Asymptotics for Hankel Determinants Associated to a Hermite Weight with a Varying Discontinuity

Theorem 2.1. Let $\alpha \in (-1, \infty)$ and $t \in (-1, \infty)$ . As $n \to \infty$ , we have <span id="page-5-0"></span> $$\log \frac{H_n(\sqrt{2n}t, 0, \alpha)}{H_n(0, 0, \alpha)} = C_1(t)n^2 + C_2(t, \alpha)n + C_3(t, \alpha) + \mathcal{O}(n^{-1}), \tag{2.1}$$ where the coefficients are given by <span id="page-5-2"></span> $$C_1(t) = -\frac{2t^3}{27} \left( \sqrt{3+t^2} - t \right) - \left( \frac{4}{3}t^2 + \frac{5}{9}t\sqrt{3+t^2} \right) - \log\left( \frac{t + \sqrt{3+t^2}}{\sqrt{3}} \right), \tag{2.2}$$ <span id="page-5-5"></span> $$C_2(t,\alpha) = \frac{\alpha t}{3} \left( t - \sqrt{3 + t^2} \right) - \alpha \log \left( \frac{t + \sqrt{3 + t^2}}{\sqrt{3}} \right), \tag{2.3}$$ <span id="page-5-6"></span> $$C_3(t,\alpha) = \frac{1 - 3\alpha^2}{6} \log\left(\frac{t + \sqrt{3 + t^2}}{\sqrt{3}}\right) - \frac{1}{48} \log\left(\frac{3 + t^2}{3}\right) - \frac{1}{16} \left(1 - 4\alpha^2\right) \log\left(\frac{3 + 5t^2 + 4t\sqrt{3 + t^2}}{3}\right).$$ (2.4) Furthermore, the error term $\mathcal{O}(n^{-1})$ is uniform for t in a compact subset of $(-1, \infty)$ . Note that there is no critical transition in (2.1) as $t \to 1$ . Remark 2.2. The probability (1.7) with $v = \sqrt{2nt}$ can be rewritten as $$\mathbb{P}(x_{\min}^{(\sqrt{2n}t,\alpha)} \ge \sqrt{2n}t) = \frac{H_n(\sqrt{2n}t,0,\alpha)}{H_n(0,0,\alpha)} \frac{H_n(0,0,\alpha)}{H_n(0,1,\alpha)} \frac{H_n(0,1,\alpha)}{H_n(\sqrt{2n}t,1,\alpha)}.$$ Large n asymptotics of these three ratios are given up to the constant term by Theorem 2.1 (for t > -1), (1.8) and (1.6) (for $t \in (-1,1)$ ), respectively. Putting these asymptotics together, we obtain as $n \to \infty$ and for $t \in (-1,1)$ , <span id="page-5-3"></span> $$\log \mathbb{P}\left(x_{\min}^{(\sqrt{2n}t,\alpha)} \ge \sqrt{2n}t\right) = \left(C_1(t) - \frac{\log 3}{2}\right)n^2 + \left(C_2(t,\alpha) - \frac{\alpha \log 3}{2} - \alpha t^2\right)n$$ $$+ \left(\frac{\alpha^2}{4} - \frac{1}{12}\right)\log n + C_3(t,\alpha) + c_0 - \frac{\alpha^2}{8}\log\left(1 - t^2\right) + \mathcal{O}\left(\frac{\log n}{n}\right). \tag{2.5}$$ It can be checked from (2.2) that $C_1(-1) = \frac{\log 3}{2}$ and $$C_1'(t) = -\frac{8}{27} (\sqrt{3+t^2} - t) (3+5t^2+4t\sqrt{3+t^2}) < 0,$$ for $t > -1$ , which shows that the leading term in (2.5) is negative. This implies that the above probability decays super exponentially fast as $n \to \infty$ for $t \in (-1,1)$ . To observe a transition in the large n asymptotics of $H_n(\sqrt{2nt}, s, \alpha)$ when s = 0 and when s is in a compact subset of (0, 1], we couple the parameter s with n in the form $$s = e^{-\lambda n}, \qquad \lambda \ge 0.$$ Large n asymptotics of $H_n(\sqrt{2n}t, e^{-\lambda n}, \alpha)$ will depend on whether $\lambda$ is greater or smaller than a critical value $\lambda_c(t)$ , which is explicit and given by <span id="page-5-4"></span> $$\lambda_c(t) = \frac{2t}{\sqrt{3}} \sqrt{3 + t^2 + 2t\sqrt{3 + t^2}} + 2\log\left(2 + t^2 + t\sqrt{3 + t^2} + \frac{\sqrt{3 + t^2} + t}{\sqrt{3}}\sqrt{3 + t^2 + 2t\sqrt{3 + t^2}}\right).$$ (2.6)

