**Appendix A**

In this appendix, we present the proofs of the lemmas used in Section 3.

**Proof of Lemma 6.** Referring to Lemma 5, we have, for any fixed *t*,*s*, there exists a unique index set

$$I(t,s) \subseteq \{1,2\}$$

such that

$$\mathbf{g}(t,\mathbf{s}) = \begin{pmatrix} 1+\mu\_1 t, 1+\mu\_2 \mathbf{s} \end{pmatrix} \mathbf{l}(t,\mathbf{s}) \begin{pmatrix} \Sigma\_{\mathrm{fr}} \end{pmatrix}\_{\mathbf{l}(t,\mathbf{s}),\mathbf{l}(t,\mathbf{s})}^{-1} \begin{pmatrix} 1+\mu\_1 t, 1+\mu\_2 \mathbf{s} \end{pmatrix}\_{\mathbf{l}(t,\mathbf{s})}^{\big|} \tag{A1}$$

and

$$\left(\Sigma\_{t^{\mathfrak{s}}}\right)^{-1}\_{I(t,\mathfrak{s}),I(t,\mathfrak{s})}\left(1+\mu\_1 t, 1+\mu\_2 \mathfrak{s}\right)^{\top}\_{I(t,\mathfrak{s})} > \mathfrak{0}\_{I(t,\mathfrak{s})}.\tag{A2}$$

Since *<sup>I</sup>*(*<sup>t</sup>*,*<sup>s</sup>*) = {1}, {2} or {1, <sup>2</sup>}, we have that


Clearly, if *ρ* ≤ 0 then

$$E\_1 = E\_2 = \bigotimes, \ E\_3 = (0, \infty)^2.$$

In this case,

$$\mathfrak{g}(t,\mathfrak{s}) = \mathfrak{g}\mathfrak{z}(t,\mathfrak{s}), \quad (t,\mathfrak{s}) \in (0,\infty)^{\overline{2}}.$$

Next, we focus on the case where *ρ* > 0. We consider the regions *A* and *B* separately. Analysis on *A*. We have

$$A\_1 = \overline{A} \cap E\_1 = \{ s \le t \le f\_1(s) \}, \quad f\_1(s) = \frac{\rho - 1}{\mu\_1} + \frac{\rho \mu\_2}{\mu\_1} s,$$

$$A\_2 = \overline{A} \cap E\_2 = \{ s \le t \le f\_2(s) \}, \quad f\_2(s) = \frac{\rho s}{1 + (\mu\_2 - \rho \mu\_1)s}.$$

$$A\mathfrak{z} = \overline{A} \cap E\mathfrak{z} = \{ t \ge s, t > \max(f\_1(s), f\_2(s)) \}.$$

Next we analyse the intersection situation of the functions *f*(*s*) = *s*, *f*1(*s*), *f*2(*s*) on region *A*. Clearly, for any *s* > 0 we have *f*2(*s*) < *s*. Furthermore, *f*1(*s*) = *f*2(*s*) has a unique positive solution *s*1 given by

$$s\_1 = \frac{1 - \rho}{\rho(\mu \underline{\rho} - \rho \mu\_1)}.$$

Finally, for *ρμ*2 ≤ *μ*1 we have that *f*1(*s*) does not intersect with *f*(*s*) on (0, ∞) but for *ρμ*2 > *μ*1 the unique intersection point is given by *s*∗1 > *s*1 (cf. (14)). To conclude, we have, for *ρ* ≤ *μ*1/*μ*2,

$$\emptyset(t, s) = \emptyset\_3(t, s), \quad (t, s) \in \overline{A}\_r$$

and for *ρ* > *μ*1/*μ*2,

$$\mathcal{g}(t,s) = \begin{cases} \mathcal{g}\_3(t,s), & \text{if } (t,s) \in \overline{A} \cap \{ t \ge \max(s, f\_1(s)), t > f\_1(s) \}, \\\mathcal{g}\_2(s), & \text{if } (t,s) \in \overline{A} \cap \{ s \le t \le f\_1(s) \}. \end{cases}$$

Additionally, we have from Lemma 5 *g*3(*f*1(*s*),*<sup>s</sup>*) = *g*2(*s*) for all *s* ≥ *s*∗1.

