*3.1. Proof of Proposition 3*

We shall discuss in order the case when −1 < *ρ* < 0 and the case when 0 ≤ *ρ* < 1 in the following two subsections. In both scenarios we shall first derive the minimizers of the function *g*(*<sup>t</sup>*,*<sup>s</sup>*) on regions *A* and *B* (see (10)) separately and then look for a global minimizer by comparing the two minimum values. For clarity some scenarios are analysed in forms of lemmas.

### 3.1.1. Case −1 < *ρ* < 0

By Lemma 6, we have that

$$\mathbf{g}(t, \mathbf{s}) = \mathbf{g}\_3(t, \mathbf{s}), \quad (t, \mathbf{s}) \in (0, \infty)^2.$$

We shall derive the minimizers of *g*3(*<sup>t</sup>*,*<sup>s</sup>*) on *A*, *B* separately. Minimizers of *g*3(*<sup>t</sup>*,*<sup>s</sup>*) on *A*. We have, for any fixed *s*,

$$\frac{\partial \mathcal{g}\_3(t,s)}{\partial t} = \frac{\partial \mathcal{g}\_A(t,s)}{\partial t} = 0 \iff (\mu\_1 t + 1 - \rho - \rho \mu\_2 \mathbf{s})(\mu\_1 t - (2\mu\_1 \rho^2 - \rho \mu\_2)\mathbf{s} + \rho - 1) = 0, \quad \forall t \in (0, 1)$$

where the representation (11) is used. Two roots of the above equation are:

$$t\_1 t\_1 = t\_1(s) := \frac{\rho - 1 + \rho \mu\_2 s}{\mu\_1}, \quad t\_2 = t\_2(s) := \frac{1 - \rho + (2\mu\_1 \rho^2 - \rho \mu\_2)s}{\mu\_1}.\tag{15}$$

Note that, due to the form of the function *gA*(*<sup>t</sup>*,*<sup>s</sup>*) given in (11), for any fixed *s*, there exists a unique minimizer of *gA*(*<sup>t</sup>*,*<sup>s</sup>*) on *A* which is either an inner point *t*1 or *t*2 (the one that is larger than *s*), or a boundary

point *s*. Next, we check if any of *ti*, *i* = 1, 2, is larger than *s*. Since *ρ* < 0, *t*1 < 0 < *t*2. So we check if *t*2 > *s*. It can be shown that

$$t\_2 > s \quad \Leftrightarrow \ (\mu\_1 + \rho \mu\_2 - 2\mu\_1 \rho^2)s < 1 - \rho. \tag{16}$$

Two scenarios *μ*1 + *ρμ*2 − <sup>2</sup>*μ*1*ρ*<sup>2</sup> ≤ 0 and *μ*1 + *ρμ*2 − <sup>2</sup>*μ*1*ρ*<sup>2</sup> > 0 will be distinguished. Scenario *μ*1 + *ρμ*2 − <sup>2</sup>*μ*1*ρ*<sup>2</sup> ≤ 0. We have from (16) that

$$t\_1 < 0 < \text{s} < t\_{2\prime}$$

and thus

$$\inf\_{(t,s)\in\overline{A}}\mathbb{g}\_3(t,s) = \inf\_{s>0} f\_A(s),$$

where

$$f\_A(\mathbf{s}) := \mathbf{g}\_A(t\_2(\mathbf{s}), \mathbf{s}) = \frac{(1 + \mu\_2 \mathbf{s})^2}{\mathbf{s}} + 4\mu\_1((1 - \rho) + (\rho^2 \mu\_1 - \rho \mu\_2)\mathbf{s}).$$

