Exact asymptotics of component-wise extrema of two-dimensional Brownian motion

Abstract

We derive the exact asymptotics of

$ {\mathbb {P} \left \{ \underset {t\ge 0}{\sup } \left (X_{1}(t) - \mu _{1} t\right )> u, \ \underset {s\ge 0}{\sup } \left (X_{2}(s) - \mu _{2} s\right )> u \right \} },\ \ u\to \infty , $

where (X₁(t), X₂(s))_{t, s≥ 0} is a correlated two-dimensional Brownian motion with correlation ρ ∈ [− 1,1] and μ₁, μ₂ > 0. It appears that the play between ρ and μ₁, μ₂ leads to several types of asymptotics. Although the exponent in the asymptotics as a function of ρ is continuous, one can observe different types of prefactor functions depending on the range of ρ, which constitute a phase-type transition phenomena.

Appendix A: Proofs of Lemmas 4.1–5.1

In this section we give proofs of Lemmas 4.1–5.1 that are the building blocks of the proofs of Theorems 2.1 and 2.3.

We begin with the analysis of the local behaviour of function $g(t,s), (t,s)\in (0,\infty )$ at its minimizer in scenarios (i)–(iv) of Proposition 3.1, respectivelly.

Lemma A.1

Assume that μ₁ < μ₂. We have

• If $-1< \rho <\hat \rho _{1}$, then as (t, s) → (0, 0),

  $$ \begin{array}{@{}rcl@{}} g(t_A+t,s_A+s)=g_A(t_A,s_A)&+&\frac{a_1}{2}t^2(1+o(1)) - a_2 ts(1+o(1))\\ &+&\frac{a_3}{2} s^2(1+o(1)), \end{array} $$

  where, with h(ρ) := μ₂ − 2(μ₁ + μ₂)ρ + 3μ₁ρ² > 0,

  $$ \begin{array}{@{}rcl@{}} a_1:&=&\frac{2\mu_1^3(\mu_2-2\mu_1\rho)}{h(\rho)}>0, \ \ a_2:=\frac{-2\rho\mu_1^2(\mu_2-2\mu_1\rho)^2}{h(\rho)}, \\ a_3:&=&\frac{2(\mu_2-2\mu_1\rho)^4(1-2\rho)}{h(\rho)}>0. \end{array} $$

• If $\hat \rho _{1} < \rho < \hat \rho _{2}$, then

  • (ii.1), as (t, s) → (0, 0), with s < t (i.e., (t^∗ + t, s^∗ + s) ∈ A),

    $$ \begin{array}{@{}rcl@{}} g(t^*+t,s^*+s)=g_L(t^*) + b_1(t-s ) (1+o(1)) + \frac{ c_1 }{2}s^2(1+o(1)), \end{array} $$

    where

    $$ \begin{array}{@{}rcl@{}} b_1&:=&\frac{(\rho-1-2\rho^2)+2\rho(\mu_2-\mu_1\rho)s^* +(1+\rho)\mu_1^2{s^*}^2}{(1-\rho)(1+\rho)^2{s^*}^2}>0,\\ c_1&:=& \frac{2 }{ {s^*}^3}\left( 1+\frac{\rho^2(\rho(1-\rho)-(\mu_2-\mu_1\rho) s^*)^2}{(1-\rho^2)^3}\right)>0. \end{array} $$

  • (ii.2), as (t, s) → (0, 0), with s > t (i.e., (t^∗ + t, s^∗ + s) ∈ B),

    $$ \begin{array}{@{}rcl@{}} g(t^*+t,s^*+s)=g_L(t^*)+b_2(s-t ) (1 + o(1)) + \frac{ c_2 }{2}t^2(1+o(1)), \end{array} $$

    where

    $$ \begin{array}{@{}rcl@{}} b_2&:=&\frac{(\rho-1-2\rho^2)+2\rho(\mu_1-\mu_2\rho)t^* +(1+\rho)\mu_2^2{t^*}^2}{(1-\rho)(1+\rho)^2{t^*}^2}>0, \\ c_2&:=& \frac{2 }{ {t^*}^3}\left( 1+\frac{\rho^2(\rho(1-\rho)-(\mu_1-\mu_2\rho) t^*)^2}{(1-\rho^2)^3}\right)>0. \end{array} $$

  • (ii.3), as (t, s) → (0, 0), with s = t (i.e., (t^∗ + t, s^∗ + s) ∈ L),

    $$ \begin{array}{@{}rcl@{}} g(t^*+t, s^*+t)=g_L(t^*)+\frac{b_0}{2}t^2(1+o(1)), \end{array} $$

    where $b_{0}:=\frac {4}{(1+\rho ){t^{*}}^{3}}.$

• If $\rho = \hat \rho _{1}$ (in this case t_A = s_A = t^∗ = s^∗), then

  • (iii.1), as (t, s) → (0, 0), with s < t,

    $$ \begin{array}{@{}rcl@{}} g(t_A+t,s_A+s)&=&g_A(t_A,s_A)+\frac{a_1}{2}t^2(1+o(1))\\ &&- a_2 ts(1+o(1)) +\frac{a_3}{2} s^2(1+o(1)), \end{array} $$

  • (iii.2), as (t, s) → (0, 0), with s > t,

    $$ \begin{array}{@{}rcl@{}} g(t^* + t,s^*+s) = g_L(t^*) + b_2(s - t ) (1 + o(1)) + \frac{ c_2 }{2}t^2(1 + o(1)). \end{array} $$

  • (iii.3), as (t, s) → (0, 0), with s = t,

    $$ \begin{array}{@{}rcl@{}} g(t^*+t, s^*+t)=g_L(t^*)+\frac{b_0}{2}t^2(1+o(1)). \end{array} $$

The proof of Lemma A.1 is tedious but only involves basic calculations using Taylor expansion, and thus it is omitted.

Next we present below a generalized version of the Bonferroni’s inequality. The proof can be found in, e.g., Hashorva and Ji (2014).

Lemma A.2

Let $({\Omega }, \mathcal {F}, \mathbb {P})$ be a probability space and A₁,⋯ , A_n and B₁,⋯ , B_m be n + m events in $\mathcal {F}$ with n, m ≥ 2. Then

$$ \begin{array}{@{}rcl@{}} &&\sum\limits_{k=1}^n\sum\limits_{l=1}^m{\mathbb{P} \left\{ A_k\cap B_l \right \} }\ge {\mathbb{P} \left\{ \underset{l=1,\ldots, m}{\underset{k=1,\ldots,n} \bigcup }(A_k\cap B_l) \right \} }\ge\sum\limits_{k=1}^n\sum\limits_{l=1}^m{\mathbb{P} \left\{ A_k\cap B_l \right \} }\\ &&\ \ \ \qquad -\sum\limits_{k=1}^n\underset{1\le l_1<l_2\le m}{\sum}{\mathbb{P} \left\{ A_k\cap B_{l_1}\cap B_{l_2} \right \} } -\sum\limits_{l=1}^m\underset{1\le k_1<k_2\le n}{\sum}{\mathbb{P} \left\{ A_{k_1}\cap A_{k_2}\cap B_l \right \} }. \end{array} $$

1.1 A.1 Proof of Lemma 4.1

Let T₀ > 0 be a fixed large constant (will be determined later). It is easily seen that

$$ \begin{array}{@{}rcl@{}} r_0(u)&\le& {\mathbb{P} \left\{ \exists_{(t,s) \in [0,T_0]^2\setminus U_1} \ X_1(t)>\sqrt u (1+\mu_1 t), X_2(s)>\sqrt u (1+\mu_2 s) \right \} }\\ &&+{\mathbb{P} \left\{ \exists_{t\ge T_0} \ X_1(t)>\sqrt u (1+\mu_1 t) \right \} }+{\mathbb{P} \left\{ \exists_{s\ge T_0} \ X_2(s)>\sqrt u (1+\mu_2 s) \right \} }. \end{array} $$

Next we consider upper bounds for each term on the right-hand side. According to Lemma 5 of Dȩbicki et al. (2019), for any fixed t, s, there exists a unique index set

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

such that

$$ \begin{array}{@{}rcl@{}} g(t,s)=(1+{\mu}_1 t, 1+{\mu}_2 s)_{I(t,s)} \ ({\Sigma}_{ts})_{{I(t,s)},{I(t,s)}}^{-1} \ (1+{\mu}_1 t, 1+{\mu}_2 s)_{I(t,s)}^{\top}, \end{array} $$

(25)

and

$$ \begin{array}{@{}rcl@{}} ({\Sigma}_{ts})_{{I(t,s)},{I(t,s)}}^{-1} \ (1+{\mu}_1 t, 1+{\mu}_2 s)_{I(t,s)}^{\top} > {\boldsymbol{0_{I(t,s)}}}. \end{array} $$

(26)

In the above, we use notation that if I ⊂{1, 2}, then for a vector ${\boldsymbol {a}} \in \mathbb {R}^{2}$ we denote by a_I = (a_i, i ∈ I) a sub-block vector of a. Similarly, if further J ⊂{1, 2}, for a matrix $M=(m_{ij})_{i,j\in \{1,2\}}\in \mathbb {R}^{2\times 2}$ we denote by M_{I, J} = (m_ij)_{i∈I, j∈J} the sub-block matrix of M determined by I and J. Furthermore, we write $M_{II}^{-1}=(M_{II})^{-1}$ for the inverse matrix of M_II whenever it exists.

Thus,

$$ \begin{array}{@{}rcl@{}} && {\mathbb{P} \left\{ \exists_{(t,s) \in [0,T_0]^2\setminus U_1} \ X_1(t)>\sqrt{u} (1+\mu_1 t), X_2(s)>\sqrt{u} (1+\mu_2 s) \right \} }\\ && \le {\mathbb{P} \left\{ \exists_{(t,s) \in [0,T_0]^2\setminus U_1} \ (1 + \mu_1 t, 1+\mu_2 s)_{I(t,s)}\ ({\Sigma}_{ts})_{{I(t,s)},{I(t,s)}}^{-1} \ (X_1(t),X_2(s))_{I(t,s)}^{\top} >\sqrt u g(t,s) \right \} } \\ &&= {\mathbb{P} \left\{ { \exists_{(t,s) \in [0,T_0]^2\setminus U_1} \ \frac{ Z(t,s)}{g(t,s)}>\sqrt{u} } \right \} }, \end{array} $$

(27)

where

$$ \begin{array}{@{}rcl@{}} Z(t,s):= (1+\mu_1 t, 1+\mu_2 s)_{I(t,s)}\ ({\Sigma}_{ts})_{{I(t,s)},{I(t,s)}}^{-1} \ (X_1(t),X_2(s))_{I(t,s)}^{\top} . \end{array} $$

(28)

Note that

$$ \begin{array}{@{}rcl@{}} \text{Var}\left( \frac{Z(t,s)}{g(t,s)}\right)=\frac{1}{g(t,s)}. \end{array} $$

(29)