Theorem 2.3. Let $s = e^{-\lambda n}$ with $\lambda \in [0, \infty)$ , we have the following asymptotic results. (1) If $t \in (-1,1)$ and $\lambda \geq \lambda_c(t)$ , then <span id="page-6-0"></span> $$\log \frac{H_n(\sqrt{2nt}, e^{-\lambda n}, \alpha)}{H_n(\sqrt{2nt}, 0, \alpha)} = \mathcal{O}(n^{-1/2} e^{-n(\lambda - \lambda_c(t))}), \quad as \quad n \to \infty,$$ (2.7) and large n asymptotics for $\log H_n(\sqrt{2n}t, 0, \alpha)$ are given by Theorem 2.1. Furthermore, the $\mathcal{O}$ term in (2.7) is uniform for t in a compact subset of (-1, 1) and for $\lambda \geq \lambda_c(t)$ . (2) If $t \in (-1,1)$ and $0 \le \lambda \le \lambda_c(t)$ are fixed, then <span id="page-6-2"></span> $$\lim_{n \to \infty} \frac{1}{n^2} \log \frac{H_n(\sqrt{2n}t, e^{-\lambda n}, \alpha)}{H_n(\sqrt{2n}t, 1, \alpha)} = -\int_0^\lambda \Omega(t, \tilde{\lambda}) d\tilde{\lambda}, \tag{2.8}$$ where $$\Omega(t,\lambda) = \int_a^b \rho(x;t,\lambda) dx, \qquad \rho(x;t,\lambda) = \frac{2}{\pi} \sqrt{c-x} \sqrt{\frac{x-b}{x-t}} \sqrt{x-a},$$ and a < b < t < c, with a, b and c depending on $\lambda$ and t, are uniquely determined by the following equations: $$t = a + b + c,$$ $$2 = a^2 + b^2 + c^2 - t^2,$$ $$\lambda = 4 \int_b^t \frac{\sqrt{c - x}}{\sqrt{t - x}} \sqrt{x - b} \sqrt{x - a} dx.$$ <span id="page-6-5"></span>Remark 2.4. In Theorem 2.3, we restrict ourselves to the case $t \in (-1,1)$ . With increasing effort, this result can be extended for $t \in (-1,\infty)$ . If $t \ge 1$ , a new region appears in the $(t,\lambda)$ plane which deserves a separate analysis (which we expect to be straightforward but long). Therefore, we decided not to proceed in this direction. Remark 2.5. Note that the denominators on the left hand sides of (2.7) and (2.8) are different. We can use Theorem 2.3 to obtain information about the large deviation of the smallest thinned $\mathrm{GUE}(\sqrt{2n}t,\alpha)$ eigenvalue as follows. By (1.9) and (1.7) with $v=\sqrt{2n}t$ and $s=e^{-\lambda n}$ , we have $$\mathbb{P}\big(y_{\min}^{(\sqrt{2n}t,e^{-\lambda n},\alpha)} \ge \sqrt{2n}t\big) = \frac{H_n\big(\sqrt{2n}t,e^{-\lambda n},\alpha\big)}{H_n\big(\sqrt{2n}t,0,\alpha\big)} \mathbb{P}\big(x_{\min}^{(\sqrt{2n}t,\alpha)} \ge \sqrt{2n}t\big).