Next, since

$$f\_A'(s) = 0 \quad \Leftrightarrow \ s\_A = s\_A(\rho) := \frac{1}{|\mu\_2 - 2\rho\mu\_1|} = \frac{1}{\mu\_2 - 2\rho\mu\_1} > 0,\tag{17}$$

the unique minimizer of *g*3(*<sup>t</sup>*,*<sup>s</sup>*) on *A* is given by (*tA*,*sA*) with

$$t\_A := t\_2(\mathbf{s}\_A) = \frac{1 - 2\rho}{\mu\_1}.$$

Scenario *μ*1 + *ρμ*2 − <sup>2</sup>*μ*1*ρ*<sup>2</sup> > 0. We have from (16) that

$$t\_1 < 0 < s < t\_2 \iff s < \frac{1 - \rho}{\mu\_1 + \rho \mu\_2 - 2\mu\_1 \rho^2} = \frac{1 - \rho}{\rho(\mu\_2 - \mu\_1 \rho) + \mu\_1(1 - \rho^2)} =: s^{\*\*}(\rho) = s^{\*\*},\tag{18}$$

and in this case,

$$\inf\_{(t,s)\in\overline{A}}\mathbb{g}\_3(t,\mathbf{s}) = \min\left(\inf\_{0$$

where *gL*(*s*) is given in (12). Note that

$$g\_L'(s) = 0 \quad \Leftrightarrow \ s^\* = s^\*(\rho) = \sqrt{\frac{2(1-\rho)}{\mu\_1^2 + \mu\_2^2 - 2\rho\mu\_1\mu\_2}}.\tag{20}$$

Next, for −1 < *ρ* < 0 we have that (recall *s*∗∗ given in (18))

$$s^{\*\*} \ge \frac{1-\rho}{\mu\_1(1-\rho^2)} > \frac{1}{\mu\_1} \ge \frac{1}{\mu\_2} > s\_{A\prime} \quad s^{\*\*} > \frac{1-\rho}{\mu\_1} > s^\*.$$

Therefore, by (19) we conclude that the unique minimizer of *g*3(*<sup>t</sup>*,*<sup>s</sup>*) on *A* is again given by (*tA*,*sA*). Consequently, for all −1 < *ρ* < 0, we have that the unique minimizer of *g*3(*<sup>t</sup>*,*<sup>s</sup>*) on *A* is given by (*tA*,*sA*), and

$$\inf\_{(t,s)\in\overline{A}}\mathbb{g}\_3(t,s) = \mathbb{g}\_A(t\_{A\prime}s\_A) = 4(\mu\_2 + (1-2\rho)\mu\_1). \tag{21}$$

Minimizers of *g*3(*<sup>t</sup>*,*<sup>s</sup>*) on *B*. Similarly, we have, for any fixed *t*,

$$\frac{\partial \mathcal{g}\_3(t,s)}{\partial s} = \frac{\partial \mathcal{g}\_B(t,s)}{\partial s} = 0 \iff (\mu\_2 s + 1 - \rho - \rho \mu\_1 t)(\mu\_2 s - (2\mu\_2 \rho^2 - \rho \mu\_1)t + \rho - 1) = 0.$$

Two roots of the above equation are:

$$s\_1 = s\_1(t) := \frac{\rho - 1 + \rho \mu\_1 t}{\mu\_2}, \quad s\_2 = s\_2(t) := \frac{1 - \rho + (2\mu\_2 \rho^2 - \rho \mu\_1)t}{\mu\_2}. \tag{22}$$

Next, we check if any of *si*, *i* = 1, 2, is greater than *t*. Again *s*1 < 0 < *s*2 as *ρ* < 0. So we check if *s*2 > *t*. It can be shown that

$$s\_2 > t \quad \Leftrightarrow \ (\mu\_2 + \rho \mu\_1 - 2\mu\_2 \rho^2)t < 1 - \rho. \tag{23}$$

Thus, for Scenario *μ*2 + *ρμ*1 − <sup>2</sup>*μ*2*ρ*<sup>2</sup> ≤ 0 we have that

$$s\_1 < 0 < \text{t } < s\_2$$

and in this case

$$\inf\_{(t,s)\in\mathbb{B}}\mathbb{g}\_3(t,s) = \inf\_{t>0} f\_\mathbb{B}(t)\_r$$

with

$$f\_B(t) := \mathcal{g}\_B(t, \mathbf{s}\_2(t)) = \frac{(1 + \mu\_1 t)^2}{t} + 4\mu\_2((1 - \rho) + (\rho^2 \mu\_2 - \rho \mu\_1)t).$$