In order to apply the Borell-TIS inequality, we first show that

$$ \underset{(t,s)\to(t^{(b)},s^{(b)})}{\limsup}\frac{{\left\lvert {Z(t,s)} \right\rvert}}{g(t,s)}<{\infty},\ \ \ \ \text{almost\ surely} $$

holds for any (t^(b), s^(b)) on the boundary {(t, s) : t ≥ 0, s = 0}∪{(t, s) : t = 0, s ≥ 0}. In fact, if the above does not hold for some boundary point (t^(b), s^(b)), then for any M > 0 there exist a sequence $\{(t_{k},s_{k})\}_{k=1}^{\infty }$ and some measurable set E such that $(t_{k},s_{k}) \to (t^{(b)},s^{(b)}), {\mathbb {P} \left \{ {E} \right \} }>0$ and

$$ \frac{\left\lvert Z(t_{k},s_{k}) \right\rvert}{g(t_{k},s_{k})}\ge M \ \ \ \text{on}\ E $$

for all large enough k. Then we have

$$ \begin{array}{@{}rcl@{}} \text{Var}\left( \frac{Z(t_k,s_k)}{g(t_k,s_k)} \right)\ge M^2 {\mathbb{P} \left\{ {E} \right \} >0}. \end{array} $$

(30)

On the other hand, by Lemma 6 of Dȩbicki et al. (2019) we have g(t, s) = g₃(t, s) for all (t, s) ∈{(t, s) : t ≥ 0, s = 0}∪{(t, s) : t = 0, s ≥ 0}, and thus by Eq. 29 and Eq. 13 we have $\lim _{k\to {\infty }} \text {Var}\left (\frac {Z(t_{k},s_{k})}{g(t_{k},s_{k})} \right ) =0$. This is a contradiction with Eq. 30. Therefore, $\frac { Z(t,s)}{g(t,s)}, (t,s) \in [0,T_{0}]^{2}\setminus U_{1}$ is almost surely bounded. Consequently, by the Borell-TIS inequality (see, e.g., Adler and Taylor (2007)) we have, for any fixed small constant 𝜃₀ > 0

$$ \begin{array}{@{}rcl@{}} {\mathbb{P} \left\{ { \exists_{(t,s) \in [0,T_0]^2\setminus U_1 } \ \frac{Z(t,s)}{g(t,s)}>\sqrt u } \right \} }\le e^{-\frac{(\sqrt u -C_0)^2}{2} \widehat g} \end{array} $$

holds for all u such that

$$ \sqrt u>C_{0}:=\mathbb{E}\left\{ \underset{(t,s)\in [0,T_{0}]^{2}\setminus U_{1} }{\sup} \frac{ Z(t,s)}{g(t,s)}\right\}. $$

Moreover, since X_i is the standard Brownian motion,

$$ \begin{array}{@{}rcl@{}} \lim_{t\to{\infty}}\frac{X_i(t)}{1+\mu_i t} = 0\ \ \ \ \mathrm{almost\ surely}, \end{array} $$

showing that the random process $\frac {X_{i}(t)}{1+\mu _{i} t}, t\ge T_{0}$ has almost surely bounded sample paths on $[T_{0},{\infty })$. Again by the Borell-TIS inequality

$$ \begin{array}{@{}rcl@{}} {\mathbb{P} \left\{ { \exists_{t\ge T_0} \ X_i(t)>\sqrt u (1+\mu_i t) } \right \} }\le e^{-\frac{(\sqrt u -C_i)^2}{2}\frac{(1+\mu_i T_0)^2}{T_0}} \end{array} $$

holds for all $\sqrt u>C_{i}:=\mathbb {E}\left \{ \sup _{t\in [T_{0},{\infty })} \frac {X_{1}(t)}{1+\mu _{i} t}\right \}$. Since for all large enough T₀ it holds that $\frac {(1+\mu _{i} T_{0})^{2}}{T_{0}} >\widehat g,$ the claim for r₀(u) is established.

Below we consider r₁(u). Since (t_A, s_A) ∈ A, we have from Proposition 3.1 that for any chosen small 𝜃₀

$$ \begin{array}{@{}rcl@{}} g(t,s)=g_A(t,s), \ \ (t,s)\in U_1 \subset A, \end{array} $$

and further (cf. Eq. 28)

$$ \begin{array}{@{}rcl@{}} Z(t,s)&=& (1+\mu_1 t, 1+\mu_2 s)\ {\Sigma}_{ts}^{-1} \ (X_1(t),X_2(s))^{\top}\\&& =: h_1(t,s) X_1(t) + h_2(t,s) X_2(s), \ \ (t,s)\in U_1, \end{array} $$

with

$$ \begin{array}{@{}rcl@{}} h_1(t,s) =\frac{(1+\mu_1 t)s-\rho s (1+\mu_2 s))}{ts-\rho^2 s ^2},\ \ \ h_2(t,s) =\frac{(1+\mu_2 s)t-\rho s (1+\mu_1 t))}{ts-\rho^2 s ^2}. \end{array} $$

Thus, similarly to Eq. 27 we conclude that

$$ \begin{array}{@{}rcl@{}} r_1(u)\le {\mathbb{P} \left\{ { \exists_{(t,s) \in U_1 \setminus \triangle^{(1)}_u \times \triangle_u^{(2)} } \ \frac{ Z(t,s)}{g_A(t,s)}>\sqrt u } \right \} } . \end{array} $$

(31)

Since h₁(t, s), h₂(t, s), g_A(t, s), (t, s) ∈ U₁ are all smooth functions and

$$ \begin{array}{@{}rcl@{}} \mathbb{E}\left\{ (X_i(t_1)-X_i(t_2))^2\right\}={\left\lvert t_1-t_2 \right\rvert}, \ \ i=1,2 \end{array} $$

one can check that, for all (t₁, s₁), (t₂, s₂) ∈ U₁,

$$ \begin{array}{@{}rcl@{}} \mathbb{E}\left\{ \left( \frac{Z(t_1,s_1)}{g_A(t_1,s_1)}-\frac{Z(t_2,s_2)}{g_A(t_2,s_2)}\right)^2\right\}\le \text{Const}\cdot ({\left\lvert {t_1-t_2} \right\rvert}+{\left\lvert {s_1-s_2} \right\rvert}). \end{array} $$

Therefore, an application of the Piterbarg’s inequality in Dȩbicki et al. (2017)[Lemma 5.1] (see also Piterbarg (1996) [Theorem 8.1] or Piterbarg (2016) [Theorem 3]) yields that

$$ \begin{array}{@{}rcl@{}} r_1(u)\le {\mathbb{P} \left\{ { \exists_{(t,s) \in U_1 \setminus \triangle^{(1)}_u \times \triangle_u^{(2)} } \ \frac{ Z(t,s)}{g_A(t,s)}>\sqrt u } \right \} }\le C_3 u^{3/2} e^{-\frac{ u }{2} \widetilde g_u}, \end{array} $$

(32)

where C₃ > 0 is some constant which does not depend on u and

$$ \begin{array}{@{}rcl@{}} \widetilde g_u : = \underset{(t,s)\in U_1 \setminus \triangle^{(1)}_u \times \triangle_u^{(2)} }{\inf} g_A(t,s). \end{array} $$

Moreover, we have from (i) of Lemma A.1 that for all (t_A + t, s_A + s) ∈ U₁

$$ \begin{array}{@{}rcl@{}} g_A(t_A+t,s_A + s)&\!\ge\!& g_A(t_A,s_A)+\frac{a_1}{2}(1 - \varepsilon)\left( (1 - \rho^2) t^2+\left( \rho t+\frac{\mu_2-2\mu_1\rho}{\mu_1}s\right)^2 \right. \\ &&\left.\ \ \ \ \ +\left( \frac{\mu_2-2\mu_1\rho}{\mu_1}\right)^2\left( \frac{\mu_2-2\mu_1\rho}{\mu_1} (1-2\rho)-1\right) s^2 \right) \end{array} $$

(33)

holds with some small ε > 0, where for all $-1< \rho <\hat \rho _{1}$ (see also the proof of (b).(i) in Lemma 9 of Dȩbicki et al. (2019) for ρ > 0)

$$ \begin{array}{@{}rcl@{}} \frac{\mu_2-2\mu_1\rho}{\mu_1} (1-2\rho)-1>0. \end{array} $$

Thus

$$ \begin{array}{@{}rcl@{}} \widetilde g_u \ge g_A(t_A,s_A) +\frac{a_1}{2}(1 - \varepsilon)\min\left( (1-\rho^2), \left( \frac{\mu_2-2\mu_1\rho}{\mu_1}\right)^2\left( \frac{\mu_2-2\mu_1\rho}{\mu_1} (1 - 2\rho) - 1\right) \right)\frac{(\ln(u))^2}{u}. \end{array} $$

Inserting the above to Eq. 31 completes the proof.

1.2 A.2 Proof of Lemma 4.2

We first analyze the summand p_{j, l; u}. We set

$$ \begin{array}{@{}rcl@{}} \boldsymbol{b}_{j,l;u} =(a_{j;u}, b_{l;u})^{\top}, \ \ \ \ a_{j;u}=1+\mu_1 (t_A+\frac{jT}{u}), \ \ b_{l;u}=1+\mu_2 (s_A+\frac{lS}{u}). \end{array} $$

(34)

It follows that

$$ \begin{array}{@{}rcl@{}} p_{j,l;u}& = &{\mathbb{P} \left\{ { \underset{s\in[0,S]}{\exists_{t \in [0, T]}} \begin{array}{lll} X_{1}(t_{A}+\frac{jT}{u}+\frac{t}{u})> a_{j;u} \sqrt{u} +\frac{\mu_{1}}{\sqrt u} t \\ X_{2}(s_{A}+\frac{lS}{u}+\frac{s}{u})> b_{l;u} \sqrt{u} +\frac{\mu_{2}}{\sqrt u} s \end{array} } \right \} } \\ & = &{\mathbb{P} \left\{ { \underset{s\in[0,S]}{\exists_{t \in [0, T]}} \begin{array}{lll} X_{1}(t_{A}+\frac{jT}{u})+ X_{1}(t_{A}+\frac{jT}{u}+\frac{t}{u})- X_{1}(t_{A} + \frac{jT}{u})> a_{j;u} \sqrt{u} +\frac{\mu_{1}}{\sqrt u} t \\ X_{2}(s_{A}+\frac{lS}{u})+ X_{2}(s_{A}+\frac{lS}{u}+\frac{s}{u})- X_{2}(s_{A}+\frac{lS}{u})> b_{l;u} \sqrt{u} +\frac{\mu_{2}}{\sqrt u} s \end{array} } \right \} } . \end{array} $$

(35)

Since $(t_{A}+\frac {jT}{u}, s_{A}+\frac {lS}{u})\in A$ for all large u, the covariance matrix of $\boldsymbol {Z}_{j,l;u}:=(X_{1}(t_{A}+\frac {jT}{u}), X_{2}(s_{A}+\frac {lS}{u}))^{\top }$ is given by

$$ \ {\Sigma}_{j,l;u}= \left( \begin{array}{ll} t_{A}+\frac{jT}{u} & \rho\ (s_{A}+\frac{lS}{u})\\ \rho\ (s_{A}+\frac{lS}{u}) &\quad s_{A}+\frac{lS}{u} \end{array} \right). $$

