$ is bounded as follows.

$$\begin{aligned}&\quad {\mathbb {E}}\left[ \Vert G({\mathbf {x}},\xi )\Vert ^2\right] = \int _{{\mathbb {R}}^{d}} \Vert G({\mathbf {x}},\xi )\Vert ^2 {\tilde{p}}(\xi ) d\xi \nonumber \\&= \sum _{i=1}^{d} \int _{\varXi _i({\mathbf {x}})} \Vert G({\mathbf {x}},\xi )\Vert ^2 {\tilde{p}}(\xi ) d\xi + \int _{\varXi _{d+1}({\mathbf {x}})} \Vert \underbrace{G({\mathbf {x}},\xi )}_{ \ = \ \mathbf{0}}\Vert ^2 {\tilde{p}}(\xi ) d\xi \nonumber \\&\quad + \int _{\varXi _0({\mathbf {x}})} \Vert G({\mathbf {x}},\xi )\Vert ^2 {\tilde{p}}(\xi ) d\xi \nonumber \\&= \sum _{i=1}^{d}\int _{\varXi _i({\mathbf {x}})} \Vert G({\mathbf {x}},\xi )\Vert ^2 {\tilde{p}}(\xi ) d\xi , \end{aligned}$$

(41)

where the last equality follows from observing that $G({\mathbf {x}},\xi ) = 0$ for $\xi \in \varXi _{d+1}({\mathbf {x}})$ and the integral in (41) is zero because $\varXi _0({\mathbf {x}})$ is a measure zero set. It follows that

$$\begin{aligned} {\mathbb {E}}&\left[ \Vert G(x,\xi )\Vert ^2 \right] \\&=\sum _{i=1}^{d}\int _{\varXi _i(x)} 4\mathcal {C}^2(2\pi \sigma ^2)^d \frac{\Vert T_{i,\bullet }\Vert ^2}{(T_{i,\bullet } {\mathbf {x}}-\mu _i)^2} \left( \frac{\xi _k}{{T_{i,\bullet } {\mathbf {x}}-\mu _i}}\right) ^4 \\&\qquad e^{-\frac{2(\xi _i)^2}{(T_{i,\bullet } {\mathbf {x}}-\mu _i)^2}+\frac{\Vert \xi \Vert _{2}^2}{\sigma ^2}}\left( \tfrac{1}{(2\pi \sigma ^2)^{d/2}} e^{-\tfrac{\Vert \xi \Vert _{2}^2}{2\sigma ^2}}\right) d\xi ,\\&\le \sum _{i=1}^{d}\int _{\varXi _i(x)} 4\mathcal {C}^2(2\pi \sigma ^2)^d \frac{\Vert T_{i,\bullet }\Vert ^2}{\delta ^2}\left( \frac{\xi _k}{{T_{i,\bullet } {\mathbf {x}}-\mu _i}} \right) ^4 e^{-\frac{2(\xi _i)^2}{(T_{i,\bullet } {\mathbf {x}}-\mu _i)^2}+\frac{\Vert \xi \Vert _{2}^2}{\sigma ^2}}\left( \tfrac{1}{(2\pi \sigma ^2)^{d/2}} e^{-\tfrac{\Vert \xi \Vert _{2}^2}{2 \sigma ^2}}\right) d\xi \\&\le \sum _{i=1}^{d}\int _{\varXi _i(x)} 4\mathcal {C}^2(2\pi \sigma ^2)^d \frac{\Vert T_{i,\bullet }\Vert ^2}{\delta ^2}\left( \frac{\xi _i}{{T_{i,\bullet } {\mathbf {x}}-\mu _i}}\right) ^4\\&\qquad e^{-\left( 2-\tfrac{\alpha ^2}{\sigma ^2}\right) \frac{(\xi _i)^2}{(T_{i,\bullet } {\mathbf {x}}-\mu _i)^2}}\left( \tfrac{1}{(2\pi \sigma ^2)^{d/2}} ^{-\tfrac{\Vert \xi \Vert _{2}^2}{2\sigma ^2}}\right) d\xi \end{aligned}$$

where the first inequality follows from ${T_{i,\bullet } {\mathbf {x}}-\mu _i \ge \delta >0 }$ for all i, and the second inequality follows from $\xi \in \varXi _i({\mathbf {x}})$. It follows from Lemma 3 that given any $\alpha$, by choosing the variance $\sigma ^2$ of the random variable $\xi$ such that $\sigma ^2 = \alpha ^2$ leads to the bound ${{\mathbb {E}} \left[ \Vert G({\mathbf {x}},\xi )\Vert ^2 \right] \le 16\mathcal {C}^2(2\pi \sigma ^2)^d {\sum _{i=1}^d} \frac{\Vert T_{i,\bullet }\Vert ^2}{\delta ^2 e^2}}.$ $\square$