Next, note that

$$f\_B^{\prime}(t) = 0 \quad \Leftrightarrow \quad t\_B = t\_B(\rho) := \frac{1}{|\mu\_1 - 2\rho\mu\_2|} = \frac{1}{\mu\_1 - 2\rho\mu\_2} > 0. \tag{24}$$

Therefore, the unique minimizer of *g*3(*<sup>t</sup>*,*<sup>s</sup>*) on *B* is given by (*tB*,*sB*) with

$$s\_B := s\_2(t\_B) = \frac{1 - 2\rho}{\mu\_2}, \quad \inf\_{(t,s)\in \overline{B}} g\_3(t,s) = g\_B(t\_B, s\_B) = 4(\mu\_1 + (1 - 2\rho)\mu\_2).$$

For Scenario *μ*2 + *ρμ*1 − <sup>2</sup>*μ*2*ρ*<sup>2</sup> > 0 we have from (23) that

$$s\_1 < 0 < t < s\_2 \iff t < \frac{1 - \rho}{\mu\_2 + \rho \mu\_1 - 2\mu\_2 \rho^2} = \frac{1 - \rho}{\rho(\mu\_1 - \rho \mu\_2) + \mu\_2(1 - \rho^2)} =: t^{\*\*}(\rho) = t^{\*\*}.\tag{25}$$

> In this case,

$$\inf\_{(t,s)\in\mathbb{Z}}\operatorname{g}\_{\mathbb{S}}(t,s) = \min\left(\inf\_{0\le t\le t^{\text{res}}} f\_B(t), \inf\_{t\ge t^{\text{res}}} \operatorname{g}\_L(t)\right).$$

Though it is not easy to determine explicitly the optimizer, we can conclude that the minimizer should be taken at (*tB*,*sB*), (*t*<sup>∗</sup>, *t*<sup>∗</sup>) or (*t*∗∗, *<sup>t</sup>*∗∗), where *t*∗ = *<sup>t</sup>*<sup>∗</sup>(*ρ*) = *<sup>s</sup>*<sup>∗</sup>(*ρ*). Further, we have from the discussion in (19) that

$$\lg\_A(t\_{A'}\mathfrak{s}\_A) < \lg\_L(\mathfrak{s}^\*) = \lg\_L(t^\*) = \min(\lg\_L(t^\*), \lg\_L(t^{\*\*})),$$

and

*(d)* 

$$\mathcal{g}\_B(t\_B, \mathbf{s}\_B) = 4(\mu\_1 + (1 - 2\rho)\mu\_2) \ge 4(\mu\_2 + (1 - 2\rho)\mu\_1) = \mathcal{g}\_A(t\_A, \mathbf{s}\_A).$$

Combining the above discussions on *A*, *B*, we conclude that Proposition 3 holds for −1 < *ρ* < 0.

3.1.2. Case 0 ≤ *ρ* < 1

We shall derive the minimizers of *g*(*<sup>t</sup>*,*<sup>s</sup>*) on *A*, *B* separately. We start with discussions on *B*, for which we give the following lemma. Recall *<sup>t</sup>*<sup>∗</sup>(*ρ*) = *s*<sup>∗</sup>(*ρ*) defined in (20) (see also (6)), *tB*(*ρ*) defined in (24), *<sup>t</sup>*∗∗(*ρ*) defined in (25) and *s*∗1 (*ρ*) defined in (14) for *μ*1/*μ*2 < *ρ* < 1. Note that where it applies, 1/0 is understood as +∞ and 1/∞ is understood as 0.

**Lemma 7.** *We have:*


$$t\_B(\rho) \ge t^{\*\*}(\rho), \quad t^\*(\rho) \ge t^{\*\*}(\rho),$$

*where both equalities hold only when ρ* = 0 *and μ*1 = *μ*2. *Itholdsthat*

$$t^\*(\not p\_2) = t\_\mathbb{B}(\not p\_2) = s\_1^\*(\not p\_2) = t^{\*\*}(\not p\_2) = \frac{1}{\mu\_2}.\tag{26}$$

*Moreover, for μ*1/*μ*2 < *ρ* < 1 *we have*


Recall that by definition *gL*(*s*) = *gA*(*<sup>s</sup>*,*<sup>s</sup>*) = *gB*(*<sup>s</sup>*,*<sup>s</sup>*),*<sup>s</sup>* > 0 (cf. (12)). For the minimum of *g*(*<sup>t</sup>*,*<sup>s</sup>*) on *B* we have the following lemma.