$$ Therefore, for $t \in (-1,1)$ and $\lambda \geq \lambda_c(t)$ , by (2.7), as $n \to \infty$ we have <span id="page-6-3"></span> $$\log \mathbb{P}\left(y_{\min}^{(\sqrt{2n}t,e^{-\lambda n},\alpha)} \ge \sqrt{2n}t\right) = \log \mathbb{P}\left(x_{\min}^{(\sqrt{2n}t,\alpha)} \ge \sqrt{2n}t\right) + \mathcal{O}\left(n^{-1/2}e^{-n(\lambda-\lambda_c(t))}\right),\tag{2.9}$$ and large n asymptotics of $\log \mathbb{P}(x_{\min}^{(\sqrt{2n}t,\alpha)} \geq \sqrt{2n}t)$ are given by (2.5). In the regime $t \in (-1,1)$ and $0 \leq \lambda \leq \lambda_c(t)$ , (2.8) implies <span id="page-6-4"></span> $$\log \mathbb{P}\left(y_{\min}^{(\sqrt{2n}t,e^{-\lambda n},\alpha)} \ge \sqrt{2n}t\right) = \left(-\int_0^\lambda \Omega(t,\tilde{\lambda})d\tilde{\lambda}\right)n^2 + o(n^2). \tag{2.10}$$ Since (2.9) and (2.10) are both valid for $\lambda = \lambda_c(t)$ , by equalling the leading term, we have <span id="page-6-6"></span> $$-\int_0^{\lambda_c(t)} \Omega(t,\lambda) d\lambda = C_1(t) - \frac{\log 3}{2}.$$ (2.11) We will give an independent and more direct proof of this formula at the end of Section 7. Remark 2.6. Note that the limit [\(2.8\)](#page-6-2) is independent of α. The subleading terms in the large n asymptotics of ^Hn( √ 2nt,e−λn,α) Hn( √ 2nt,1,α) are expected to depend on α and to be oscillatory and described in terms of elliptic θ-functions. These functions appear in our analysis (see, e.g., [\(6.12\)](#page-30-0)). This heuristic is also supported by the analogy of our situation with [\[6\]](#page-41-0), where the authors obtained θ-functions in the subleading terms.

Lemma 4.1. We have the following differential identities $$\partial_t \log H_n(\sqrt{2nt}, 0, \alpha) = 4nU_{1,11},\tag{4.5}$$ $$s\partial_s \log H_n(\sqrt{2n}t, s, \alpha) = \int_{-\infty}^t \frac{\widetilde{w}(x)}{2\pi i} [U^{-1}(x)U'(x)]_{21} dx.$$ (4.6) Proof. The differential identity (4.5) is obtained by substituting (4.4) and (4.3) into (3.6). Similarly, using (3.7) and (4.1), the differential identity (3.9) can be rewritten as $$s\partial_s \log H_n(\sqrt{2nt}, s, \alpha) = \int_{-\infty}^{\sqrt{2nt}} \frac{w(x)}{2\pi i} [Y^{-1}(x)Y'(x)]_{21} dx,$$ which gives (4.6) after using (4.4) and a change of variables. <span id="page-11-4"></span>Remark 4.2. Note that $[Y^{-1}(z)Y'(z)]_{21}$ only involves the first column of Y, which is entire (see (4.1) or equivalently (4.2)). Thus $[Y_+^{-1}(x)Y_+'(x)]_{21} = [Y_-^{-1}(x)Y_-'(x)]_{21}$ for $x \in \mathbb{R}$ , and we simply denote it by $[Y^{-1}(x)Y'(x)]_{21}$ without ambiguity. The same remark holds for U.

Lemma 7.1. As $n \to \infty$ , we have <span id="page-33-5"></span> $$U_{+}(x) \begin{pmatrix} 1 \\ 0 \end{pmatrix} = e^{\frac{n\ell}{2}} e^{ng_{+}(x)} e^{-\frac{n\ell}{2}\sigma_{3}} \begin{pmatrix} \mathcal{O}\left(n^{\frac{1}{2} + \max(\alpha, 0)}\right) \\ \mathcal{O}\left(n^{\frac{1}{2} + \max(\alpha, 0)}\right) \end{pmatrix}, \tag{7.7}$$ <span