Analysis on *B*. The two scenarios *ρ* ≤ *μ*1/*μ*2 and *ρ* > *μ*1/*μ*2 will be considered separately. For *ρ* ≤ *μ*1/*μ*2, we have

$$\begin{aligned} B\_1 &= B \cap E\_1 = \{ t < s \le h\_1(t) \}, \quad h\_1(t) = \frac{\rho t}{1 + (\mu\_1 - \rho \mu\_2)t}, \\ B\_2 &= B \cap E\_2 = \{ t < s \le h\_2(t) \}, \quad h\_2(t) = \frac{\rho - 1}{\mu\_2} + \frac{\rho \mu\_1}{\mu\_2}t, \\ B\_3 &= B \cap E\_3 = \{ s > \max(t, h\_1(t), h\_2(t)) \}. \end{aligned}$$

It is easy to check that

$$t > h\_1(t), \quad t > h\_2(t), \quad \forall t > 0,$$

and thus

$$\mathbf{g}(t, \mathbf{s}) = \mathbf{g}\_3(t, \mathbf{s}), \quad (t, \mathbf{s}) \in B.$$

For *ρ* > *μ*1/*μ*2, we have

$$\begin{aligned} B\_1 &= B \cap E\_1 = \{ w\_1(s) \le t < s \}, \quad w\_1(s) = \frac{s}{\rho + (\rho \mu\_2 - \mu\_1)s}, \\ B\_2 &= B \cap E\_2 = \{ w\_2(s) \le t < s \}, \quad w\_2(s) = \frac{1 - \rho}{\mu\_1 \rho} + \frac{\mu\_2}{\mu\_1 \rho} s, \\ B\_3 &= B \cap E\_3 = \{ t < \min(s, w\_1(s), w\_2(s)) \}. \end{aligned}$$

Next we analyze the intersection situation of the functions *w*(*s*) = *s*, *<sup>w</sup>*1(*s*), *<sup>w</sup>*2(*s*) on region *B*. Clearly, for any *s* > 0, *<sup>w</sup>*2(*s*) > *s*. *<sup>w</sup>*1(*s*) and *<sup>w</sup>*2(*s*) do not intersect on (0, <sup>∞</sup>). *w*(*s*) and *<sup>w</sup>*1(*s*) has a unique intersection point *s*∗1(cf. (14)).

To conclude, we have, for *ρ* ≤ *μ*1/*μ*2,

$$\mathbf{g}(t, \mathbf{s}) = \mathbf{g}\_3(t, \mathbf{s})\_\prime \quad (t, \mathbf{s}) \in B\_\prime$$

and for *ρ* > *μ*1/*μ*2,

$$\mathcal{g}(t,s) = \begin{cases} \mathcal{g}\_3(t,s), & \text{if } (t,s) \in \mathcal{B} \cap \{t < \min(s, w\_1(s))\}, \\\mathcal{g}\_2(s), & \text{if } (t,s) \in \mathcal{B} \cap \{w\_1(s) \le t < s\}. \end{cases}$$

Additionally, it follows from Lemma 5 that *g*3(*<sup>w</sup>*1(*s*),*<sup>s</sup>*) = *g*2(*s*) for all *s* ≥ *s*∗1. Consequently, the claim follows by a combination of the above results. This completes the proof.

**Proof of Lemma 7.** (a) The claim for *<sup>t</sup>*<sup>∗</sup>(*ρ*) follows by noting its following representation:

$$t^\*(\rho) = s^\*(\rho) = \sqrt{\frac{2(1-\rho)}{\mu\_1^2 + \mu\_2^2 - 2\mu\_1\mu\_2 + 2\mu\_1\mu\_2 - 2\rho\mu\_1\mu\_2}} = \sqrt{\frac{2}{\frac{(\mu\_1-\mu\_2)^2}{1-\rho} + 2\mu\_1\mu\_2}}.$$

.