**Lemma 8.** *We have*

*(i) If* 0 ≤ *ρ* < *ρ*ˆ2*, then*

$$\inf\_{(t,s)\in\overline{\mathcal{B}}}\mathcal{g}(t,s) = \mathcal{g}\_L(t^\*) = \frac{2}{1+\rho}(\mu\_1 + \mu\_2 + 2/t^\*)\_{\mathcal{A}}$$

*where* (*t*<sup>∗</sup>, *t*<sup>∗</sup>) *is the unique minimizer of g*(*<sup>t</sup>*,*<sup>s</sup>*) *on B.(ii) If ρ* = *ρ*ˆ2*, then <sup>t</sup>*<sup>∗</sup>(*ρ*ˆ2) = *s*<sup>∗</sup>(*ρ*ˆ2) = 1/*μ*2 *and*

$$\inf\_{(t,s)\in\overline{\mathbb{B}}}\mathbb{g}(t,s) = \lg\_{\mathbb{L}}(1/\mu\_2) = \lg\_{\mathbb{L}}(1/\mu\_2) = 4\mu\_{2\prime}$$

*where the minimum of g*(*<sup>t</sup>*,*<sup>s</sup>*) *on B is attained at* (1/*μ*2, 1/*μ*2)*, with g*3(1/*μ*2, 1/*μ*2) = *g*2(1/*μ*2) *and* 1/*μ*2 *is the unique minimizer of g*2(*s*),*<sup>s</sup>* ∈ (0, <sup>∞</sup>).

*(iii) If ρ*ˆ2 < *ρ* < 1*, then*

$$\inf\_{(t,s)\in\mathbb{B}}\lg(t,s) = \inf\_{(t,s)\in D\_2}\lg(s) = \lg(1/\mu\_2) = 4\mu\_{2,s}$$

*where the minimum of g*(*<sup>t</sup>*,*<sup>s</sup>*) *on B is attained when g*(*<sup>t</sup>*,*<sup>s</sup>*) = *g*2(*s*) *on D*2 *(see Figure 1).*

Next we consider the minimum of *g*(*<sup>t</sup>*,*<sup>s</sup>*) on *A*. Recall *s*<sup>∗</sup>(*ρ*) defined in (20), *sA*(*ρ*) defined in (17) and *s*∗∗(*ρ*) defined in (18). We first give the following lemma.

**Lemma 9.** *We have*


$$s\_A(\mathfrak{p}\_1) = s^{\ast \ast}(\mathfrak{p}\_1) = s^\ast(\mathfrak{p}\_1),$$

*and*

$$\begin{array}{llll} \text{(i)} & \text{s}\_{\text{A}}(\rho) < \text{s}^{\ast \ast}\_{}(\rho) \text{ for all } \rho \in [0, \hat{\rho}\_{1}), & \text{s}\_{\text{A}}(\rho) > \text{s}^{\ast \ast}\_{}(\rho) \text{ for all } \rho \in (\hat{\rho}\_{1}, 1), \\\text{(ii)} & \text{s}^{\ast}(\rho) < \text{s}^{\ast \ast}(\rho) \text{ for all } \rho \in [0, \hat{\rho}\_{1}), & \text{s}^{\ast}(\rho) > \text{s}^{\ast \ast}(\rho) \text{ for all } \rho \in (\hat{\rho}\_{1}, 1). \end{array}$$

*(c) For all μ*1/*μ*2 < *ρ* < 1*, it holds that s*∗∗(*ρ*) < *s*∗1 (*ρ*)*.*

For the minimum of *g*(*<sup>t</sup>*,*<sup>s</sup>*) on *A* we have the following lemma.