Proof of Lemma 10:

If ${{\tilde{G}}}({\mathbf {x}}_k,\xi ) \triangleq G({\mathbf {x}}_k,\xi ) - {\mathbb {E}}[G({\mathbf {x}}_k,\xi )]$, by the conditional independence of ${\tilde{G}}({\mathbf {x}}_k,\xi _j)$ and ${\tilde{G}}({\mathbf {x}}_k,\xi _{\ell })$ for $j \ne \ell$, we have

$$\begin{aligned}&{\mathbb {E}}\left[ \Vert {\bar{w}}_{G,k}\Vert ^2 \mid \mathcal {F}_k \right] = \frac{1}{N^2_k}{\mathbb {E}}\left[ \left\| \sum _{j=1}^{N_k} {\tilde{G}} ({\mathbf {x}}_k,\xi _j)\right\| ^2 \Bigg | \, \mathcal {F}_k \right] \nonumber \\&\quad = \frac{1}{N_k^2} {\mathbb {E}}\left[ \left[ \sum _{j=1}^{N_k} \Vert {\tilde{G}}({\mathbf {x}}_k,\xi _j)\Vert ^2 + \sum _{\ell \ne j} 2{\tilde{G}}({\mathbf {x}}_k,\xi _{\ell })^T {\tilde{G}}({\mathbf {x}}_k,\xi _j) \right] \, \Bigg | \, \mathcal {F}_k \right] \nonumber \\&\quad = \frac{1}{N_k}\left( {\mathbb {E}}\left[ \Vert {G}({\mathbf {x}}_k,\xi )\Vert ^2\, \left| \, \mathcal {F}_k \right. \right] \right. \nonumber \\&\qquad \left. + \Vert {\mathbb {E}}\left[ G({\mathbf {x}}_k,\xi ) \, \left| \, \mathcal {F}_k \right. \right] \Vert ^2 - 2{\mathbb {E}}\left[ G({\mathbf {x}}_k,\xi )\, \left| \, \mathcal {F}_k \right. \right] ^T{\mathbb {E}}\left[ G({\mathbf {x}}_k,\xi ) \, \left| \, \mathcal {F}_k \right. \right] \right) \nonumber \\&\quad = \frac{1}{N_k}\left( {\mathbb {E}}\left[ \Vert {G}({\mathbf {x}}_k,\xi ) \Vert ^2\, \left| \, \mathcal {F}_k \right. \right] - \Vert {\mathbb {E}} \left[ G({\mathbf {x}}_k,\xi ) \, \left| \, \mathcal {F}_k \right. \right] \Vert ^2 \right) \nonumber \\&\quad \le \frac{1}{N_k} {\mathbb {E}} \left[ \Vert {G}({\mathbf {x}}_k,\xi )\Vert ^2 \mid \mathcal {F}_k\right] . \end{aligned}$$

(42)

By (42) and Prop. 2, ${\mathbb {E}}\left[ \Vert {\bar{w}}_{G,k}\Vert ^2 \, \left| \, \mathcal {F}_k \right. \right] \ \le \ \frac{\mathcal {C}^2_{{\mathcal {K}}}(2\pi )^n}{eN_k} {\mathbb {E}}_{{{\tilde{p}}}}\left[ \Vert \xi \Vert ^2\right]$ for Setting A. Similarly, for Setting B, by Lemma. 8,

$$\begin{aligned} {{\mathbb {E}}[\Vert {\bar{w}}_{G,k}\Vert ^2 \mid \mathcal {F}_k ] \le 16\mathcal {C}^2(2\pi \sigma ^2)^d {\sum _{i=1}^d} \frac{\Vert T_{i,\bullet }\Vert ^2}{\delta ^2 e^2 N_k} }. \end{aligned}$$