The claims for *tB*(*ρ*) and *s*∗1 (*ρ*) follow directly from their definition.

(b) First note that

$$t^{\ast\ast}(0) = t^{\ast\ast}(\not p\_2) = \frac{1}{\mu\_2}.$$

Next it is calculated that

$$\frac{\partial t^{\ast\ast}(\rho)}{\partial \rho} = \frac{-2\mu\_2\rho^2 + 4\mu\_2\rho - \mu\_1 - \mu\_2}{(\mu\_2 + \rho\mu\_1 - 2\mu\_2\rho^2)^2}.$$

Thus, the claim of (b) follows by analysing the sign of *∂t*∗∗(*ρ*) *∂ρ* over (0, *ρ*<sup>ˆ</sup>). (c) For any 0 ≤ *ρ* ≤ *μ*1/*μ*2 we have |*μ*1 − <sup>2</sup>*ρμ*2| ≤ *μ*1 and thus

$$t\_B(\rho) \ge \frac{1}{\mu\_1} \ge \frac{1}{\mu\_2} \ge \frac{1 - \rho}{\mu\_2(1 - \rho^2)} \ge \frac{1 - \rho}{\rho(\mu\_1 - \rho \mu\_2) + \mu\_2(1 - \rho^2)} = t^{\ast \ast}(\rho).$$

Further, since

$$
\mu\_1^2 + \mu\_2^2 - 2\rho\mu\_1\mu\_2 = \mu\_1(\mu\_1 - \rho\mu\_2) + \mu\_2(\mu\_2 - \rho\mu\_1) \\
\leq \mu\_2(\mu\_1 - \rho\mu\_2) + \mu\_2(\mu\_2 - \rho\mu\_1) \\
\leq 2\mu\_2^2(1 - \rho),
$$
 
$$
\text{is } \epsilon \text{allow that}
$$

it follows that

$$t^\*(\rho) \ge \frac{1}{\mu\_2} \ge t^{\*\*}(\rho).$$

(d) It is easy to check that (26) holds. For (i) we have

$$t^\*(\rho) - s\_1^\*(\rho) = (1 - \rho) \left( \frac{1}{f\_1(\rho)} - \frac{1}{f\_2(\rho)} \right),$$

where

$$\begin{aligned} f\_1(\rho) &= \sqrt{\frac{(1-\rho)(\mu\_1^2 + \mu\_2^2 - 2\rho\mu\_1\mu\_2)}{2}} = \sqrt{\mu\_1\mu\_2\rho^2 - \frac{(\mu\_1 + \mu\_2)^2}{2}\rho + \frac{\mu\_1^2 + \mu\_2^2}{2}}, \\ f\_2(\rho) &= -\rho\mu\_2 - \mu\_1. \end{aligned}$$

> Analysing the properties of the above two functions, we have *f*1(*ρ*) is strictly decreasing on [0, 1] with

$$f\_1(0) = \sqrt{\frac{\mu\_1^2 + \mu\_2^2}{2}} > -\mu\_1 = f\_2(0), \quad f\_1(1) = 0 \le \mu\_2 - \mu\_1 = f\_2(1).$$

and thus there is a unique intersection point of the two curves *t* <sup>∗</sup>(*ρ*) and *s*<sup>∗</sup> 1 (*ρ*) which is *ρ* = *ρ*ˆ2. Therefore, the claim of (i) follows. Similarly, the claim of (ii) follows since

$$t\_B(\rho) - s\_1^\*(\rho) = \frac{- (\mu\_1 + \mu\_2)\rho + 2\mu\_2\rho^2}{(\rho\mu\_2 - \mu\_1)(2\rho\mu\_2 - \mu\_1)}.$$

Finally, the claims of (iii), (iv) and (v) follow easily from (a), (b) and (26). This completes the proof.