Thus, the density function of Z_{j, l; u} is given by

$$ \begin{array}{@{}rcl@{}} \phi_{{\Sigma}_{j,l;u}}(\boldsymbol{w})=\frac{1}{\sqrt{(2\pi)^{2}\left\lvert {\Sigma}_{j,l;u} \right\rvert}}\exp\left( -\frac{1}{2}\boldsymbol{w}^{\top}({\Sigma}_{j,l;u})^{-1} \boldsymbol{w} \right),\ \ \ \boldsymbol{w}=(w_1,w_2)^{\top}. \end{array} $$

By conditioning on the value of Z_{j, l; u} we rewrite Eq. 35 as

$$ \begin{array}{@{}rcl@{}} p_{j,l;u} = {\int}_{\mathbb{R}^{2}}\phi_{{\Sigma}_{j,l;u} }(\boldsymbol{w}) \mathbb{P} \left\{\left. \underset{s\in[0,S]}{\exists_{t \in [0, T]}} \begin{array}{lll} X_1(t_A+\frac{jT}{u} + \frac{t}{u})- X_1(t_A + \frac{jT}{u})> a_{j;u} \sqrt{u} +\frac{\mu_1}{\sqrt u} t - w_1\\ X_2(s_A+\frac{lS}{u} + \frac{s}{u})- X_2(s_A + \frac{lS}{u})> b_{l;u} \sqrt{u} + \frac{\mu_2}{\sqrt u} s -w_2 \end{array} \right| \boldsymbol{Z}_{j,l;u} = {\boldsymbol{w}} \right \} d{\boldsymbol{ w}}, \end{array} $$

Using change of variables $\boldsymbol {w} ={\sqrt {u}} \boldsymbol {b}_{j,l;u}-\boldsymbol {x}/ {\sqrt {u}}$ we further obtain

$$ \begin{array}{@{}rcl@{}} p_{j,l;u}= u^{-1}{\int}_{\mathbb{R}^{2}}\phi_{ {\Sigma}_{j,l;u} }(\sqrt{u} \boldsymbol{b}_{j,l;u}-\boldsymbol{x}/ {\sqrt{u}} ) P_{j,l;u}(\boldsymbol{x}) d{\boldsymbol{x}}, \end{array} $$

where

$$ \begin{array}{@{}rcl@{}} P_{j,l;u}({\boldsymbol{x}}) :=\mathbb{P} \left\{\left. \underset{s\in[0,S]}{\exists_{t \in [0, T]}} \begin{array}{lll} X_1(t_A+\frac{jT}{u}+\frac{t}{u})- X_1(t_A+\frac{jT}{u})> \frac{\mu_1}{\sqrt u} t+\frac{x_1}{\sqrt u}\\ X_2(s_A+\frac{lS}{u}+\frac{s}{u})- X_2(s_A+\frac{lS}{u})> \frac{\mu_2}{\sqrt u} s +\frac{x_2}{\sqrt u} \end{array} \right | \boldsymbol{Z}_{j,l;u} = {\sqrt{u}} \boldsymbol{b}_{j,l;u} - \frac{\boldsymbol{x}}{ {\sqrt{u}}} \right \} . \end{array} $$

Now, we analyse P_{j, l; u}(x). Due to the fact that (t_A, s_A) ∈ A, we have for all t ∈ [0, T], s ∈ [0, S], and large enough u

$$ \begin{array}{@{}rcl@{}} t_A+\frac{jT}{u}+\frac{t}{u}\ge t_A+\frac{jT}{u}>s_A+\frac{lS}{u}+\frac{s}{u}\ge s_A+\frac{lS}{u}. \end{array} $$

Thus, by the properties of Brownian motion

$$ \begin{array}{@{}rcl@{}} P_{j,l;u}(\boldsymbol{x}) &= &\mathbb{P} \left\{ \exists_{t\in[0,T]}\ X_1(t)-\mu_1 t> x_1 \right \} \\ && \times \mathbb{P} \left\{ \left. \exists_{s\in[0,S]} \ X_2(s_A+\frac{lS}{u} + \frac{s}{u})- X_2(s_A + \frac{lS}{u})> \frac{\mu_2}{\sqrt u} s + \frac{x_2}{\sqrt u} \right | \boldsymbol{Z}_{j,l;u}= {\sqrt{u}} \boldsymbol{b}_{j,l;u} - \frac{\boldsymbol{x}}{ {\sqrt{u}}} \right \} , \end{array} $$

Next we have

$$ \phi_{{\Sigma}_{j,l;u} }({\sqrt{u}} \boldsymbol{b}_{j,l;u}\!-\boldsymbol{x}/ {\sqrt{u}} ) = \frac{1}{\sqrt{(2\pi)^{2}\left\lvert {\Sigma}_{j,l;u} \right\rvert}}\exp\left( - \frac{1}{2} ({\sqrt{u}} \boldsymbol{b}_{j,l;u}-\boldsymbol{x}/ {\sqrt{u}} )^{\top}({\Sigma}_{j,l;u})^{-1} ({\sqrt{u}} \boldsymbol{b}_{j,l;u} - \boldsymbol{x}/ {\sqrt{u}} ) \right), $$

where the exponent can be rewritten as

$$ \begin{array}{@{}rcl@{}} && ({\sqrt{u}} \boldsymbol{b}_{j,l;u}-\boldsymbol{x}/ {\sqrt{u}} )^{\top}({\Sigma}_{j,l;u})^{-1} ({\sqrt{u}} \boldsymbol{b}_{j,l;u}-\boldsymbol{x}/ {\sqrt{u}} ) \\ &&=u (\boldsymbol{b}_{j,l;u})^{\top}{\Sigma}_{j,l;u}^{-1} \boldsymbol{b}_{j,l;u} - 2 \boldsymbol{x}^{\top} {\Sigma}_{j,l;u}^{-1} \boldsymbol{b}_{j,l;u} + \frac{1}{u } \boldsymbol{x}^{\top} {\Sigma}_{j,l;u}^{-1} \boldsymbol{x} \\ &&= u g_A(t_A+\frac{jT}{u}, s_A+\frac{lS}{u}) - 2 \boldsymbol{x}^{\top} {\Sigma}_{j,l;u}^{-1} \boldsymbol{b}_{j,l;u} + \frac{1}{u} \boldsymbol{x}^{\top} {\Sigma}_{j,l;u}^{-1} \boldsymbol{x} . \end{array} $$

Define

$$ \begin{array}{@{}rcl@{}} f_{j,l;u}(\boldsymbol{x}) := \exp\left( \boldsymbol{x}^{\top}{\Sigma}_{j,l;u}^{-1} \boldsymbol{b}_{j,l;u} - \frac{1}{2u } \boldsymbol{x}^{\top}{\Sigma}_{j,l;u}^{-1} \boldsymbol{x} \right),\ \ \ \ \boldsymbol{x} \in \mathbb{R}^2. \end{array} $$

Thus, it follows that

$$ \begin{array}{@{}rcl@{}} p_1(u) = \frac{u^{-1}}{ 2\pi} \sum\limits_{j=-N_u^{(1)}}^{N_u^{(1)}} \sum\limits_{l=-N_u^{(2)}}^{N_u^{(2)}} \frac{1}{\sqrt{\left\lvert {\Sigma}_{j,l;u} \right\rvert}} \exp\left( -\frac{1}{2} u g_A(t_A + \frac{jT}{u}, s_A+\frac{lS}{u}) \right) {\int}_{\mathbb{R}^{2}} f_{j,l;u}(\boldsymbol{x})P_{j,l;u}(\boldsymbol{x}) d\boldsymbol{x}. \end{array} $$

Further, we obtain from (i) of Lemma A.1 that, for all large enough u,

$$ \begin{array}{@{}rcl@{}} g_A(t_A+\frac{jT}{u}, s_A+\frac{lS}{u}) \sim g_A(t_A,s_A)+\frac{1}{2}\left( a_1 \left( \frac{jT}{u}\right)^2-2a_2 \left( \frac{jT}{u}\right)\left( \frac{lS}{u}\right) +a_3 \left( \frac{lS}{u}\right)^2\right) \end{array} $$

holds uniformly for $-N_{u}^{(1)}\le j\le {N}_{u}^{(1)}, {-N}_{u}^{(2)}\le l\le {N}_{u}^{(2)}.$

Consequently, by Lemma A.3 below we obtain

$$ \begin{array}{@{}rcl@{}} \underset{u \to \infty}{\lim} \frac{p_1(u)}{ \exp\left( - g_A(t_A, s_A)u /{2} \right) } = \frac{1}{2\pi\sqrt{\left\lvert {\Sigma}_{0} \right\rvert}} \frac{\mathcal{H} (\mu_1;T)\mathcal{H}(\mu_2-2\mu_1\rho;S)}{TS} {\int}_{\mathbb{R}^2} e^{\frac{- (a_1x_1^2-2a_2 x_1x_2+a_3 x_2^2)}{4}} d \boldsymbol{x}, \end{array} $$

which gives the result for p₁(u). The claim for p₂(u) follows with the same arguments. $\Box$

Lemma A.3

For any T, S > 0

$$ \begin{array}{@{}rcl@{}} \lim\limits_{u\to{\infty}}{\int}_{\mathbb{R}^{2}} f_{j,l;u}(\boldsymbol{x})P_{j,l;u}(\boldsymbol{x}) d \boldsymbol{x}=\mathcal{H} (\mu_1;T)\mathcal{H}(\mu_2-2\mu_1\rho;S) \end{array} $$

holds uniformly for $-{N}_{u}^{(1)}\le j\le N_{u}^{(1)}, -N_{u}^{(2)}\le l\le N_{u}^{(2)}.$

We omit the tedious proof of Lemma A.3 since its idea is standard, i.e., it is based on finding a uniform integrable bound for the integrand and then using the dominated convergence theorem.