In addition, for Setting A, ${\mathbb {E}}\left[ \Vert {\bar{w}}_{f,k}\Vert ^2 \, \left| \, \mathcal {F}_k \right. \right] \le \frac{2(\mathcal {C}_{{\mathcal {K}}}^2(2\pi )^n+1)}{N_k}$ while for Setting B, we obtain that ${\mathbb {E}}\left[ \, \Vert {\bar{w}}_{f,k}\Vert ^2 \, \left| \, \mathcal {F}_k \right. \right] \le \frac{\mathcal {C}^2(2\pi \sigma ^2)^d}{N_k}.$ $\square$

Proof of Lemma 11:

(Setting A) Consider ${\bar{w}}_k$, where ${\bar{w}}_k$ is defined as ${\bar{w}}_k \triangleq \frac{-(G_k+{\bar{w}}_{G,k})}{(f({\mathbf {x}}_k) + {\bar{w}}_{f,k})^{2}+\epsilon _k}- \frac{-G_k}{(f({\mathbf {x}}_k))^{2}}.$ We have that

$$\begin{aligned} \Vert {\bar{w}}_k\Vert ^2&= \left\| \frac{-(G_k+{\bar{w}}_{G,k})}{(f({\mathbf {x}}_k) + {\bar{w}}_{f,k})^{2}+\epsilon _k}- \frac{-G_k}{(f({\mathbf {x}}_k))^{2}} \right\| ^2 \\&= \left\| \frac{-(G_k+{\bar{w}}_{G,k})}{(f({\mathbf {x}}_k) + {\bar{w}}_{f,k})^{2} +\epsilon _k}-\frac{{-}(G_k+{\bar{w}}_{G,k})}{(f({\mathbf {x}}_k))^{2}+\epsilon _k} +\frac{{-}(G_k+{\bar{w}}_{G,k})}{(f({\mathbf {x}}_k))^{2}+\epsilon _k}\right. \\&\quad \left. - \frac{{-}G_k}{(f({\mathbf {x}}_k))^{2}+\epsilon _k} + \frac{{-}G_k}{(f({\mathbf {x}}_k))^{2}+\epsilon _k}- \frac{{-}G_k}{(f({\mathbf {x}}_k))^{2}} \right\| ^2 \\&\le 3\left\| G_k - G_k+{\bar{w}}_{G,k}\right\| ^2 \frac{1}{((f({\mathbf {x}}_k))^{2} +\quad \epsilon _k)^2}\\&\quad +3\left\| G_k+{\bar{w}}_{G,k}\right\| ^2 \left\| \frac{1}{(f({\mathbf {x}}_k))^{2}+\epsilon _k} -\frac{1}{(f({\mathbf {x}}_k) + {\bar{w}}_{f,k})^{2}+\epsilon _k}\right\| ^2 \\&\quad + 3\left\| G_k\right\| ^2 \left\| \frac{1}{(f({\mathbf {x}}_k))^{2}} -\frac{1}{(f({\mathbf {x}}_k))^{2} +\epsilon _k}\right\| ^2 \\&\le 3\left\| {\bar{w}}_{G,k}\right\| ^2 \frac{1}{((f({\mathbf {x}}_k))^{2}+\epsilon _k)^2}\\&\quad +3\left\| G_k+{\bar{w}}_{G,k}\right\| ^2 \left\| \frac{(2f({\mathbf {x}}_k)+{\bar{w}}_{f,k}) {\bar{w}}_{f,k}}{((f({\mathbf {x}}_k))^{2}+\epsilon _k)((f({\mathbf {x}}_k) + {\bar{w}}_{f,k})^{2} +\epsilon _k)}\right\| ^2 \\&\quad + 3\left\| G_k\right\| ^2 \left\| \frac{\epsilon _k}{(f({\mathbf {x}}_k))^{2}((f({\mathbf {x}}_k))^{2} +\epsilon _k)}\right\| ^2 \\&\le 3\left\| {\bar{w}}_{G,k}\right\| ^2 \frac{1}{\epsilon _f^{{4}}}+{3}\left\| G_k+{\bar{w}}_{G,k}\right\| ^2 \left\| \frac{(2f({\mathbf {x}}_k)+{\bar{w}}_{f,k})}{\epsilon _f^{2} \epsilon _k}\right\| ^2\Vert {\bar{w}}_{f,k}\Vert ^2 + 3\left\| G_k\right\| ^2 \left( \frac{\epsilon _k^2}{\epsilon _f^{{8}}}\right) \\&\le 3\left\| {\bar{w}}_{G,k}\right\| ^2 \frac{1}{\epsilon _f^4}+{3} \left\| G_k+{\bar{w}}_{G,k}\right\| ^2 \frac{(8f^2({\mathbf {x}}_k)\Vert {\bar{w}}_{f,k}\Vert ^2+2\Vert {\bar{w}}_{f,k}\Vert ^4)}{\epsilon _f^{4} \epsilon ^2_k} + \left\| G_k\right\| ^2 \left( \frac{\epsilon _k^2}{\epsilon _f^{8}}\right) , \end{aligned}$$