**Proof of Lemma 8.** Consider first the case where 0 ≤ *ρ* ≤ *μ*1/*μ*2. Recall (22). We check if any of *si*, *i* = 1, 2, is greater than *t*. Clearly, *s*1 ≤ *t*. Next, we check whether *s*2 > *t*. It is easy to check that

$$s\_2 > t \iff t < t^{\*\*}$$

where (recall (25))

$$t^{\ast \ast} = t^{\ast \ast}(\rho) = \frac{1 - \rho}{\rho(\mu\_1 - \mu\_2 \rho) + \mu\_2(1 - \rho^2)} > 0.$$

Then

$$\inf\_{(t,s)\in\overline{\mathcal{B}}}\mathcal{g}\_3(t,s) = \min\left(\inf\_{0$$

Consequently, it follows from (c) of Lemma 7 the claim of (i) holds for 0 ≤ *ρ* ≤ *μ*1/*μ*2.

Next, we consider *μ*1/*μ*2 < *ρ* < 1. Recall the function *<sup>w</sup>*1(*s*) defined in (13). Denote the inverse function of *<sup>w</sup>*1(*s*) by

$$\vartheta\_1(t) = \frac{\rho t}{1 - (\rho \mu\_2 - \mu\_1)t}, \quad t \ge s\_1^\*.$$

We have from Lemma 6 that

$$
\lg\_B(t, \mathfrak{d}\_1(t)) = \mathfrak{g}\_2(t), \quad t \ge s\_1^\*.
$$

Further note that 1/*μ*2 is the unique minimizer of *g*2(*s*),*<sup>s</sup>* > 0. For *μ*1/*μ*2 < *ρ* < *ρ*ˆ2, we have from (d) in Lemma 7 that

$$\inf\_{s\_1^\* \le s} \mathcal{g}\_2(s) = \mathcal{g}\_2(s\_1^\*) = \mathcal{g}\_L(s\_1^\*) > \mathcal{g}\_L(t^\*).$$

and further

$$\begin{aligned} \inf\_{(t,s)\in\overline{\mathbb{B}}}\mathcal{g}(t,s)&=\min\limits\_{0$$

where (*t* ∗, *t* ∗) is the unique minimizer of *g*(*<sup>t</sup>*,*<sup>s</sup>*) on *B*. Therefore, the claim for *μ*1/*μ*2 < *ρ* < *ρ*ˆ2 is established.

> For *ρ* = *ρ*ˆ2, because of (26) we have

$$\begin{aligned} \inf\_{(t,s)\in\overline{\mathbb{B}}} \operatorname{g}(t,s) &= \min\limits\_{0$$

and the unique minimum of *g*(*<sup>t</sup>*,*<sup>s</sup>*) on *B* is attained at (1/*μ*2, 1/*μ*2). Moreover, for all *ρ*ˆ2 < *ρ* < 1 we have

$$s\_2(t\_B) = \wp\_1(t\_B) = \frac{1}{\mu\_2} > s\_1^\*.$$

Thus,

$$\begin{aligned} \inf\_{(t,s)\in\overline{\mathbb{B}}}\mathsf{g}(t,s)&=\min\limits\_{0$$

and the unique 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. This completes the proof.

**Proof of Lemma 9.** (a) The claim for *s*<sup>∗</sup>(*ρ*) has been shown in the proof of (a) in Lemma 7. Next, we show the claim for *<sup>s</sup>*∗∗(*ρ*), for which it is sufficient to show that *∂s*∗∗(*ρ*) *∂ρ*< 0 for all *ρ* ∈ [0, 1]. In fact, we have

$$\frac{\partial s^{\ast \ast}(\rho)}{\partial \rho} = \frac{-2\mu\_1\rho^2 + 4\mu\_1\rho - \mu\_1 - \mu\_2}{(\mu\_1 + \rho\mu\_2 - 2\mu\_1\rho^2)^2} < 0.$$

(b) In order to prove (i), the following two scenarios will be discussed separately:

$$(\text{S1}). \quad \mu\_2 < 2\mu\_1; \quad \text{ (S2)}. \quad \mu\_2 \ge 2\mu\_1.$$

