Moderate deviations for parameters estimation in a geometrically ergodic Heston process

Abstract

We establish a moderate deviation principle for the maximum likelihood estimator of the four parameters of a geometrically ergodic Heston process. We also obtain moderate deviations for the maximum likelihood estimator of the couple of dimensional and drift parameters of a generalized squared radial Ornstein–Uhlenbeck process. We restrict ourselves to the most tractable case where the dimensional parameter satisfies $a>2$ and the drift coefficient is such that $b<0$. In contrast to the previous literature, parameters are estimated simultaneously.

Appendices

Appendix 1: Changes of parameters

To compute the limits of the cumulant generating functions (11) and (15), which is the aim of “Appendix 2”, we recall some changes of probability formulas. We denote by $\mathbb {P}_{c,d}^{a,b}$ the distribution of the solution of (1) associated with parameters a, b, c and d, and by $\mathbb {E}_{c,d}^{a,b}$ the corresponding expectation. At first, we change both parameters a and b of the first equation of (1), which corresponds to the CIR process. Applying Girsanov’s formula given e.g. in Theorem 1.12 of Kutoyants (2004), we have

$$\begin{aligned} \begin{array}{lcl} \displaystyle \frac{\mathrm {d}\mathbb {P}_{c,d}^{a,b}}{\mathrm {d}\mathbb {P}_{c,d}^{\alpha ,\beta }}&{}=&{} \displaystyle \exp \left( \int _0^T{\frac{a+bX_t-(\alpha +\beta X_t)}{4X_t} \, \mathrm {d}X_t} - \frac{1}{2} \int _0^T{\frac{(a+bX_t)^2 -(\alpha +\beta X_t)^2}{4X_t} \, \mathrm {d}t} \right) \\ &{}=&{} \displaystyle \exp \left( \frac{a-\alpha }{4} \int _0^T{X_t^{-1} \, \mathrm {d}X_t}+\frac{b-\beta }{4} \int _0^T{ \mathrm {d}X_t} \right. \\ &{} {} &{} \displaystyle \left. - \frac{1}{8} \int _0^T{\left( (a^2-\alpha ^2) X_t^{-1}+(b^2-\beta ^2) X_t + 2(ab-\alpha \beta )\right) \, \mathrm {d}t} \right) \\ &{}=&{} \displaystyle X_T^{\frac{a-\alpha }{4}} \exp \left( -\frac{a-\alpha }{4} \left( \log X_0 +bT\right) + \frac{b-\beta }{4}(X_T-X_0-\alpha T) \right) \\ &{} {} &{} \, \displaystyle \times \exp \left( - \frac{T}{8} \left( (b^2-\beta ^2) S_T + \left( 4 (\alpha -a) - \alpha ^2 + a^2\right) \Sigma _T\right) \right) . \end{array} \end{aligned}$$

(21)

For the last equality, we use It’s formula applied to $\log X_T$, which gives

$$\begin{aligned} \int _0^T{X_t^{-1} \, \mathrm {d}X_t}= \log X_T- \log X_0 +2 \int _0^T{X_t^{-1} \, \mathrm {d}t}. \end{aligned}$$

We also need to change parameters c and d of the second equation in (1). We rewrite this equation with new parameters $\gamma$ and $\delta$:

$$\begin{aligned} \mathrm {d}Y_t=(\gamma +\delta X_t) \, \mathrm {d}t + 2 \sqrt{X_t}\,\left[ \rho \,\mathrm {d}B_t + \sqrt{1-\rho ^2} \mathrm {d}\widetilde{W}_t\right] \end{aligned}$$

where

$$\begin{aligned} \mathrm {d}\widetilde{W}_t=\mathrm {d}W_t + \frac{\mathrm {d}t}{2\sqrt{1-\rho ^2}} \left( \left( c-\gamma \right) X_t^{-1/2} + \left( d-\delta \right) X_t^{1/2} \right) . \end{aligned}$$