where $f({\mathbf {x}}_k) \ge \epsilon _f$ for every ${\mathbf {x}}_k \in \mathcal {X}.$ Taking conditional expectations and recalling the independence of ${\bar{w}}_{f,k}$ and ${\bar{w}}_{G,k}$ conditional on $\mathcal {F}_k$, the following bound emerges.

$$\begin{aligned} {\mathbb {E}}&\left[ \Vert {\bar{w}}_k\Vert ^2 \, \bigg | \, \mathcal {F}_k \, \right] \le 3{\mathbb {E}}\left[ \left\| {\bar{w}}_{G,k}\right\| ^2 \, \bigg | \, \mathcal {F}_k\, \right] \frac{1}{\epsilon _f^2} \\&+{3}{\mathbb {E}}\left[ \left\| G_k+{\bar{w}}_{G,k}\right\| ^2 \frac{(8f^2({\mathbf {x}}_k)\Vert {\bar{w}}_{f,k}\Vert ^2+2\Vert {\bar{w}}_{f,k}\Vert ^4)}{\epsilon _f^{{4}} \epsilon _k^2}\, \bigg | \, \mathcal {F}_k \right] + 3{\mathbb {E}}\left[ \left\| G_k\right\| ^2 \left( \frac{\epsilon _k^2}{\epsilon _f^{{8}}}\right) \, \bigg | \, \mathcal {F}_k \right] \\&\le 3\frac{\nu _G^2}{\epsilon _f^2N_k} +{3}{\mathbb {E}}\left[ \left\| G_k+{\bar{w}}_{G,k}\right\| ^2 \, \bigg | \, \mathcal {F}_k \right] {\mathbb {E}}\left[ \frac{(8f^2({\mathbf {x}}_k)\Vert {\bar{w}}_{f,k}\Vert ^2+2\Vert {\bar{w}}_{f,k}\Vert ^4)}{\epsilon _f^{{4}} \epsilon _k^2}\, \bigg | \, \mathcal {F}_k \right] \\&\quad + \left( \frac{3\epsilon _k^2M_G^2}{\epsilon _f^{{8}}}\right) \\&\le 3\frac{\nu _G^2}{\epsilon _f^2N_k}+{3} M_G^2 \frac{8f^2({\mathbf {x}}_k)\nu _f^2}{\epsilon _f^{4} \epsilon _k^2N_k}+{3}M_G^2{\mathbb {E}}\left[ \frac{\Vert {\bar{w}}_{f,k}\Vert ^4}{\epsilon _f^{4} \epsilon _k^2}\, \bigg | \, \mathcal {F}_k \right] + 3\left( \frac{\epsilon _k^2M_G^2}{\epsilon _f^{{8}}}\right) , \end{aligned}$$

where $\Vert G_k\Vert ^2 = \Vert {\mathbb {E}}\left[ G({\mathbf {x}}_k,\xi ) \, \left| \, \mathcal {F}_k \right. \right] \Vert ^2 \le {\mathbb {E}}\left[ \Vert G({\mathbf {x}}_k,\xi )\Vert ^2 \, \left| \, \right. \mathcal {F}_k\right] \le M_G^2$ by Jensen’s inequality. From Prop. 2(b,c), $| F({\mathbf {x}},\xi )| \le {M_F}$ for any ${\mathbf {x}}, \xi$, implying that