1.3 A.3 Proof of Lemma 4.3

Let us begin with Π₁(u). It follows that

$$ \begin{array}{@{}rcl@{}} {\Pi}_1(u)& = &\sum\limits_{j=-N_u^{(1)}}^{N_u^{(1)}}\underset{ -N_u^{(2)}\le l_1< l_2\le N_u^{(2)}}{\sum}p_{j,l_1,l_2;u}\\ &=&\sum\limits_{j=-N_u^{(1)}}^{N_u^{(1)}} \sum\limits_{l=-N_u^{(2)}}^{N_u^{(2)}} p_{j,l,l+1;u}+ \sum\limits_{j=-N_u^{(1)}}^{N_u^{(1)}} \sum\limits_{l_1=-N_u^{(2)}}^{N_u^{(2)}} \sum\limits_{l_2=l_1+2}^{N_u^{(2)}} p_{j,l_1,l_2;u} =: {\Pi}_{11}(u)+{\Pi}_{12}(u). \end{array} $$

In order to deal with Π₁₁(u) we note that

$$ \begin{array}{@{}rcl@{}} p_{j,l,l+1;u}=p_{j,l;u}+p_{j,l+1;u}-\widetilde p_{j,l;u}, \end{array} $$

where

$$ \widetilde p_{j,l;u}=\mathbb{P} \left\{ \exists_{(t,s)\in\triangle^{(1)}_{j;u}\times (\triangle_{l;u}^{(2)}\cup \triangle_{l+1;u}^{(2)})} \ X_{1}(t)> \sqrt{u}(1 + \mu_{1} t ), X_{2}(s)> \sqrt{u}(1 + \mu_{2} s ) \right \} . $$

Then we have

$$ \begin{array}{@{}rcl@{}} {\Pi}_{11}(u)=\sum\limits_{j=-N^{(1)}_u}^{N_u^{(1)}} \sum\limits_{l=-N^{(2)}_u}^{N_u^{(2)}} (p_{j,l;u}+p_{j,l+1;u}-\widetilde p_{j,l;u}). \end{array} $$

Using the same arguments as in the proof of Lemma 4.2 we obtain

$$ \begin{array}{@{}rcl@{}} \underset{u\to{\infty}}{\lim}\frac{{\Pi}_{11}(u)}{ e^{- g_A(t_A,s_A) u /2 } }= \frac{1}{\mu_1 (\mu_2-2\mu_1\rho)} \left( \frac{2\mathcal{H}(\mu_1;T)\mathcal{H}(\mu_2-2\mu_1\rho;S)}{T S} -\frac{\mathcal{H}(\mu_1;T)\mathcal{H}(\mu_2-2\mu_1\rho;2S)}{T S} \right), \end{array} $$

which gives that

$$ \begin{array}{@{}rcl@{}} \underset{S\to{\infty}}{\limsup}\underset{T\to{\infty}}{\limsup} \underset{u\to{\infty}}{\lim}\frac{{\Pi}_{11}(u)}{ e^{- g_A(t_A,s_A) u /2 } }= 0. \end{array} $$

Next we consider Π₁₂(u) which is more involved. We have (recall Eq. 34 for a_{j; u}, b_{l; u})

$$ \begin{array}{@{}rcl@{}} p_{j,l_{1},l_{2};u}&=&\mathbb{P} \left\{ \underset{ s_{2}\in[0,S]} {\underset{ s_{1} \in [0, S]} {\underset{ t \in [0, T]} \exists}} \begin{array}{lll} X_{1}(t_{A}+\frac{jT}{u}+\frac{t}{u})> a_{j;u} \sqrt{u} +\frac{\mu_{1}}{\sqrt{u}} t \\ X_{2}(s_{A}+\frac{l_{1}S}{u}+\frac{s_{1}}{u})> b_{l_{1};u} \sqrt{u} +\frac{\mu_{2}}{\sqrt {u}} s_{1}\\ X_{2}(s_{A}+\frac{l_{2}S}{u}+\frac{s_{2}}{u})> b_{l_{2};u} \sqrt{u} +\frac{\mu_{2}}{\sqrt u} s_{2} \end{array} \right \} \\ &\quad\le & \mathbb{P} \left\{ \underset{ s_{2}\in[0,S]} {\underset{ s_{1} \in [0, S]} {\underset{ t \in [0, T]} \exists}} \begin{array}{ll} X_{1}(t_{A}+\frac{jT}{u}+\frac{t}{u})> a_{j;u} \sqrt{u} +\frac{\mu_{1}}{\sqrt u} t \\ \frac{1}{2}\left( X_{2}(s_{A}+\frac{l_{1}S}{u}+\frac{s_{1}}{u})+ X_{2}(s_{A}+\frac{l_{2}S}{u}+\frac{s_{2}}{u})\right)> b_{l_{1},l_{2};u} \sqrt{u} +\frac{\mu_{2}}{2\sqrt u} (s_{1}+s_{2}) \end{array} \right \} \\&=&: P_{j,l_{1},l_{2};u}, \end{array} $$

(36)

with

$$ \begin{array}{@{}rcl@{}} b_{l_1,l_2;u} =1+\mu_2\left( s_A+\frac{l_1S}{u}+\frac{(l_2-l_1)S}{2u}\right). \end{array} $$

For notational simplicity, we shall denote

$$ \begin{array}{@{}rcl@{}} \widetilde{t_A}=t_A+\frac{jT}{u}, \ \ \widetilde{s_A}=s_A+\frac{l_1S}{u},\ \ \overline{s_A}=\widetilde{s_A}+\frac{(l_2-l_1)S}{2 u}, \ \ \widehat{s_A}=\widetilde{s_A}+\frac{(l_2-l_1)S}{4 u}. \end{array} $$

Again by conditioning on the event

$$ \begin{array}{@{}rcl@{}} E_{j,l_1,l_2;u}(x_1,x_2):=\left\{X_1(\widetilde{t_A}) = a_{j;u} \sqrt{u}-\frac{x_1}{\sqrt {u}}, \ \ \ \frac{1}{2}\left( X_2(\widetilde{s_A})+ X_2(s_A+\frac{l_2S}{u} )\right)=b_{l_1,l_2;u} \sqrt{u} -\frac{x_2}{ \sqrt u} \right\}, \end{array} $$

we have

$$ \begin{array}{@{}rcl@{}} P_{j,l_1,l_2;u}= u^{-1}{\int}_{\mathbb{R}^{2}}\phi_{{\Sigma}_{j,l_1,l_2;u} }({\sqrt{u}} \boldsymbol{b}_{j,l_1,l_2;u}-\boldsymbol{x}/ {\sqrt{u}} ) F(j,l_1,l_2;u,\boldsymbol{x}) d \boldsymbol{x}, \end{array} $$

where

$$ \begin{array}{@{}rcl@{}} {\Sigma}_{j,l_1,l_2;u} = \left( \begin{array}{ll} \widetilde{t_A} & \rho\ \overline{s_A}\\ \rho\ \overline{s_A} & \widehat{s_A} \end{array} \right), \ \ \ \boldsymbol{b}_{j,l_1,l_2;u}=(a_{j;u}, b_{l_1,l_2;u})^{\top} \end{array} $$

and

$$ \begin{array}{@{}rcl@{}} F(j,l_1,l_2;u,\boldsymbol{x}) := \mathbb{P} \left\{ \exists_{t \in [0, T]}\ X_1(t) - \mu_1 t> x_1 \right \} \ \mathbb{P} \left\{ \left. \underset{ s_2\in[0,S]} {\underset{ s_1 \in [0, S]} { \exists}} Y_{j,l_1,l_2;u}(s_1,s_2) \!>\!x_2\ \right | E_{j,l_1,l_2;u}(x_1,x_2) \right \} , \end{array} $$

where

$$ \begin{array}{@{}rcl@{}} Y_{j,l_1,l_2;u}(s_1,s_2) = \frac{\sqrt u}{2} \left( X_2(\widetilde{s_A} + \frac{s_1}{u})- X_2(\widetilde{s_A})+ X_2(s_A + \frac{l_2S}{u}+\frac{s_2}{u})- X_2(s_A + \frac{l_2S}{u}) \right)-\frac{\mu_2}{2}(s_1+s_2). \end{array} $$

Similarly as in the proof of Lemma 4.2, we obtain

$$ \begin{array}{@{}rcl@{}} \phi_{{{\Sigma}}_{j,l_1,l_2;u} }({\sqrt{u}} \boldsymbol{b}_{j,l_1,l_2;u} - \boldsymbol{x}/ {\sqrt{u}} ) = \frac{1}{\sqrt{(2\pi)^{2}\left\lvert {\Sigma}_{j,l_1,l_2;u} \right\rvert}}\exp\left( - \frac{1}{2} u \ (\boldsymbol{b}_{j,l_1,l_2;u})^{\top}{\Sigma}_{j,l_1,l_2;u}^{-1} \boldsymbol{b}_{j,l_1,l_2;u} \right)\ f_{j,l_1,l_2;u}(\boldsymbol{x}), \end{array} $$

where

$$ \begin{array}{@{}rcl@{}} f_{j,l_1,l_2;u}(\boldsymbol{x}):= \exp\left( \boldsymbol{x}^{\top} {\Sigma}_{j,l_1,l_2;u}^{-1} \boldsymbol{b}_{j,l_1,l_2;u} - \frac{1}{2u } \boldsymbol{x}^{\top} {\Sigma}_{j,l_1,l_2;u}^{-1} \boldsymbol{x} \right). \end{array} $$

Next, some elementary calculations give that

$$ \begin{array}{@{}rcl@{}} (\boldsymbol{b}_{j,l_1,l_2;u})^{\top}{\Sigma}_{j,l_1,l_2;u}^{-1} \boldsymbol{b}_{j,l_1,l_2;u} = g_A(t_A + \frac{jT}{u}, s_A + \frac{l_1S}{u}+\frac{(l_2-l_1)S}{2 u}) + \frac{ \widetilde{t_A} g_A(\widetilde{t_A}, \overline{s_A} ) -a_{j;u} ^2 }{4 (\widetilde{t_A} \widehat{s_A}-\rho^2{\overline{s_A}}^2)}\ \frac{(l_2-l_1)S}{u}. \end{array} $$

Further, note that

$$ \begin{array}{@{}rcl@{}} g_A(t_A + \frac{jT}{u}, s_A + \frac{l_1S}{u} + \frac{(l_2-l_1)S}{2 u}) = g_A(t_A + \frac{jT}{u}, s_A + \frac{l_1S}{u})+\frac{\partial g_A(t,s)}{\partial s} \mid_{(\widetilde{t_A}, \widetilde{s_A}+\theta_{l_1,l_2;u}\frac{(l_2-l_1)S}{2 u})} \frac{(l_2-l_1)S}{2 u} \end{array} $$

holds for some $\theta _{l_{1},l_{2};u}\in (0,1)$ and

$$ \begin{array}{@{}rcl@{}} \frac{\partial g_A(t,s)}{\partial t} \mid_{(\widetilde{t_A}, \widetilde{s_A}+\theta_{l_1,l_2;u}\frac{(l_2-l_1)S}{2 u})} \ \ \to\ 0, \ \ \ \ u\to{\infty} \end{array} $$

holds uniformly for j, l₁, l₂ (hereafter when we write j, l₁, l₂ we mean ${-N}_{u}^{(1)}\le j \le {N}_{u}^{(1)}, {-N}_{u}^{(2)}\le l_{1},l_{2}\le {N}_{u}^{(2)}$)). Consequently

$$ \begin{array}{@{}rcl@{}} \exp\left( -\frac{1}{2} u \ (\boldsymbol{b}_{j,l_1,l_2;u})^{\top}{\Sigma}_{j,l_1,l_2;u}^{-1} \boldsymbol{b}_{j,l_1,l_2;u} \right) \sim \exp\left( -\frac{1}{2} u \ g_A(t_A+\frac{jT}{u}, s_A+\frac{l_1S}{u}) \right) e^{- Q_0 (l_2-l_1)S} \end{array} $$

(37)

holds uniformly for j, l₁, l₂ as $u\to {\infty }$, where (by (b).(i) of Lemma 9 of Dȩbicki et al. (2019) or Lemma A.1.(i) with a₁ > 0)

$$ \begin{array}{@{}rcl@{}} Q_0=\frac{t_A g_A(t_A,s_A) -(1+\mu_1 t_A)^2}{8(t_A s_A-\rho^2s_A^2)} >0. \end{array} $$

Next, we consider the uniform, in j, l₁, l₂, limit of the following:

$$ \begin{array}{@{}rcl@{}} \mathbb{P} \left\{ \underset{ s_2\in[0,S]} {\underset{ s_1 \in [0, S]} { \exists}} Y_{j,l_1,l_2;u}(s_1,s_2) >x_2\ \Bigg{|} E_{j,l_1,l_2;u}(x_1,x_2) \right \} , \ \ \ u\to{\infty} \end{array} $$