id="page-33-7"></span> $$U'_{+}(x)\begin{pmatrix}1\\0\end{pmatrix} = e^{\frac{n\ell}{2}}e^{ng_{+}(x)}e^{-\frac{n\ell}{2}\sigma_{3}}\begin{pmatrix}\mathcal{O}\left(n^{\frac{5}{2}+\max(\alpha,0)}\right)\\\mathcal{O}\left(n^{\frac{5}{2}+\max(\alpha,0)}\right)\end{pmatrix},\tag{7.8}$$ uniformly for $x \in (-\infty, t) \cap D_t$ . Proof. For $z \in D_t$ , z outside the lenses, we have <span id="page-33-1"></span> $$U(z)\begin{pmatrix} 1\\0 \end{pmatrix} = e^{\frac{n\ell}{2}} e^{ng(z)} e^{-\frac{n\ell}{2}\sigma_3} R(z) P(z) \begin{pmatrix} 1\\0 \end{pmatrix}. \tag{7.9}$$ If furthermore, $\Im z > 0$ , by (5.19) and (B.4) we have <span id="page-33-0"></span> $$P(z)\begin{pmatrix} 1\\0 \end{pmatrix} = e^{\frac{\pi i \alpha}{2}} (z-t)^{-\frac{\alpha}{2}} e^{-n\xi(z)} E(z) \begin{pmatrix} I_{\alpha}(2n\sqrt{-f(z)})\\ -2\pi i n\sqrt{-f(z)} I'_{\alpha}(2n\sqrt{-f(z)}) \end{pmatrix}. \tag{7.10}$$ Let $x \in (-\infty, t) \cap D_t$ . Note that $\tilde{\xi}(x) > 0$ ( $\tilde{\xi}$ is defined in (5.16)) and thus $\sqrt{-f(x)}_+ = \frac{1}{2}\tilde{\xi}(x)$ . Inserting (7.10) into (7.9), we can take the limit $z \to x$ , this gives <span id="page-33-2"></span> $$U_{+}(x) \begin{pmatrix} 1 \\ 0 \end{pmatrix} = (-1)^{n} e^{\frac{n\ell}{2}} e^{ng_{+}(x)} e^{-n\tilde{\xi}(x)} (t-x)^{-\frac{\alpha}{2}} e^{-\frac{n\ell}{2}\sigma_{3}}$$ $$\times R(x) E(x) \begin{pmatrix} I_{\alpha}(n\tilde{\xi}(x)) \\ -\pi i n\tilde{\xi}(x) I'_{\alpha}(n\tilde{\xi}(x)) \end{pmatrix}. \tag{7.11}$$ Since E is analytic in $D_t$ , one has from (5.21) that as $n \to \infty$ <span id="page-33-4"></span> $$E(z) = \mathcal{O}(1)n^{\frac{\sigma_3}{2}}, \qquad E'(z) = \mathcal{O}(1)n^{\frac{\sigma_3}{2}}, \qquad \text{uniformly for } z \in D_t.$$ (7.12) To obtain a uniform bound from (7.11), we distinguish three cases. Let M > 0 be an arbitrary large but fixed constant and let m > 0 be an arbitrary small but fixed constant. Case (a): $n\tilde{\xi}(x) \geq M$ as $n \to \infty$ . In this case we need large $\zeta$ asymptotics for $I_{\alpha}(\zeta)$ and $I'_{\alpha}(\zeta)$ . From (B.2), we have <span id="page-33-3"></span> $$I_{\alpha}(\zeta) = \frac{e^{\zeta}}{\sqrt{2\pi\zeta}} \left( 1 + \mathcal{O}(\zeta^{-1}) \right), \qquad I_{\alpha}'(\zeta) = \frac{e^{\zeta}}{\sqrt{2\pi\zeta}} \left( 1 + \mathcal{O}(\zeta^{-1}) \right), \qquad \text{as} \quad \zeta \to \infty.$$ (7.13) If we insert (7.13) into (7.11), the result follows for Case (a) from (5.26), (7.12) and from the fact that $(t-x)^{-\frac{\alpha}{2}} = \mathcal{O}(n^{\max(\alpha,0)})$ . Case (b): $m \leq n\tilde{\xi}(x) \leq M$ as $n \to \infty$ . In this case we have $I_{\alpha}(n\tilde{\xi}(x)) = \mathcal{O}(1)$ , $n\tilde{\xi}(x)I'_{\alpha}(n\tilde{\xi}(x)) = \mathcal{O}(1)$ , $e^{-n\tilde{\xi}(x)} = \mathcal{O}(1)$ , $(t-x)^{-\frac{\alpha}{2}} = \mathcal{O}(n^{\alpha})$ . Again from (5.26) and (7.12), we obtain <span