$$\begin{aligned} \Vert {\bar{w}}_{f,k}\Vert ^2 = \left\| \frac{\sum _{j=1}^{N_k} F({\mathbf {x}}_k,\xi _j)}{N_k} - f({\mathbf {x}}_k)\right\| ^2&\le 2 \left\| \frac{\sum _{j=1}^{N_k} F({\mathbf {x}}_k,\xi _j)}{N_k}\right\| ^2 + 2f^2({\mathbf {x}}_k) \\&\le 2({M_F}^2+1). \end{aligned}$$

Consequently, by recalling that $\epsilon _k = 1/N_k^{1/4}$, the following holds a.s.

$$\begin{aligned} {\mathbb {E}}\left[ \Vert {\bar{w}}_k\Vert ^2 \, \bigg | \, \mathcal {F}_k \right]&\le \frac{3\nu _G^2}{\epsilon _f^2N_k}+{24}M_G^2 \frac{f^2({\mathbf {x}}_k)\nu _f^2}{\epsilon _f^{{4}} \epsilon _k^2N_k} +{3}{\mathbb {E}}\left[ \frac{\Vert {\bar{w}}_{f,k}\Vert ^4}{\epsilon _f^4 \epsilon _k^2}\, \bigg | \, {\mathbf {x}}_k \right] + \left( \frac{\epsilon _k^2M_G^2}{\epsilon _f^8}\right) \\&\le \frac{\nu _G^2}{\epsilon _f^2N_k}+{24}M_G^2 \frac{f^2({\mathbf {x}}_k)\nu _f^2}{\epsilon _f^{{4}} \epsilon _k^2N_k}+\frac{{6}({M_F}^2+1)M_G^2\nu _f^2}{\epsilon _f^4 \epsilon _k^2 N_k}\\&\le \frac{3\nu _G^2}{\epsilon _f^2\sqrt{N_k}}+M_G^2 \frac{{24}f^2({\mathbf {x}}_k)\nu _f^2}{\epsilon _f^{{4}} \sqrt{N_k}}+\frac{{6}(M_F^2+1)\nu _f^2}{\epsilon _f^{{4}} \sqrt{N_k}} + \left( \frac{3M_G^2}{\epsilon _f^{{8}}\sqrt{N_k}}\right) \\&\triangleq \frac{\nu ^2}{\sqrt{N_k}}, \text{ where } \nu ^2 \triangleq \frac{3\nu _G^2}{\epsilon _f^2}+M_G^2 \frac{{24}\nu _f^2}{\epsilon _f^{{4}} }+\frac{{6}({M_F}^2+1)\nu _f^2}{\epsilon _f^{{4}}} + \left( \frac{3M_G^2}{\epsilon _f^{{8}}}\right) . \end{aligned}$$

(Setting B) Since ${\bar{w}}_k \triangleq {\frac{{-}(G_{k}+{\bar{w}}_{G,k})}{(f({\mathbf {x}}_k) + {\bar{w}}_{f,k})+\epsilon _k}+ \frac{G_{k}}{f({\mathbf {x}}_k)}}$ and