For the conditional mean we can derive that

$$ \begin{array}{@{}rcl@{}} \mathbb{E}\left\{ Y_{j,l_1,l_2;u}(s_1,s_2) { \mid} E_{j,l_1,l_2;u}(x_1,x_2) \right\}&=& -\frac{\mu_2}{2}(s_1+s_2) \\ &&+ \left( \frac{\rho (s_1+s_2)}{2\sqrt u}, \frac{s_1}{4\sqrt u}\right)\ {\Sigma}_{j,l_1,l_2;u}^{-1} (\boldsymbol{b}_{j,l_1,l_2;u} - \boldsymbol{x}/ {\sqrt{u}} ), \end{array} $$

which further gives that

$$ \begin{array}{@{}rcl@{}} &&\mathbb{E}{ Y_{j,l_1,l_2;u}(s_1,s_2) { \mid} E_{j,l_1,l_2;u}(x_1,x_2) } = - \frac{\mu_2}{2}(s_1 + s_2) + \frac{2\rho a_{j;u} \widehat{s_A} - \rho a_{j;u} \overline{s_A} -2\rho^2 b_{j,l_1,l_2;u} \overline{s_A} + b_{j,l_1,l_2;u} \widetilde{t_A}}{4(\widetilde{t_A} \widehat{s_A}-\rho^2 {\overline{s_A}}^2)} s_1\\ && \ \ \ \ \ \ \ \ \ \ \ +\frac{ \rho a_{j;u} \widehat{s_A}-\rho^2 b_{j,l_1,l_2;u} \overline{s_A} }{2(\widetilde{t_A} \widehat{s_A}-\rho^2 {\overline{s_A}}^2)} s_2 + \frac{ \rho \overline{s_A} s_1 -2\rho \widehat{s_A} (s_1+s_2) }{4(\widetilde{t_A} \widehat{s_A}-\rho^2 {\overline{s_A}}^2)}\ \frac{x_1}{u} + \frac{ 2\rho ^2 \overline{s_A} (s_1+s_2) - \widetilde{t_A}s_1 }{4(\widetilde{t_A} \widehat{s_A}-\rho^2 {\overline{s_A}}^2)}\ \frac{x_2}{u}\\ && \ \ \ \ \ \ \ \ \ \ \quad to -\frac{1}{2}\left( \mu_2-2\mu_1\rho\right) s_2, \ \ \ \ u\to{\infty}. \end{array} $$

For the conditional variance of the increments we have

$$ \begin{array}{@{}rcl@{}} &&\text{Var}\left\{ Y_{j,l_1,l_2;u}(s_1,s_2) - Y_{j,l_1,l_2;u}(s^{\prime}_1,s^{\prime}_2) { \mid} E_{j,l_1,l_2;u}(x_1,x_2) \right\} = \frac{\left\lvert s_1-{s}_1^{\prime} \right\rvert+\left\lvert s_2-{s}_2^{\prime} \right\rvert}{4}\\ &&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\left( \frac{\rho(s_1-s^{\prime}_1 + s_2-s^{\prime}_2)}{2\sqrt u}, \frac{s_1-s^{\prime}_1}{4 \sqrt u}\right) {\Sigma}_{j,l_1,l_2;u}^{-1} \left( \frac{\rho(s_1 - s^{\prime}_1 + s_2 - s^{\prime}_2)}{2\sqrt u}, \frac{s_1 - s^{\prime}_1}{4 \sqrt u}\right)^{\top}\\ &&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \to \frac{\left\lvert s_1-s^{\prime}_1 \right\rvert+\left\lvert s_2-s^{\prime}_2 \right\rvert}{4}, \ \ \ u\to{\infty}. \end{array} $$

Therefore, similarly as in Lemma A.3 we can show that as $u\to {\infty }$

$$ \begin{array}{@{}rcl@{}} &&\mathbb{P} \left\{\left. \underset{ s_2\in[0,S]} {\underset{ s_1 \in [0, S]} { \exists}} Y_{j,l_1,l_2;u}(s_1,s_2) >x_2\ \right | E_{j,l_1,l_2;u}(x_1,x_2) \right \} \ \\&&\to \mathbb{P} \left\{ \underset{ s_2\in[0,S]} {\underset{ s_1 \in [0, S]} { \exists}} \frac{1}{2}(B_1(s_1)+B_2(s_2)) -\frac{1}{2}\left( \mu_2-2\mu_1\rho\right) s_2 >x_2 \right \} . \end{array} $$

Consequently, the dominated convergence theorem gives

$$ \begin{array}{@{}rcl@{}} &&{\int}_{\mathbb{R}^{2}} f_{j,l_1,l_2;u}(\boldsymbol{x}) F(j,l_1,l_2;u,\boldsymbol{x}) d \boldsymbol{x} \\ && \to {\int}_{\mathbb{R}}e^{2\mu_1 x_1} \mathbb{P} \left\{ \exists_{t\in[0,T]}\ X_1(t)-\mu_1 t> x_1 \right \} dx_1 \\ &&\ \ \ \ \ \times {\int}_\mathbb{R} e^{2(\mu_2-2\mu_1\rho) x_2} \mathbb{P} \left\{ \underset{ s_2\in[0,S]} {\underset{ s_1 \in [0, S]} { \exists}} \frac{1}{2}(B_1(s_1)+B_2(s_2)) -\frac{1}{2}\left( \mu_2-2\mu_1\rho\right) s_2 >x_2 \right \} dx_2\\ &&=: \mathcal{H}(\mu_1; T)\ \mathcal{H}(\mu_1,\mu_2;S) \end{array} $$

(38)

holds uniformly for j, l₁, l₂, as $u\to {\infty }$.

Next we derive a useful upper bound for ${\mathscr{H}}(\mu _{1},\mu _{2};S), S>0$:

$$ \begin{array}{@{}rcl@{}} \mathcal{H}(\mu_1,\mu_2;S) \le (\lfloor S\rfloor)^2 e^{Q_0 S} \mathcal{H}(\mu_1,\mu_2;1)<{\infty}. \end{array} $$

(39)

In order to prove (39), by taking j = l₁ = 0, l₂ = 1 we arrive at

$$ \begin{array}{@{}rcl@{}} P_{0,0,1;u}&=&\mathbb{P} \left\{ \underset{ s_2\in[0,S]} {\underset{ s_1 \in [0, S]} {\underset{ t \in [0, T]} \exists}} \begin{array}{ll} X_1(t_A+\frac{t}{u})> a_{0;u} \sqrt{u} +\frac{\mu_1}{\sqrt u} t \\ \frac{1}{2}(X_2(s_A+\frac{s_1}{u})+ X_2(s_A+\frac{S}{u}+\frac{s_2}{u}))> b_{0,1;u} \sqrt{u} +\frac{\mu_2}{2\sqrt u} (s_1+s_2) \end{array} \right \} \\ &\sim & \frac{u^{-1}}{\sqrt{(2\pi)^{2}\left\lvert {\Sigma}_{0,0,0;u} \right\rvert}} \exp\left( -\frac{1}{2} u \ g_A(t_A, s_A) \right) e^{- Q_0 S} \mathcal{H}(\mu_1; T)\mathcal{H}(\mu_1,\mu_2;S). \end{array} $$

(40)

Define, for any integers 0 ≤ m, n ≤⌊S⌋,

$$ \begin{array}{@{}rcl@{}} q_{m,n;u} := \mathbb{P} \left\{ \underset{ s_2\in[0,1]} {\underset{ s_1 \in [0, 1]} {\underset{ t \in [0, T]} \exists}} \begin{array}{cc} X_1(t_A+\frac{t}{u})> a_{0;u} \sqrt{u} +\frac{\mu_1}{\sqrt u} t \\ \frac{1}{2}(X_2(s_A+\frac{m}{u}+\frac{s_1}{u})+ X_2(s_A+\frac{S+n}{u}+\frac{s_2}{u}))> \widetilde b_{m,n;u} \sqrt{u} +\frac{\mu_2}{2\sqrt u} (s_1+s_2) \end{array} \right \} \end{array} $$

with

$$ \begin{array}{@{}rcl@{}} \widetilde b_{m,n;u}=1+\mu_2\left( s_0+\frac{m}{u}+\frac{ S+n-m}{2u}\right). \end{array} $$

Using the same arguments as in the derivation of Eq. 40 one can show that

$$ \begin{array}{@{}rcl@{}} q_{m,n;u} \sim \frac{u^{-1}}{\sqrt{(2\pi)^{2}\left\lvert {\Sigma}_{0,0,0;u} \right\rvert}} \exp\left( -\frac{1}{2} u \ g_A(t_A, s_A) \right) e^{- Q_0 (S+n-m)} \mathcal{H}(\mu_1; T)\mathcal{H}(\mu_1,\mu_2;1). \end{array} $$

(41)

Comparing Eqs. 40 and 41 we derive

$$ \begin{array}{@{}rcl@{}} \mathcal{H}(\mu_1,\mu_2;S) &\le& \sum\limits_{m=0}^{\lfloor S\rfloor-1} \sum\limits_{n=0}^{\lfloor S\rfloor-1} e^{- Q_0 (n-m)} \mathcal{H}(\mu_1,\mu_2;1)\\ &\le & (\lfloor S\rfloor)^2 e^{Q_0 S} \mathcal{H}(\mu_1,\mu_2;1). \end{array} $$

The finiteness of ${\mathscr{H}}(\mu _{1},\mu _{2};1)$ can be proved by using the Borell-TIS inequality. This justifies bound (39).