id="page-33-6"></span> $$U_{+}(x)\begin{pmatrix} 1\\0 \end{pmatrix} = e^{\frac{n\ell}{2}} e^{ng_{+}(x)} e^{-\frac{n\ell}{2}\sigma_{3}} \begin{pmatrix} \mathcal{O}\left(n^{\frac{1}{2}+\alpha}\right)\\ \mathcal{O}\left(n^{-\frac{1}{2}+\alpha}\right) \end{pmatrix}, \tag{7.14}$$ which is even slightly better than (7.7). Case (c): $n\xi(x) \leq m$ as $n \to \infty$ . From [36, formula (10.25.2)], we have $$I_{\alpha}(\zeta) = \left(\frac{\zeta}{2}\right)^{\alpha} \left(\frac{1}{\Gamma(1+\alpha)} + \mathcal{O}(\zeta^2)\right),$$ $$I'_{\alpha}(\zeta) = \left(\frac{\zeta}{2}\right)^{\alpha-1} \left(\frac{\alpha}{\Gamma(1+\alpha)} + \mathcal{O}(\zeta^2)\right), \quad \text{as} \quad \zeta \to 0.$$ From the above expansion, we have for Case (c) that $$\frac{I_{\alpha}(n\tilde{\xi}(x))}{(t-x)^{\frac{\alpha}{2}}} = \mathcal{O}(n^{\alpha}), \qquad \frac{n\tilde{\xi}(x)I_{\alpha}'(n\tilde{\xi}(x))}{(t-x)^{\frac{\alpha}{2}}} = \mathcal{O}(n^{\alpha})$$ and $e^{-n\tilde{\xi}(x)} = \mathcal{O}(1)$ . Thus, from (5.26) and (7.12) we obtain again (7.14), which finishes the proof of (7.7). We now turn to the proof of (7.8). From (7.11), we have $U'_{+}(x) \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \tilde{U}_{1}(x) + \tilde{U}_{2}(x)$ , where $$\widetilde{U}_{1}(x) = n(g'_{+}(x) - \widetilde{\xi}'(x))U_{+}(x)\begin{pmatrix} 1\\0 \end{pmatrix} + (-1)^{n}e^{\frac{n\ell}{2}}e^{ng_{+}(x)}e^{-n\widetilde{\xi}(x)}(t-x)^{-\frac{\alpha}{2}}e^{-\frac{n\ell}{2}\sigma_{3}} \\ \times (R'(x)E(x) + R(x)E'(x))\begin{pmatrix} I_{\alpha}(n\widetilde{\xi}(x))\\ -\pi i n\widetilde{\xi}(x)I'_{\alpha}(n\widetilde{\xi}(x)) \end{pmatrix}, \\ \widetilde{U}_{2}(x) = (-1)^{n}e^{\frac{n\ell}{2}}e^{ng_{+}(x)}e^{-n\widetilde{\xi}(x)}e^{-\frac{n\ell}{2}\sigma_{3}}R(x)E(x)\begin{pmatrix} \left(\frac{I_{\alpha}(n\widetilde{\xi}(x))}{(t-x)^{\frac{\alpha}{2}}}\right)'\\ \left(\frac{-\pi i n\widetilde{\xi}(x)I'_{\alpha}(n\widetilde{\xi}(x))}{(t-x)^{\frac{\alpha}{2}}}\right)' \end{pmatrix}.$$ The analysis of $\widetilde{U}_1(x)$ and $\widetilde{U}_2(x)$ can be done very similarly to the first part of the proof and we do not provide here all the details. From (5.4), one has $g'_+(x) - \widetilde{\xi}'(x) = 2x$ and thus by (5.26), (7.7) and (7.12), $$\widetilde{U}_1(x) = e^{\frac{n\ell}{2}} e^{ng_+(x)} e^{-\frac{n\ell}{2}\sigma_3} \begin{pmatrix} \mathcal{O}(n^{\frac{3}{2} + \max(\alpha, 0)}) \\ \mathcal{O}(n^{\frac{3}{2} + \max(\alpha, 0)}) \end{pmatrix}.