First consider (S1). If 0 ≤ *ρ* < *μ*2 2*μ*1 , then

$$\begin{split} s\_A(\rho) - s^{\ast \ast}(\rho) &= \frac{(\mu\_1 + \rho \mu\_2 - 2\mu\_1 \rho^2) - (1 - \rho)(\mu\_2 - 2\rho \mu\_1)}{(\mu\_2 - 2\rho \mu\_1)(\mu\_1 + \rho \mu\_2 - 2\mu\_1 \rho^2)} \\ &= \frac{f(\rho)}{(\mu\_2 - 2\rho \mu\_1)(\mu\_1 + \rho \mu\_2 - 2\mu\_1 \rho^2)} \end{split}$$

where

$$f(\rho) = -4\mu\_1 \rho^2 + 2(\mu\_2 + \mu\_1)\rho - \mu\_2 + \mu\_1.$$

Analysing the function *f* , we conclude that

$$f(\rho) < 0, \quad \text{for } \rho \in [0, \hat{\rho}\_1), \quad f(\rho) > 0, \quad \text{for } \rho \in (\hat{\rho}\_1, \frac{\mu\_2}{2\mu\_1}).$$

Further, for *μ*2 2*μ*1 ≤ *ρ* < 1 we have

$$s\_A(\rho) - s^{\ast \ast}(\rho) = \frac{\mu\_1 + \mu\_2 - 2\mu\_1\rho}{(2\rho\mu\_1 - \mu\_2)(\mu\_1 + \rho\mu\_2 - 2\mu\_1\rho^2)} > 0.$$

> Thus, the claim in (i) is established for (S1). Similarly, the claim in (i) is valid for (S2) . Next, note that

$$s^\*(\rho) - s^{\*\*}(\rho) = (1 - \rho) \left( \frac{1}{f\_1(\rho)} - \frac{1}{f\_2(\rho)} \right)$$

with

$$\begin{aligned} f\_1(\rho) &= \sqrt{\frac{(1-\rho)(\mu\_1^2 + \mu\_2^2 - 2\rho\mu\_1\mu\_2)}{2}} = \sqrt{\mu\_1\mu\_2\rho^2 - \frac{(\mu\_1 + \mu\_2)^2}{2}\rho + \frac{\mu\_1^2 + \mu\_2^2}{2}}, \\ f\_2(\rho) &= -\mu\_1 + \rho\mu\_2 - 2\mu\_1\rho^2. \end{aligned}$$

Analysing the properties of the above two functions, we have *f*1(*ρ*) is strictly decreasing on [0, 1] with

$$f\_1(0) = \sqrt{\frac{\mu\_1^2 + \mu\_2^2}{2}} \ge \mu\_1 = f\_2(0), \quad f\_1(1) = 0 \le \mu\_2 - \mu\_1 = f\_2(1),$$

and thus there is a unique intersection point *ρ* ∈ (0, 1) of *s*<sup>∗</sup>(*ρ*) and *<sup>s</sup>*∗∗(*ρ*). It seems not clear at the moment whether this unique point is *ρ*ˆ1 or not, since we have to solve a polynomial equation of order 4. Instead, it is sufficient to show that

$$s\_A(\mathfrak{d}\_1) = s^\*(\mathfrak{d}\_1). \tag{A3}$$

In fact, basic calculations show that the above is equivalent to

$$(2\mu\_1\flat\_1 - (\mu\_1 + \mu\_2))f(\flat\_1) = 0$$

which is valid due to the fact that *f*(*ρ*ˆ1) = 0. Finally, the claim in (c) follows since

$$
\rho \mu\_2 - \mu\_1 < \mu\_1 + \rho \mu\_2 - 2\rho^2 \mu\_1.
$$

This completes the proof.

**Proof of Lemma 10.** Two cases *ρ*ˆ1 ≤ *μ*1/*μ*2 and *ρ*ˆ1 > *μ*1/*μ*2 should be distinguished. Since the proofs for these two cases are similar, we give below only the proof for the more complicated case *ρ*ˆ1 ≤ *μ*1/*μ*2.