Now, we are ready to analyse the triple sum Π₁₂(u). We have

$$ \begin{array}{@{}rcl@{}} {\Pi}_{12}(u)& = &\sum\limits_{j=-N_u^{(1)}}^{N_u^{(1)}} \sum\limits_{l_1=-N_u^{(2)}}^{N_u^{(2)}} \sum\limits_{l_2=l_1+2}^{N_u^{(2)}}\frac{u^{-1}}{\sqrt{(2\pi)^{2}\left\lvert {\Sigma}_{j,l_1,l_2;u} \right\rvert}} \\ && \ \ \ \ \ \ \ \times\exp\left( -\frac{1}{2} u \ (\boldsymbol{b}_{j,l_1,l_2;u})^{\top}{\Sigma}_{j,l_1,l_2;u}^{-1} \boldsymbol{b}_{j,l_1,l_2;u} \right) {\int}_{\mathbb{R}^{2}} f_{j,l_1;u}(\boldsymbol{x}) F(j,l_1,l_2;u,\boldsymbol{x}) d\boldsymbol{x}. \end{array} $$

Therefore, we can derive from Eqs. 37- 38 and 39 that

$$ \begin{array}{@{}rcl@{}} \underset{u\to{\infty}}{\lim}\frac{ {\Pi}_{12}(u)}{\exp(-ug_A(t_A,s_A)/2)} \le \text{Const} \ \sum\limits_{k=1}^{{\infty}} e^{- kQ_0 S} \frac{{\mathcal{H}(\mu_1;T) \mathcal{H}(\mu_1,\mu_2;1)(\lfloor S\rfloor)^2}}{TS} . \end{array} $$

Consequently, the above implies that

$$ \begin{array}{@{}rcl@{}} \underset{S\to{\infty}}{\limsup} \underset{T\to{\infty}}{\limsup}\underset{u\to{\infty}}{\lim}\frac{{\Pi}_{12}(u)}{\exp(-ug_A(t_A,s_A)/2)}=0. \end{array} $$

Thus, the claim for Π₁(u) is established. Using similar arguments, one can further show that the claim for Π₂(u) holds. $\Box$

1.4 A.4 Proof of Lemma 4.4

The claim for r₁(u) follows from the same arguments as that for r₀(u) of Lemma 4.1. Next, as in the proof of Lemma 4.1, using the Piterbarg’s inequality we can show that

$$ \begin{array}{@{}rcl@{}} r_2(u)\le C_2 u^{3/2} e^{-\frac{ u }{2} \widetilde g_u}, \end{array} $$

where C₂ > 0 is some constant which does not depend on u, and thus the claim for r₂(u) follows since

$$ \begin{array}{@{}rcl@{}} \widetilde g_u&=& \underset{(t,s)\in D_1\cup D_2 }{\inf} g(t,s)=\underset{s \in[s_0-\theta_0, s_0-\ln(u)/\sqrt u]\cup[ s_0+\ln(u)/\sqrt u, s_0+\theta_0]}{\inf} g_L(s)\\ &\ge& g_L(s^*) +\frac{b_0}{2}(1-\varepsilon) \frac{(\ln(u))^2}{u}, \end{array} $$

where the last inequality follows by (ii.3) of Lemma A.1. Finally, the claim for r₃(u) can be proved similarly, by using Piterbarg’s inequality and (ii.1)-(ii.2) of Lemma A.1. $\Box$

1.5 A.5: Proof of Lemma 4.5

We first analyse the summand p_{j; u}. Let

$$ \begin{array}{@{}rcl@{}} \boldsymbol b_{j;u}=(a_{j;u}, b_{j,u})^{\top}, \ \ \ a_{j;u}=1+\mu_1 (t^*+\frac{jT}{u}), \ \ \ b_{j;u}=1+\mu_2 (s^*+\frac{jT}{u}). \end{array} $$

Then

$$ \begin{array}{@{}rcl@{}} p_{j;u} = \mathbb{P} \left\{ \underset{ (t,s) \in \triangle_{T,S} }\exists \begin{array}{lll} X_{1}(t^{*}+\frac{jT}{u}+\frac{t}{u})> a_{j;u} \sqrt{u} +\frac{\mu_{1}}{\sqrt {u}} t \\ X_{2}(s^{*}+\frac{jT}{u}+\frac{s}{u})> b_{j;u} \sqrt{u} +\frac{\mu_{2}}{\sqrt {u}} s \end{array} \right \} . \end{array} $$

Define $\boldsymbol {Z}_{j;u}:=(X_{1}(t^{*}+\frac {jT}{u}), X_{2}(s^{*}+\frac {jT}{u}))^{\top }$, whose density function is given by

$$ \begin{array}{@{}rcl@{}} \phi_{{\Sigma}_{j;u}}(\boldsymbol {w})=\frac{1}{\sqrt{(2\pi)^{2}\left\lvert {\Sigma}_{j;u} \right\rvert}}\exp\left( -\frac{1}{2}\boldsymbol {w}^{\top}({\Sigma}_{j;u})^{-1} \boldsymbol{w} \right),\ \ \ \boldsymbol{w}=(w_1,w_2)^{\top}, \end{array} $$

with the covariance matrix given by

$$ \ {\Sigma}_{j;u}= \left( \begin{array}{ll} t^{*}+\frac{jT}{u} & \rho\ (t^{*}+\frac{jT}{u})\\ \rho\ (t^{*}+\frac{jT}{u}) &\quad s^{*}+\frac{jT}{u} \end{array} \right). $$

By conditioning on the value of Z_{j; u} and using change of variables $\boldsymbol {w} ={\sqrt {u}} \boldsymbol {b}_{j;u}-\boldsymbol {x}/ {\sqrt {u}}$, we further obtain

$$ \begin{array}{@{}rcl@{}} p_{j;u}= u^{-1}{\int}_{\mathbb{R}^{2}}\phi_{ {\Sigma}_{j;u} }({\sqrt{u}} \boldsymbol{b}_{j;u}-\boldsymbol{x}/ {\sqrt{u}} ) \mathbb{P} \left\{ \underset{ (t,s) \in \triangle_{T,S} }\exists \begin{array}{lll} X_1(t)- \mu_1 t>x_1\\ X_2(s)- \mu_2 s>x_2 \end{array} \right \} d \boldsymbol{x}. \end{array} $$

Consequently, similar arguments as in the proof of Lemma 4.2 yield

$$ \begin{array}{@{}rcl@{}} p_1(u) &\sim& p_2(u)\ \sim\ \sum\limits_{j=-N_u^{(1)}}^{N_u^{(1)}} p_{j;u} \sim \frac{\mathcal{H}(T,S)u^{-1}}{\sqrt{(2\pi t^*)^2(1-\rho^2)}} \sum\limits_{j=-N_u^{(1)}}^{N_u^{(1)}} e^{-\frac{u}{2} g_L(t^*+\frac{jT}{u})}\\ &\sim& \frac{\mathcal{H}(T,S)u^{-1/2}}{T\sqrt{(2\pi t^*)^2(1-\rho^2)}} e^{-\frac{u}{2} g_L(t^*)}{\int}_{\mathbb{R}}e^{-\frac{b_0}{4}x^2} dx. \end{array} $$

This completes the proof. $\Box$

1.6 A.6: Proof of Lemma 4.6

First note that

$$ \begin{array}{@{}rcl@{}} &&\mathbb{P} \left\{ \exists_{(t,s) \in (t^*,s^*)+u^{-1}\triangle_{T_1+T_2,S}} \ X_1(t)> \sqrt{u}(1+ \mu_1 t ), X_2(s)> \sqrt{u}(1+ \mu_2 s ) \right \} \\ &&\ \ \le \mathbb{P} \left\{ \exists_{(t,s) \in (t^*,s^*)+u^{-1}\triangle_{T_1,S}} \ X_1(t)> \sqrt{u}(1+ \mu_1 t ), X_2(s)> \sqrt{u}(1+ \mu_2 s ) \right \} \\ &&\ \ \ \ \!+ \mathbb{P} \left\{ \exists_{(t,s) \in (t^*+\frac{T_1}{u},s^*+\frac{T_1}{u})+u^{-1}\triangle_{T_2,S}} \ X_1(t)\!>\! \sqrt{u}(1 + \mu_1 t ), X_2(s)\!>\! \sqrt{u}(1 + \mu_2 s ) \right \} . \end{array} $$

Using the same arguments as the proof of Lemma 4 5, we conclude the sub-additivity of ${\mathscr{H}}(T,S), T>0.$ Thus

$$ \begin{array}{@{}rcl@{}} \underset{T\to{\infty}}{\lim}\frac{1}{T}\mathcal{H}(T,S) = \underset{T> 0}{\inf}\frac{1}{T}\mathcal{H}(T,S) <{\infty} \end{array} $$

follows directly from Fekete’s lemma. Moreover, since by definition

$$ \begin{array}{@{}rcl@{}} \mathcal{H}(T,S)\ge {\int}_{\mathbb{R}^2}e^{\boldsymbol{x}^{\top} {\Sigma}_{*}^{-1} \boldsymbol{b}_{*} }\mathbb{P} \left\{ {\underset{ t \in [0,T]}\exists } \begin{array}{lll} X_1(t)- \mu_1 t > x_1 \\ X_2(t)- \mu_2 t >x_2 \end{array} \right \} dx_1 dx_2, \end{array} $$

the positive lower bound follows from Lemma 4.7 in Dȩbicki et al. (2018). This completes the proof. $\Box$

1.7 A.7 Proof of Lemma 4.7

We begin with the analysis of Π₁(u). We first look at p_{j, l; u}. Denote

$$ \begin{array}{@{}rcl@{}} &&\boldsymbol {b}_{u}=\boldsymbol {b}_{j,l,m,n;u}:=(a_{j,m;u}, b_{j,l,m,n;u})^{\top}, \\ && a_{j,m;u}=1+\mu_1 (t^*+\frac{jT+m}{u}), \ \ b_{j,l,m,n;u}=1+\mu_2 (t^*+\frac{jT+m}{u}+\frac{lS+n}{u}). \end{array} $$

It is derived that

$$ \begin{array}{@{}rcl@{}} p_{j,l;u} &\le &\sum\limits_{m=0}^{\lfloor T \rfloor-1}\sum\limits_{n=0}^{\lfloor S \rfloor-1} \mathbb{P} \left\{ \underset{s-t\in[\frac{lS}{u}+\frac{n}{u}, \frac{lS}{u}+\frac{n+1}{u} ]}{\underset{ t \in [t^*+\frac{jT}{u}+\frac{m}{u}, t^*+\frac{jT}{u}+\frac{m+1}{u}] }\exists } \ X_1(t)> \sqrt{u}(1+ \mu_1 t ), X_2(s)> \sqrt{u}(1+ \mu_2 s ) \right \} \\ &=& \sum\limits_{m=0}^{\lfloor T \rfloor-1}\sum\limits_{n=0}^{\lfloor S \rfloor-1} \mathbb{P} \left\{ \underset{s\in[0,1]}{\underset{ t \in [0, 1] }\exists } \begin{array}{lll} X_1(t^*+\frac{jT+m}{u}+\frac{t}{u}) > a_{j,m;u} \sqrt{u} +\frac{\mu_1}{\sqrt u} t \\ X_2(t^*+\frac{jT+m}{u}+\frac{lS+n}{u}+\frac{t+s}{u}) > b_{j,l,m,n;u} \sqrt{u} +\frac{\mu_2}{\sqrt u} (t+s) \end{array} \right \} \\ &=:& \sum\limits_{m=0}^{\lfloor T \rfloor-1}\sum\limits_{n=0}^{\lfloor S \rfloor-1} p_{j,l,m,n;u} . \end{array} $$

Next, we look at p_{j, l, m, n; u}. We define

$$ \begin{array}{@{}rcl@{}} \boldsymbol{Z}_{u} &:=& \left( X_1(t^*+\frac{jT+m}{u}), \ X_2(s^*+\frac{jT+m}{u}+\frac{lS+n}{u})\right)^{\top},\\ Y_{1;u}(t) &:= &\left( X_1(t^*+\frac{jT+m}{u}+\frac{t}{u})- X_1(t^*+\frac{jT+m}{u})\right) {\sqrt u} -\mu_1 t, \\ Y_{2;u}(t,s) &:=&\left( X_2(t^*+\frac{jT+m}{u}+\frac{lS+n}{u} + \frac{t+s}{u})- X_2(t^* + \frac{jT+m}{u}+\frac{lS+n}{u})\right) {\sqrt u} -\mu_2 (t+s). \end{array} $$

Consider the conditional process

$$ \begin{array}{@{}rcl@{}} \boldsymbol {W}_{u}(t,s):=(Y_{1;u}(t), Y_{2;u}(t,s))^{\top} \mid \boldsymbol{Z}_{u}= {\sqrt{u}} \boldsymbol{b}_{u}-\frac{\boldsymbol{x}}{ {\sqrt{u}}}. \end{array} $$