$$ Again, by splitting the analysis into the same three cases as in the first part of the proof, we obtain the estimates $$\left(\frac{I_{\alpha}(n\tilde{\xi}(x))}{(t-x)^{\frac{\alpha}{2}}}\right)' = \mathcal{O}\left(n^2 \left(\frac{I_{\alpha}(n\tilde{\xi}(x))}{(t-x)^{\frac{\alpha}{2}}}\right)\right),$$ $$\left(\frac{-i\pi n\tilde{\xi}(x)I'_{\alpha}(n\xi(x))}{(t-x)^{\frac{\alpha}{2}}}\right)' = \mathcal{O}\left(n^2 \left(\frac{-i\pi n\tilde{\xi}(x)I'_{\alpha}(n\xi(x))}{(t-x)^{\frac{\alpha}{2}}}\right)\right),$$ which yields $$\widetilde{U}_2(x) = e^{\frac{n\ell}{2}} e^{ng_+(x)} e^{-\frac{n\ell}{2}\sigma_3} \begin{pmatrix} \mathcal{O}(n^{\frac{5}{2} + \max(\alpha, 0)}) \\ \mathcal{O}(n^{\frac{5}{2} + \max(\alpha, 0)}) \end{pmatrix}$$ and finishes the proof. Note that $g_+(x) + g_-(x) - 2x^2 + \ell$ is continuous on $\mathbb{R}$ and equal to 0 at x = t by (4.7). Thus, from (4.24) and (4.8) and the fact that V(x) has a jump discontinuity at x = t, we have <span id="page-34-0"></span> $$\lim_{\substack{x \to t \\ r < t}} g_{+}(x) + g_{-}(x) - V(x) + \ell = -\lambda < -\lambda_{c} < 0.$$ (7.15) Therefore, by using first Lemma 7.1 and then (7.15), there exists $c \in (0, \lambda_c)$ such that <span id="page-34-1"></span> $$\frac{\tilde{w}(x)}{2\pi i} \left[ U^{-1}(x)U'(x) \right]_{21} = |x - t|^{\alpha} e^{n(g_{+}(x) + g_{-}(x) - V(x) + \ell)} \mathcal{O}\left(n^{3 + 2\max(\alpha, 0)}\right) = |x - t|^{\alpha} \mathcal{O}\left(e^{-(\lambda - c)n}\right),$$ (7.16) as n → ∞ uniformly for x ∈ D^t ∩ (−∞, t). Now, we will split the integral of the dif ferential identity [\(4.6\)](#page-10-3) into two parts: $$s\partial_s \log H_n(\sqrt{2n}t, s) = I_1(s) + I_2(s),$$ $$I_1(s) = \int_{(-\infty, t) \setminus D_t} \frac{\widetilde{w}(x)}{2\pi i} [U^{-1}(x)U'(x)]_{21} dx,$$ $$I_2(s) = \int_{(-\infty, t) \cap D_t} \frac{\widetilde{w}(x)}{2\pi i} [U^{-1}(x)U'(x)]_{21} dx.$$ The first integral can be evaluated using [\(7.6\)](#page-32-4). By [\(5.1\)](#page-15-3), [\(5.2\)](#page-16-5), [\(5.4\)](#page-16-0) and [\(5.7\)](#page-16-1) (see also the comment just after), we have g+(b) + g−(b) + ` − V (b) = −(λ − λc) and $$(g_{+}(x) + g_{-}(x) + \ell - V(x))'\big|_{x=\bar{b}} = 0, \qquad (g_{+}(x) + g_{-}(x) + \ell - V(x)''\big|_{x=\bar{b}} < 0.$$ Therefore, we obtain $$|I_1(s = e^{-\lambda n})| = \mathcal{O}(n^{-1/2}e^{-n(\lambda - \lambda_c)}), \quad \text{as} \quad n \to \infty.$$ On the other hand, from [\(7.16\)](#page-34-1), it immediately follows that $$|I_2(s=e^{-\lambda n})| = \mathcal{O}(e^{-(\lambda-c)n}), \quad \text{as} \quad n \to \infty,$$ where c ∈ (0, λc). Therefore, the dif ferential identity becomes $$\partial_s \log H_n(v, s, \alpha) \big|_{s=e^{-\lambda n}} = \mathcal{O}(n^{-1/2}e^{n\lambda_c}), \quad \text{as} \quad n \to \infty,$$ where in the above expression the O term is uniform for t in a compact subset of (−1, ∞) and for λ ≥ λc(t). Thus, we can integrate it from s = 0 to s = e ⁻λn, and it gives $$\log H_n(\sqrt{2n}t, e^{-\lambda n}, \alpha) = \log H_n(\sqrt{2n}t, 0, \alpha) + \mathcal{O}(n^{-1/2}e^{-n(\lambda - \lambda_c(t))}), \quad \text{as} \quad n \to \infty,$$ which is the claim [\(2.7\)](#page-6-0).