**3. Proofs of Main Results**

As discussed in the previous section, Proposition 3 combined with (8), straightforwardly implies the thesis of Theorem 1. In what follows, we shall focus on the proof of Proposition 3, for which we need to find the dominating points *t*0 by solving the two-layer minimization problem (9).

The solution of quadratic programming problem of the form (7) (inner minimization problem of (9)) has been well understood; for example, Hashorva (2005); Hashorva and Hüsler (2002) (see also Lemma 2.1 of De¸bicki et al. (2018)). For completeness and for reference, we present below Lemma 2.1 of De¸bicki et al. (2018) for the case where *d* = 2.

We introduce some more notation. If *I* ⊂ {1, <sup>2</sup>}, then for a vector *a* ∈ R<sup>2</sup> we denote by *aI* = (*ai*, *i* ∈ *I*) a sub-block vector of *a*. Similarly, if further *J* ⊂ {1, <sup>2</sup>}, for a matrix *M* = (*mij*)*<sup>i</sup>*,*j*∈{1,2} ∈ R2×<sup>2</sup> we denote by *MI J*= *MI*,*<sup>J</sup>* = (*mij*)*<sup>i</sup>*∈*I*,*j*∈*<sup>J</sup>* the sub-block matrix of *M* determined by *I* and *J*. Further, write *M*−<sup>1</sup> *I I* = (*MI I*)−<sup>1</sup> for the inverse matrix of *MI I* whenever it exists.

**Lemma 5.** *Let M* ∈ R2×<sup>2</sup> *be a positive definite matrix. If b* ∈ R<sup>2</sup> \ (−∞, <sup>0</sup>]<sup>2</sup>*, then the quadratic programming problem*

> *PM*(*b*) : *Minimise x*'*M*−1*x under the linear constraint x* ≥ *b*

*has a unique solution b and there exists a unique non-empty index set I* ⊆ {1, 2} *such that*

$$\begin{aligned} \tilde{\mathbf{b}}\_{I} &= \mathbf{b}\_{I} \neq \mathbf{0}\_{I\prime} & \mathcal{M}\_{II}^{-1} \mathbf{b}\_{I} &> \mathbf{0}\_{I\prime} \\ \text{and} & \quad \text{if } l^{\mathbb{C}} = \{1, 2\} \; \mid \; I \neq \mathcal{O}, \; then \; \tilde{\mathbf{b}}\_{I^{\mathbb{C}}} = \mathcal{M}\_{I^{\mathbb{C}}} \mathcal{M}\_{II}^{-1} \mathbf{b}\_{I} \ge \mathbf{b}\_{I^{\mathbb{C}}}. \end{aligned}$$

*Furthermore,* 
$$\begin{array}{rclcrcl}\min\_{\mathbf{x}\geq\mathbf{b}}\mathbf{x}^{\top}M^{-1}\mathbf{x}&=&\widetilde{\mathbf{b}}^{\top}M^{-1}\widetilde{\mathbf{b}}=\mathbf{b}\_{I}^{\top}M\_{II}^{-1}\mathbf{b}\_{I}>0,\\ &\mathbf{x}^{\top}M^{-1}\widetilde{\mathbf{b}}&=&\mathbf{x}\_{I}^{\top}M\_{II}^{-1}\widetilde{\mathbf{b}}\_{I}=\mathbf{x}\_{I}^{\top}M\_{II}^{-1}\mathbf{b}\_{I}&\forall\mathbf{x}\in\mathbb{R}^{2}.\end{array}$$

For the solution of the quadratic programming problem (7) a suitable representation for *g*(*<sup>t</sup>*,*<sup>s</sup>*) is worked out in the following lemma.

For 1 > *ρ* > *μ*1/*μ*2, let *D*2 = {(*<sup>t</sup>*,*<sup>s</sup>*) : *<sup>w</sup>*1(*s*) ≤ *t* ≤ *f*1(*s*)} and *D*1 = (0, ∞)<sup>2</sup> \ *D*2 , with boundary functions given by

$$f\_1(\mathbf{s}) = \frac{\rho - 1}{\mu\_1} + \frac{\rho \mu\_2}{\mu\_1} \mathbf{s}\_\prime \ w\_1(\mathbf{s}) = \frac{\mathbf{s}}{\rho + (\rho \mu\_2 - \mu\_1)\mathbf{s}'} \quad \mathbf{s} \ge \mathbf{0},\tag{13}$$

and the unique intersection point of *f*1(*s*), *<sup>w</sup>*1(*s*),*<sup>s</sup>* ≥ 0, given by

$$\mathbf{s}\_1^\* = \mathbf{s}\_1^\*(\rho) := \frac{1-\rho}{\rho\mu\_2 - \mu\_1},\tag{14}$$

as depicted in Figure 1.

**Figure 1.** Partition of (0, ∞)<sup>2</sup> into *D*1, *D*2.

### **Lemma 6.** *Let g*(*<sup>t</sup>*,*<sup>s</sup>*), *t*,*<sup>s</sup>* > 0 *be given as in* (7)*. We have:*

*(i) If* −1 < *ρ* ≤ *μ*1/*μ*2, *then*

$$\lg(t,s) = \gcd(t,s), \quad (t,s) \in (0,\infty)^2.$$

*(ii) If* 1 > *ρ* > *μ*1/*μ*2, *then*

$$\mathcal{g}(t,s) = \begin{cases} \mathcal{g}^{\mathfrak{z}}(t,s), & \acute{\mathfrak{z}}'(t,s) \in D\_1 \\ \mathcal{g}\_2(s), & \acute{\mathfrak{z}}'(t,s) \in D\_2. \end{cases}$$

*Moreover, we have g*3(*f*1(*s*),*<sup>s</sup>*) = *g*3(*<sup>w</sup>*1(*s*),*<sup>s</sup>*) = *g*2(*s*) *for all s* ≥ *s*∗1*.*