We have that $(Y_{1;u}(t), Y_{2;u}(t,s), \boldsymbol {Z}_{u}^{\top })$ is a normally distributed random vector, with mean

$$ \begin{array}{@{}rcl@{}} \boldsymbol {\widehat\mu}(t,s) := (-\mu_1 t, -\mu_2(t+s), 0, 0)^{\top} \end{array} $$

and covariance matrix given by (suppose S > 1)

$$ \begin{array}{@{}rcl@{}} \widehat{\Sigma}_{u}(t,s):= \left( \begin{array}{cccr} t&0& 0 & \rho\frac{t}{\sqrt u}\\ 0 & t+s & 0 & 0\\ 0 & 0& t^*+\frac{jT+m}{u} & \rho\ (t^*+\frac{jT+m}{u})\\ \rho\frac{t}{\sqrt u} & 0& \rho\ (t^*+\frac{jT+m}{u}) & t^*+\frac{jT+m}{u} +\frac{lS+n}{u} \end{array} \right). \end{array} $$

Thus, for the mean

$$ \begin{array}{@{}rcl@{}} \mathbb{E}\left\{ \boldsymbol {W}_{u}(t,s)\right\}& = & (-\mu_1 t, - \mu_2(t+s)) + \left( \begin{array}{cc} 0 & \rho\frac{t}{\sqrt u}\\ 0&0 \end{array} \right) \left( \begin{array}{cc} t^*+\frac{jT+m}{u} & \rho\ (t^*+\frac{jT+m}{u})\\ \rho\ (t^* + \frac{jT+m}{u}) & t^*+\frac{jT+m}{u} + \frac{lS+n}{u} \end{array} \right)^{-1} \left( {\sqrt{u}} \boldsymbol{b}_{u} - \frac{\boldsymbol{x}}{ {\sqrt{u}}}\right)\\ & = & (-\mu_1 t, -\mu_2(t+s))+ \left( \frac{\rho (b_{j,l,m,n;u} - \rho a_{j,m;u} ) t - \rho t \frac{x_2-\rho x_1}{u}}{(t^*+\frac{jT+m}{u}+\frac{lS+n}{u}) -\rho^2(t^*+\frac{jT+m}{u})} , 0\right) \\ &\quad\to & \left( - \nu(\rho) \ t, -\mu_2 (t+s) \right)^{\top}, \ \ \nu(\rho):=\frac{\rho^2-\rho+(\mu_1-\mu_2\rho)t^*}{t^*(1-\rho^2)}, \end{array} $$

as $u\to {\infty }$, where the convergence is uniform for $-N_{u}^{(1)}\le j\le N_{u}^{(1)}, 1\le l\le N_{u}^{(2)}$. Similarly, we can derive that, for any t₁, t₂ ∈ [0, T], s₁, s₂ ∈ [−S, S],

$$ \begin{array}{@{}rcl@{}} \text{Cov}(\boldsymbol {W}_{u}(t_1,s_1) -\boldsymbol {W}_{u}(t_2,s_2)) &\to &\text{Cov}((B_1(t_1)-B_1(t_2), B_2(t_1+s_1)-B_2(t_2+s_2))^{\top}) \end{array} $$

as $u\to {\infty }$, uniformly for $-N_{u}^{(1)}\le j\le N_{u}^{(1)}, 1\le l\le N_{u}^{(2)}$. Consequently, we have, as $u\to {\infty },$

$$ \begin{array}{@{}rcl@{}} &&\mathbb{P} \left\{ \left. \underset{s\in[0,1]}{\underset{ t \in [0, 1] }\exists } \begin{array}{lll} Y_{1;u}(t)>x_1 \\ Y_{2;u}(t,s) > x_2 \end{array} \right | \boldsymbol{Z}_{u}= {\sqrt{u}} \boldsymbol{b}_{u}-\frac{\boldsymbol{x}}{ {\sqrt{u}}} \right\} \\ && \to \ \mathbb{P} \left\{ \underset{s\in[0,1]}{\underset{ t \in [0, 1] }\exists } \begin{array}{lll} B_1(t)- \nu(\rho) t > x_1 \\ B_2(t+s) - \mu_2 (t+s) >x_2 \end{array} \right \} . \end{array} $$

Similar arguments as those in the proof of Lemma 4.2 gives that

$$ \begin{array}{@{}rcl@{}} p_{j,l,m,n;u} \sim \frac{\widehat{\mathcal{H}}(1,1)u^{-1}}{\sqrt{(2\pi)^{2}\lvert {\Sigma}_{*} \rvert}} e^{-\frac{u}{2} g_B(t^{*}+\frac{jT+m}{u}, s^{*}+\frac{jT+m}{u}+\frac{lS+n}{u})}, \end{array} $$

where (recall notation in Eq. 7)

$$ \begin{array}{@{}rcl@{}} \widehat{\mathcal{H}}(1,1)={\int}_{\mathbb{R}^2} e^{\boldsymbol{x}^{\top} {\Sigma}_{*}^{-1} \boldsymbol{b}_{*} } \mathbb{P} \left\{ \underset{s\in[0,1]}{\underset{ t \in [0, 1] }\exists } \begin{array}{lll} B_1(t)- \nu(\rho) t > x_1 \\ B_2(t+s) - \mu (t+s) >x_2 \end{array} \right \} dx_1 dx_2\in(0,{\infty}). \end{array} $$

It follows further from (ii.2) of Lemma A.1 that there exists some ε > 0 such that, for all t < s small,

$$ \begin{array}{@{}rcl@{}} g_B(t^*+ t,t^*+ s) \ge g_L(t^*)+b_2(1-\varepsilon)(s-t )+\frac{c_2 (1-\varepsilon) }{2} t^2, \end{array} $$

thus, for u sufficiently large

$$ \begin{array}{@{}rcl@{}} g_B(t^* + \frac{jT+m}{u}, t^* + \frac{jT+m}{u} + \frac{lS+n}{u}) \ge g_L(t^*) + b_2(1 - \varepsilon)\frac{lS}{u} + \frac{c_2(1-\varepsilon)}{2} \left( \frac{\widehat j T}{u}\right)^2 \end{array} $$

holds for all $-N_{u}^{(1)}\le j\le N_{u}^{(1)}, 1\le l\le N_{u}^{(2)}, 0\le m\le \lfloor T\rfloor -1, 0\le n\le \lfloor S\rfloor -1$, where $\widehat j=j I_{\{j\ge 0\}} +(j+1) I_{\{j<0\}}$. This implies that, for u large

$$ \begin{array}{@{}rcl@{}} e^{-\frac{u}{2} g_A(t^*+\frac{jT+m}{u}, s^*+\frac{jT+m}{u}+\frac{lS+n}{u})}\le e^{-\frac{u}{2} g_L(t^*)} \frac{\sqrt u}{T} \left( e^{- \frac{c_2(1-\varepsilon)}{4} \left( \frac{\widehat jT}{\sqrt u}\right)^2} \frac{T}{\sqrt u}\right) e^{-\frac{b_2(1-\varepsilon)}{2} lS}. \end{array} $$

Based on the above discussions we obtain

$$ \begin{array}{@{}rcl@{}} \underset{u\to{\infty}}{\lim}\frac{{\Pi}_1(u)}{u^{-1/2} \exp(- g_L(t^*){u}/{2}) } \le \frac{\widehat{\mathcal{H}}(1,1)}{\sqrt{(2\pi)^2\left\lvert {\Sigma}_{*} \right\rvert}} \ \frac{\lfloor T \rfloor \lfloor S \rfloor}{T} \underset{ l\ge 1 }{\sum} e^{-\frac{b_2(1-\varepsilon)}{2} lS} {\int}_{\mathbb{R}} e^{-\frac{c_2(1-\varepsilon)}{4} x^2}dx. \end{array} $$

Similar bounds can be found for $\overline {\Pi }_{1}(u)$, and thus the first claim follows.

Next we consider Π₂₁(u). For any j₂ > j₁ + 1 we have

$$ \begin{array}{@{}rcl@{}} q_{j_1,j_2;u}&=&\mathbb{P} \left\{ \begin{array}{cc} \exists_{(t,s) \in (t^*+\frac{j_1 T}{u},s^*+\frac{j_1 T}{u})+u^{-1}\triangle_{T,S}} \ X_1(t)> \sqrt{u}(1+ \mu_1 t ), X_2(s)\!>\! \sqrt{u}(1+ \mu_2 s ) \\ \exists_{(t,s) \in (t^*+\frac{j_2 T}{u},s^*+\frac{j_2 T}{u})+u^{-1}\triangle_{T,S}} \ X_1(t)> \sqrt{u}(1+ \mu_1 t ), X_2(s)\!>\! \sqrt{u}(1+ \mu_2 s ) \end{array} \right \} \\ &\le & u^{-1}{\int}_{\mathbb{R}^{2}}\phi_{{\Sigma}_{j_1,j_2;u} }({\sqrt{u}} \boldsymbol{b}_{j_1,j_2;u}-\boldsymbol{x}/ {\sqrt{u}} ) \overline F(j_1,j_2;u,\boldsymbol{x}) d \boldsymbol{x}=:Q_{j_1,j_2;u}, \end{array} $$

where, with $a_{j;u}=1+\mu _{1}(t^{*}+\frac {j_{1}T}{u}), b_{j;u}=1+\mu _{2}(s^{*}+\frac {j_{1}T}{u})$,

$$ \begin{array}{@{}rcl@{}} {\Sigma}_{j_1,j_2;u} = \left( t^*+\frac{j_1T}{u}+\frac{(j_2-j_1)S}{4u} \right) \left( \begin{array}{cc} 1 & \rho \\ \rho & 1 \end{array} \right), \ \ \boldsymbol{b}_{j_1,j_2;u} =\left( \frac{a_{j_1;u}+a_{j_2;u}}{2}, \frac{b_{j_1;u}+b_{j_2;u}}{2}\right), \end{array} $$

$$ \begin{array}{@{}rcl@{}} \overline F(j_1,j_2;u,\boldsymbol{x}):= \mathbb{P} \left\{\left. \underset{(t^{\prime},s^{\prime})\in\triangle_{T,S}}{\underset{ (t,s)\in\triangle_{T,S} }\exists } \begin{array}{ccc} Y_{1;u}(t,t^{\prime}) > x_1 \\ Y_{2;u}(s,s^{\prime}) >x_2 \end{array} \right| \begin{array}{lll} Y_{3;u} = \frac{a_{j_1;u}+a_{j_2;u}}{2}\sqrt u -\frac{ x_1}{\sqrt u} \\ Y_{4;u} =\frac{a_{j_1;u}+a_{j_2;u}}{2}\sqrt u -\frac{ x_2}{\sqrt u} \end{array} \right \} , \end{array} $$

with

$$ \begin{array}{@{}rcl@{}} Y_{1;u}(t,t^{\prime})&=& \frac{\sqrt u}{2} \left( X_1(t^* + \frac{j_1T}{u} + \frac{t}{u}) - X_1(t^* + \frac{j_1T}{u}) + X_1(t^* + \frac{j_2T}{u}+\frac{t^{\prime}}{u}) - X_1(t^*+\frac{j_2T}{u})\right) - \frac{\mu_1}{2}(t + t^{\prime}), \\ Y_{2;u}(s,s^{\prime})& = & \frac{\sqrt u}{2} \left( X_2(s^*+\frac{j_1T}{u} + \frac{s}{u}) - X_2(s^* + \frac{j_1T}{u}) \!+ \! X_2(s^* + \frac{j_2T}{u} + \frac{s^{\prime}}{u})- X_2(s^* + \frac{j_2T}{u})\right) - \frac{\mu_2}{2}(s + s^{\prime}), \\ Y_{3;u} &=&\frac{1}{2} \left( X_1(t^*+\frac{j_1T}{u})+ X_1(t^*+\frac{j_2T}{u})\right), \ \ \ \ Y_{4;u} = \frac{1}{2} \left( X_2(s^*+\frac{j_1T}{u})+X_2(s^*+\frac{j_2T}{u})\right). \end{array} $$