$$\begin{aligned} \Vert {\bar{w}}_{k}\Vert ^2&= \left\| \frac{{-}(G_{k}+{\bar{w}}_{G,{k}})}{(f({\mathbf {x}}_k) + {\bar{w}}_{f,{k}})+\epsilon _k}- \frac{{-}G_k}{(f({\mathbf {x}}_k))} \right\| ^2 \\&= \left\| \frac{{-}(G_{k}+{\bar{w}}_{G,{k}})}{(f({\mathbf {x}}_k) + {\bar{w}}_{f,k}) +\epsilon _k}-\frac{{-}(G_{k}+{\bar{w}}_{G,{k}})}{f({\mathbf {x}}_k)+\epsilon _k} +\frac{{-}(G_{k}+{\bar{w}}_{G,k})}{f({\mathbf {x}}_k)+\epsilon _k}\right. \\&\quad \left. - \frac{{-}G_{k}}{f({\mathbf {x}}_k)+\epsilon _k} + \frac{{-}G_{k}}{f({\mathbf {x}}_k)+\epsilon _k}- \frac{{-}G_{k}}{f({\mathbf {x}}_k)} \right\| ^2 \\&\le 3\left\| G_{k} - G_{k}+{\bar{w}}_{G,{k}}\right\| ^2 \frac{1}{{(f({\mathbf {x}}_k)+\epsilon _k)^2}} \\&\quad +3\left\| G_{k}+{\bar{w}}_{G,{k}}\right\| ^2 \left\| \frac{1}{f({\mathbf {x}}_k)+\epsilon _k} -\frac{1}{(f({\mathbf {x}}_k) + {\bar{w}}_{f,{k}})+\epsilon _k}\right\| ^2 \\&\quad + 3\left\| G_{k}\right\| ^2 \left\| \frac{1}{f({\mathbf {x}}_k)} -\frac{1}{f({\mathbf {x}}_k)+\epsilon _k} \right\| ^2 \\&\le 3\left\| {\bar{w}}_{G,{k}}\right\| ^2 \frac{1}{{(f({\mathbf {x}}_k)+\epsilon _k)^2}} +3\left\| G_{k}+ {\bar{w}}_{f,{k}}\right\| ^2 \\&\qquad \left\| \frac{{\bar{w}}_{f,k}}{(f({\mathbf {x}}_k) +\epsilon _k)(\underbrace{(f({\mathbf {x}}_k) + {\bar{w}}_{f,{k}})}_{{\ge 0, F({\mathbf {x}}_k,\xi ) \ge 0}} +\epsilon _k)}\right\| ^2 \\&\quad + 3\left\| G_{k}\right\| ^2 \left\| \frac{\epsilon _k}{f({\mathbf {x}}_k)(f({\mathbf {x}}_k)+\epsilon _k)} \right\| ^2 \\&\le 3\left\| {\bar{w}}_{G,{k}}\right\| ^2 \frac{1}{{\epsilon _f^2}} +3\left\| G_k+{\bar{w}}_{G,{k}}\right\| ^2 \left\| \frac{1}{\epsilon _f \epsilon _k}\right\| ^2\Vert {\bar{w}}_{f,{k}}\Vert ^2 + 3\left\| G_{k}\right\| ^2 \left( \frac{\epsilon _k^2}{\epsilon _f^{4}}\right) \\&\le 3\left\| {\bar{w}}_{G,{k}}\right\| ^2 \frac{1}{\epsilon _f^2}+3\left\| G_{k}+{\bar{w}}_{G,{k}}\right\| ^2 \frac{\Vert {\bar{w}}_{f,{k}}\Vert ^2}{\epsilon _f^{2} \epsilon ^2_k} + \left\| G_{k}\right\| ^2 \left( \frac{\epsilon _k^2}{\epsilon _f^{4}}\right) , \end{aligned}$$

where $f({\mathbf {x}}_k) \ge \epsilon _f$ and for every ${\mathbf {x}}_k \in \mathcal {X}.$ Taking expectations conditioned on $\mathcal {F}_k$ and recalling the independence of ${\bar{w}}_{f,k}$ and ${\bar{w}}_{G,k}$ conditional on $\mathcal {F}_k$, we have the following bound.