Note that, for 0 ≤ *ρ* ≤ *μ*1/*μ*2, as in (19),

$$\inf\_{(t,s)\in\overline{\mathcal{A}}}\mathcal{g}(t,s) = \inf\_{(t,s)\in\overline{\mathcal{A}}}\mathcal{g}\_{\mathcal{S}}(t,s) = \min\left(\inf\_{0$$

and thus the claim for 0 ≤ *ρ* ≤ *μ*1/*μ*2 follows directly from (i)–(ii) of (b) in Lemma 9.

Next, we consider the case *μ*1/*μ*2 < *ρ* < *ρ*ˆ2 (note here *ρ*ˆ1 < *μ*1/*μ*2 < *ρ*). We have by (i) of (d) in Lemma 7 and (i)–(ii) of (b) in Lemma 9 that

$$s^{\ast\ast}(\rho) < s^{\ast}(\rho) = t^{\ast}(\rho) < s\_1^{\ast}(\rho), \quad s\_1^{\ast}(\rho) > \frac{1}{\mu\_2}, \quad s\_A(\rho) > s^{\ast\ast}(\rho).$$

> Thus, it follows from Lemma 6 that

$$\begin{aligned} \inf\_{(t,s)\in\overline{\mathcal{X}}}\operatorname{g}(t,s)&=\min\left(\inf\_{0$$

and (*t*<sup>∗</sup>,*s*<sup>∗</sup>) ∈ *L* is the unique minimizer of *g*(*<sup>t</sup>*,*<sup>s</sup>*) on *A*. Here we used the fact that

$$\inf\_{s\_1^\* < s} \mathbb{g}\_A(f\_1(s), s) = \inf\_{s\_1^\* < s} \mathbb{g}\_2(s) = \mathbb{g}\_A(f\_1(s\_1^\*), s\_1^\*) = \mathbb{g}\_2(s\_1^\*) > \mathbb{g}\_L(s^\*).$$

Next, if *ρ* = *ρ*ˆ2, then

$$s\_1^\*(\mathfrak{h}\_2) = s^\*(\mathfrak{h}\_2) = \frac{1}{\mu\_2},$$

and thus

$$\begin{split} \inf\_{(t,s)\in\overline{A}}\mathsf{g}(t,s) &= \min\left(\inf\_{0$$

Furthermore, the unique minimum of *g*(*<sup>t</sup>*,*<sup>s</sup>*) on *A* is attained at (1/*μ*2, 1/*μ*2), with *g*3(1/*μ*2, 1/*μ*2) = *g*2(1/*μ*2).

Finally, for *ρ*ˆ2 < *ρ* < 1, we have

$$s^{\ast\ast}(\rho) < s\_1^{\ast}(\rho) < s^{\ast}(\rho) < \frac{1}{\mu\_2}, \quad s\_A(\rho) > s^{\ast\ast}(\rho)\_{\prime\prime}$$

and thus

inf (*<sup>t</sup>*,*<sup>s</sup>*)∈*<sup>A</sup> g*(*<sup>t</sup>*,*<sup>s</sup>*) = min inf 0<*s*<*s*∗∗ *gA*(*<sup>t</sup>*2(*s*),*<sup>s</sup>*), inf *<sup>s</sup>*∗∗≤*s*≤*s*<sup>∗</sup>1 *gA*(*<sup>s</sup>*,*<sup>s</sup>*), inf *<sup>s</sup>*<sup>∗</sup>1<*<sup>s</sup> gA*(*f*1(*s*),*<sup>s</sup>*), inf *<sup>s</sup>*<sup>∗</sup>1<*<sup>s</sup> g*2(*s*)= *g*2(1/*μ*2),

where the unique minimum of *g*(*<sup>t</sup>*,*<sup>s</sup>*) on *A* is attained when *g*3(*<sup>t</sup>*,*<sup>s</sup>*) = *g*2(*s*) on *D*2. This completes the proof.