Next we have that $(Y_{1;u}(t,t^{\prime }), Y_{2;u}(s,s^{\prime }), Y_{3;u},Y_{4;u})$ is a normally distributed random vector, with mean

$$ \begin{array}{@{}rcl@{}} \boldsymbol{\widehat\mu}(t,t^{\prime},s,s^{\prime}) = \left( -\frac{\mu_1}{2}(t+t^{\prime}), -\frac{\mu_2}{2}(s+s^{\prime}), 0, 0\right)^{\top} \end{array} $$

and covariance matrix given by (suppose T > S)

$$ \begin{array}{@{}rcl@{}} \widehat{\Sigma}_{u}(t,s)= \left( \begin{array}{cccr} \frac{t+t^{\prime}}{4}&\frac{\rho(t\wedge s+t^{\prime} \wedge s^{\prime})}{4}& \frac{t}{4 \sqrt u} & \frac{\rho t}{4 \sqrt u}\\ \frac{\rho(t\wedge s+t^{\prime} \wedge s^{\prime})}{4} & \frac{s+s^{\prime}}{4}& \frac{\rho s}{4 \sqrt u} & \frac{s}{4 \sqrt u}\\ \frac{t}{4 \sqrt u} & \frac{\rho s}{4 \sqrt u}& t^*+\frac{j_1T}{u}+\frac{(j_2-j_1)T}{4u} & \rho\ \left( t^*+\frac{j_1T}{u}+\frac{(j_2-j_1)T}{4u}\right)\\ \frac{\rho t}{4 \sqrt u} & \frac{s}{4 \sqrt u}& \rho\ \left( t^*+\frac{j_1T}{u}+\frac{(j_2-j_1)T}{4u}\right) & s^*+\frac{j_1T}{u}+\frac{(j_2-j_1)T}{4u} \end{array} \right). \end{array} $$

Similarly as before, one can get

$$ \begin{array}{@{}rcl@{}} Q_{j_1,j_2;u} \sim \frac{\widetilde{\mathcal{H}}(T,S)u^{-1}}{\sqrt{(2\pi)^2\left\lvert {\Sigma}_{*} \right\rvert}} e^{-\frac{u}{2} \left( 1+\frac{\frac{(j_2-j_1)T}{4u}}{t^*+\frac{j_1T}{u}+\frac{(j_2-j_1)T}{4u}}\right) g_L(t^*+\frac{j_1T}{u}+\frac{(j_2-j_1)T}{2u})}, \end{array} $$

as $u\to {\infty },$ where

$$ \begin{array}{@{}rcl@{}} \widetilde{\mathcal{H}}(T,S): = {\int}_{\mathbb{R}^2}e^{\boldsymbol{x}^{\top} {\Sigma}_{*}^{-1} \boldsymbol{b}_{*} }\mathbb{P} \left\{ \underset{(t^{\prime},s^{\prime})\in\triangle_{T,S}}{\underset{ (t,s)\in\triangle_{T,S} }\exists } \begin{array}{lll} \frac{1}{2}(X_1(t)+\widetilde X_1(t^{\prime}))-\frac{\mu_1t^*-1}{4t^*} t- \frac{\mu_1}{2} t^{\prime} > x_1 \\ \frac{1}{2}(X_2(s)+\widetilde X_2(s^{\prime})) -\frac{\mu_2s^*-1}{4s^*} s- \frac{\mu_2}{2}s^{\prime} >x_2 \end{array} \right \} dx_1 dx_2\in(0,{\infty}), \end{array} $$

with $(\widetilde X_{1}, \widetilde X_{2})$ an independent copy of (X₁, X₂). Particularly, letting j₁ = 0, j₂ = 2 we can show, similarly as in Eq. 39, that

$$ \begin{array}{@{}rcl@{}} \widetilde{\mathcal{H}}(T,S) \le \widetilde{\mathcal{H}}(1,S) (\lfloor T \rfloor )^2 e^{\frac{g_L(t^*)}{8t^*} T}. \end{array} $$

Therefore, as $u\to {\infty },$

$$ \begin{array}{@{}rcl@{}} {\Pi}_{21} (u) &\lesssim & \frac{\widetilde{\mathcal{H}}(T,S)u^{-1/2}}{T \sqrt{(2\pi)^2\left\lvert {\Sigma}_{*} \right\rvert}} e^{-\frac{u}{2}g_L(t^*)} \left( \sum\limits_{j_1=-N_u^{(1)}}^{N_u^{(1)}} e^{-\frac{b_0}{4}\left( \frac{j_1 T}{\sqrt u}\right)^2} \frac{T}{\sqrt u}\right) \left( \sum\limits_{j_2=j_1+2}^{N_u^{(1)}} e^{-\frac{g_L(t^*)}{8t^*} (j_2-j_1)T}\right)\\ &\lesssim& \frac{\widetilde{\mathcal{H}}(1,S) \lfloor T\rfloor^2 u^{-1/2}}{T \sqrt{(2\pi)^2\left\lvert {\Sigma}_{*} \right\rvert}} e^{-\frac{u}{2}g_L(t^*)} \ {\int}_{\mathbb{R}}e^{-\frac{b_0}{4} x^2} dx \ \sum\limits_{k=1}^{{\infty}} e^{-\frac{g_L(t^*)}{8t^*} kT}. \end{array} $$

Finally, we consider Π₂₂(u). Note that

$$ \begin{array}{@{}rcl@{}} {\Pi}_{22}(u) = \sum\limits_{j=-{N}_u^{(1)}}^{{N}_u^{(1)}} p_{j;u}+p_{j+1;u}-\widetilde{p}_{j;u}, \end{array} $$

where

$$ \begin{array}{@{}rcl@{}} \widetilde{p}_{j;u}=\mathbb{P} \left\{ \exists_{(t,s) \in (t^*+\frac{j T}{u},t^*+\frac{j T}{u})+u^{-1}\triangle_{2T,S}} \ X_1(t)> \sqrt{u}(1+ \mu_1 t ), X_2(s)> \sqrt{u}(1+ \mu_2 s ) \right \} . \end{array} $$

Consequently, the claim for Π₂₂(u) follows directly by using Lemma 4.5. $\Box$

1.8 A.8 Proof of Lemma 5.1

Similarly as in Eq. 27 we obtain

$$ \begin{array}{@{}rcl@{}} P_{\theta_0,0}(u)&\le& \mathbb{P} \left\{ \exists_{(t_1,s_1)\in U_{11} } \ Z(t_1,s_1) >\sqrt u g(t_1,s_1),\ \ \exists_{(t_2,s_2)\in U_{12}} \ Z(t_2,s_2)>\sqrt u g(t_2,s_2) \right \} \\ &\le& \mathbb{P} \left\{ \exists_{(t_1,s_1)\in U_{11} } \ \overline{Z}(t_1,s_1)>\sqrt u\sqrt{g(t_0,s_0)},\ \ \exists_{(t_2,s_2)\in U_{12} } \ \overline{Z}(t_2,s_2)>\sqrt u \sqrt{g(s_0,t_0)} \right \} \\ &\le& \mathbb{P} \left\{ \exists_{(t_1,s_1)\in U_{11} , (t_2,s_2)\in U_{12}} \ \overline{Z}(t_1,s_1)+ \overline{Z}(t_2,s_2)> 2\sqrt u \sqrt{g(t_0,s_0)} \right \} , \end{array} $$

where, we used the fact that $g(t_{0},s_{0})=g(s_{0},t_{0}) \le \inf _{(t,s)\in (U_{11}\cup U_{12})}g(t,s)$, and

$$ \begin{array}{@{}rcl@{}} \overline{Z}(t_i,s_i):= \frac{Z(t_i,s_i)}{\sqrt{\text{Var} (Z(t_i,s_i))}}= \frac{Z(t_i,s_i)}{\sqrt{g(t_i,s_i) }}, \ \ \ i=1,2. \end{array} $$

Further note that

$$ \begin{array}{@{}rcl@{}} \mathbb{E}\left\{ Z(t_0,s_0) Z (s_0,t_0)\right\}&=&\mathbb{E}\left\{ (2\mu X_1(t_0) +2(1-2\rho)\mu X_2(s_0)) (2(1-2\rho)\mu X_1(s_0) +2\mu X_2(t_0))\right\}\\ &=& 8(1+2\rho)(1-\rho)\mu. \end{array} $$

We obtain

$$ \begin{array}{@{}rcl@{}} \mathbb{E}\left\{ (\overline{Z}(t_0,s_0)+ \overline{Z}(s_0,t_0))^2\right\}&=&2+ 2\mathbb{E}\left\{ \overline{Z}(t_0,s_0) \overline{Z}(s_0,t_0)\right\}\\ &=&2+ 2\frac{\mathbb{E}\left\{ Z(t_0,s_0) Z (s_0,t_0)\right\}}{g(t_0,s_0)}\\ &=&2+ 2 (1+2\rho)<4. \end{array} $$

Thus, for sufficiently small 𝜃₀ > 0,

$$ \begin{array}{@{}rcl@{}} \sigma ^2:=\underset{(t_2,s_2)\in U_{12}}{\underset{(t_1,s_1)\in U_{11}}{\sup}}\mathbb{E}\left\{ (\overline{Z}(t_1,s_1)+ \overline{Z}(t_2,s_2))^2\right\}<4, \end{array} $$

where we use continuity of the functions involved. Again, using the Borell-TIS inequality we obtain

$$ \begin{array}{@{}rcl@{}} \mathbb{P} \left\{ \exists_{(t_1,s_1)\in U_{11}, (t_2,s_2)\in U_{12} } \ \overline{Z}(t_1,s_1)+ \overline{Z}(t_2,s_2)> 2\sqrt u \sqrt{g(t_0,s_0)} \right \} \le e^{-\frac{(2\sqrt u \sqrt{g(t_0,s_0)}-C_0)^2}{2\sigma ^2}} \end{array} $$

holds for all large u such that

$$ \begin{array}{@{}rcl@{}} 2\sqrt u \sqrt{g(t_0,s_0)}>C_0:=\mathbb{E}\left\{ \underset{(t_2,s_2)\in U_{12}}{\underset{(t_1,s_1)\in U_{11}}{\sup}} (\overline{Z}(t_1,s_1)+ \overline{Z}(t_2,s_2)) \right\}. \end{array} $$

Thus, the claim follows.