$$\begin{aligned}&{\mathbb {E}}\left[ \Vert {\bar{w}}_{k}\Vert ^2 \, \bigg | \, \mathcal {F}_k\right] \\&\quad \le \left( 3{\mathbb {E}}\left[ \left\| {\bar{w}}_{G,{k}}\right\| ^2 \,\bigg | \,\mathcal {F}_k\right] \frac{1}{\epsilon _f^2}\right. \\&\qquad \left. +3{\mathbb {E}}\left[ \left\| G_{k}+{\bar{w}}_{G,{k}}\right\| ^2 \frac{\Vert {\bar{w}}_{f,{k}}\Vert ^2}{\epsilon _f^{2} \epsilon _k^2}\bigg | \mathcal {F}_k\right] + 3{\mathbb {E}}\left[ \left\| G_{k}\right\| ^2 \left( \frac{\epsilon _k^2}{\epsilon _f^{4}}\right) \,\bigg | \,\mathcal {F}_k\right] \right) \\&\quad \le \left( 3\frac{\nu _{G}^2}{\epsilon _f^2N_k} + 3{\mathbb {E}}\left[ \left\| G_{k}+{\bar{w}}_{G,{k}}\right\| ^2 \,\bigg | \,\mathcal {F}_k\right] {\mathbb {E}}\left[ \frac{\Vert {\bar{w}}_{f,{k}}\Vert ^2}{\epsilon _f^{2} \epsilon _k^2} \,\bigg | \,\mathcal {F}_k\right] + 3\left( \frac{\epsilon _k^2M_{G}^2}{\epsilon _f^{4}}\right) \right) \\&\quad \le \left( 3\frac{\nu _{G}^2}{\epsilon _f^2N_k}+3M_G^2 \frac{\nu _{f}^2}{\epsilon _f^{2} \epsilon _k^2N_k}+ 3\left( \frac{\epsilon _k^2M_{G}^2}{\epsilon _f^{4}}\right) \right) . \end{aligned}$$

By selecting $\epsilon _k = 1/N_k^{1/4}$, we have that

$$\begin{aligned} {\mathbb {E}}\left[ \Vert {\bar{w}}_{k}\Vert ^2 \, \bigg | \, \mathcal {F}_k\right]&\le \frac{\nu ^2}{\sqrt{N_k}}, \text{ where } \nu ^2 \triangleq \left( 3\frac{\nu _{G}^2}{\epsilon _f^2}+3M_G^2 \frac{\nu _{f}^2}{\epsilon _f^{2} }+ 3\left( \frac{M_{G}^2}{\epsilon _f^{4}}\right) \right) . \end{aligned}$$

$\square$

Proof of Proposition 7:

(i) Using the update rule of ${\mathbf {x}}_{k+1}$ and the fact that ${\mathbf {x}}^*=\varPi _{{\mathcal {X}}} [{\mathbf {x}}^*]$, for any $d_k + {\bar{w}}_{k}$ where $d_k \in \partial h({\mathbf {x}}_k)$ and $k \ge 1$,

$$\begin{aligned} {1\over 2}\Vert {\mathbf {x}}_{k+1}-{\mathbf {x}}^*\Vert ^2&{ = } {1\over 2}\Vert \varPi _{{\mathcal {X}}} ({\mathbf {x}}_k-\gamma _k {(d_k + {\bar{w}}_k)})-\varPi _{\mathcal {X}}({\mathbf {x}}^*))\Vert ^2\\&\le {1\over 2}\Vert {\mathbf {x}}_k-\gamma _k {(d_k+{\bar{w}}_k)}-{\mathbf {x}}^*\Vert ^2\\&={1\over 2}\Vert {\mathbf {x}}_k-{\mathbf {x}}^*\Vert ^2+{1\over 2}\gamma _k^2\Vert {d_k + {\bar{w}}_k}\Vert ^2-\gamma _k({\mathbf {x}}_k-{\mathbf {x}}^*)^T({d_k}+{{\bar{w}}}_{k}), \end{aligned}$$

where in the second inequality, we employ the non-expansivity of projection operator. Now by using the convexity of h, we obtain:

$$\begin{aligned} 2\gamma _k (h({\mathbf {x}}_k)-h({\mathbf {x}}^*))&\le \left( \Vert {\mathbf {x}}_k-{\mathbf {x}}^*\Vert ^2-\Vert {\mathbf {x}}_{k+1}-{\mathbf {x}}^*\Vert ^2\right) +{\Vert {d_k+{\bar{w}}_k}\Vert ^2\gamma _k^2}\\&\quad -{2}\gamma _k{{\bar{w}}}_{k}^T({\mathbf {x}}_k-{\mathbf {x}}^*)\\&\le \left( \Vert {\mathbf {x}}_k-{\mathbf {x}}^*\Vert ^2-\Vert {\mathbf {x}}_{k+1}-{\mathbf {x}}^*\Vert ^2\right) +{\Vert {d_k+{\bar{w}}_k} \Vert ^2\gamma _k^2}\\&\quad + \gamma _k^2 \Vert {\mathbf {x}}_k-{\mathbf {x}}^*\Vert ^2 + \Vert {\bar{w}}_k\Vert ^2, \end{aligned}$$

where we use $a^Tb\le {1\over 2}\Vert a\Vert ^2+{1\over 2}\Vert b\Vert ^2$. Now by summing from $k = {\widehat{K}}$ to $K-1$, where ${\widehat{K}}$ is an integer satisfying $0 \le {\widehat{K}} < K-1$, we obtain the next inequality.

$$\begin{aligned} \sum _{k={\widehat{K}}}^{K-1} 2\gamma _k (h({\mathbf {x}}_k)-h({\mathbf {x}}^*))&\le {\Vert {\mathbf {x}}_{{{\hat{K}}}}-{\mathbf {x}}^*\Vert ^2}+\sum _{k={\widehat{K}}}^{K-1} \gamma _k^2({\Vert {d_k+{\bar{w}}_k}\Vert ^2}+ \Vert {\mathbf {x}}_k-{\mathbf {x}}^*\Vert ^2) +\Vert {\bar{w}}_k\Vert ^2. \end{aligned}$$

Dividing both sides by $2\sum _{k={\widehat{K}}}^{K-1} \gamma _k$, taking expectations on both sides, and invoking Lemma 11 which leads to ${\mathbb {E}}[\Vert {{\bar{w}}}_{k}\mid {\mathcal {F}}_k\Vert ]^2\le {\nu ^2\over \sqrt{N_k}}$ and the bound of the subgradient, i.e., ${\mathbb {E}}[\Vert d_k + {\bar{w}}_k\Vert ^2]\le M_G^2$, we obtain the following bound.

$$\begin{aligned} {\mathbb {E}}&\left[ \frac{\sum _{k={\widehat{K}}}^{K-1} 2\gamma _k (h({\mathbf {x}}_k)-h({\mathbf {x}}^*))}{\sum _{k={\widehat{K}}}^{K-1} 2\gamma _k}\right] \nonumber \\&\le {\mathbb {E}}\left[ \frac{{\Vert {\mathbf {x}}_{{{\hat{K}}}}-{\mathbf {x}}^*\Vert ^2} +\sum _{k={\widehat{K}}}^{K-1} \gamma _k^2{\Vert {d_k+{\bar{w}}_k}\Vert ^2} + \sum _{k={\widehat{K}}}^{K-1} \gamma _k^2 \Vert {\mathbf {x}}_k-{\mathbf {x}}^*\Vert ^2 + \sum _{k={\widehat{K}}}^{K-1} \Vert {\bar{w}}_k\Vert ^2}{\sum _{k={\widehat{K}}}^{K-1}2 \gamma _k}\right] \end{aligned}$$

(43)

$$\begin{aligned}&\le \frac{{\mathbb {E}}[\Vert x_{{\widehat{K}}}-x^*\Vert ^2]}{\sum _{k={\widehat{K}}}^{K-1}2\gamma _k } + \frac{\sum _{k={\widehat{K}}}^{K-1}\gamma _k^2 (M_G^2 + B^2)}{\sum _{k={\widehat{K}}}^{K-1}2\gamma _k }+ \frac{\sum _{k={\widehat{K}}}^{K-1} \tfrac{\nu ^2}{\sqrt{N_k}}}{\sum _{k={\widehat{K}}}^{K-1}2\gamma _k}. \end{aligned}$$

(44)

By utilizing Jensen’s inequality, we obtain that

$$\begin{aligned} {\mathbb {E}}\left[ (h({\bar{x}}_{{\widehat{K}},K}-h({\mathbf {x}}^*))\right] \le {\mathbb {E}}\left[ \frac{\sum _{k={\widehat{K}}}^{K-1} 2\gamma _k (h({\mathbf {x}}_k)-h({\mathbf {x}}^*))}{\sum _{k={\widehat{K}}}^{K-1} 2\gamma _k}\right] , \end{aligned}$$

where ${\bar{x}}_{{\widehat{K}},K} \triangleq \tfrac{\sum _{k={\widehat{K}}}^{K-1} \gamma _k x_k}{\sum _{k={\widehat{K}}}^{K-1} \gamma _k},$ which when combined with (44) leads to (30). $\square$