Thus, we obtain that

$$\begin{aligned} \begin{array}{lcl} \displaystyle \frac{\mathrm {d}\mathbb {P}_{c,d}^{a,b}}{\mathrm {d}\mathbb {P}_{\gamma ,\delta }^{a,b}}&{}=&{} \displaystyle \exp \left( \frac{c-\gamma }{2\sqrt{1-\rho ^2}} \int _0^T X_t^{-1/2} \, \mathrm {d}W_t+\frac{d-\delta }{2\sqrt{1-\rho ^2}} \int _0^T X_t^{1/2} \, \mathrm {d}W_t\right) \\ &{} {} &{} \, \displaystyle \times \exp \left( \frac{T}{8\left( 1-\rho ^2\right) }\left[ (d-\delta )^2 S_T + \left( c-\gamma \right) ^2 \Sigma _T + 2 \left( c-\gamma \right) \left( d-\delta \right) \right] \right) . \end{array} \end{aligned}$$

(22)

Appendix 2: Proof of the pointwise limit of the cumulant generating function

We want to compute the pointwise limit of the cumulant generating function $\mathscr {L}_T$ of $\left( \lambda _T T\right) ^{-1/2} \mathscr {M}_T$. With $\mathscr {M}_T$ being replaced by its expression in $X_t$, $B_t$, $W_t$ and $\rho$, we have that, for all $u=\left( u_1,u_2,u_3,u_4\right) \in \mathbb {R}^4$,

$$\begin{aligned} E_T\left( u\right) :=\mathbb {E}_{c,d}^{a,b} \left[ e^{\sqrt{\lambda _T/T} \, \left\langle u, \mathscr {M}_T\right\rangle }\right] = \mathbb {E}_{c,d}^{a,b} \left[ \mathscr {E}_{1,T} \, \mathscr {E}_{2,T} \, \mathscr {E}_{3,T} \, \mathscr {E}_{4,T}\right] \end{aligned}$$

(23)

where

$$\begin{aligned} \mathscr {E}_{1,T}= & {} \exp \left( v_{3,T} \, \int _0^T X_t^{-1/2} \, \mathrm {d}B_t \right) , \quad \mathscr {E}_{2,T} = \exp \left( v_{4,T} \int _0^T X_t^{1/2} \, \mathrm {d}B_t \right) {,}\\ \mathscr {E}_{3,T}= & {} \exp \left( \sqrt{\frac{\lambda _T}{T}} \sqrt{1-\rho ^2} u_3 \int _0^T X_t^{-1/2} \, \mathrm {d}W_t \right) {,}\\ \mathscr {E}_{4,T}= & {} \exp \left( \sqrt{\frac{\lambda _T}{T}} \sqrt{1-\rho ^2} u_4 \int _0^T X_t^{1/2} \, \mathrm {d}W_t \right) {.} \end{aligned}$$

with $v_{3,T}=\sqrt{\lambda _T/T}(u_1 + \rho u_3)$ and $v_{4,T}=\sqrt{\lambda _T/T}(u_2+ \rho u_4)$. We use (22) to change parameters c and d in order to kill the terms involving $W_t$. We obtain that

$$\begin{aligned} \mathbb {E}_{c,d}^{a,b} \left[ \mathscr {E}_{1,T} \, \mathscr {E}_{2,T} \, \mathscr {E}_{3,T} \, \mathscr {E}_{4,T}\right] = \mathbb {E}_{\gamma _T,\delta _T}^{a,b}\left[ G_T \mathscr {E}_{1,T} \, \mathscr {E}_{2,T} \right] \end{aligned}$$

(24)

where $\gamma _T= c+ 2 u_3 \sqrt{\lambda _T/T} \left( 1-\rho ^2\right)$, $\delta _T= d + 2 u_4 \sqrt{\lambda _T/T} \left( 1-\rho ^2\right)$ and

$$\begin{aligned} G_T=\exp \left( \frac{T}{8\, (1-\rho ^2)}\left( (c-\gamma _T)^2 \Sigma _T+(d-\delta _T)^2 S_T + 2 (c-\gamma _T)(d-\delta _T)\right) \right) . \end{aligned}$$

(25)

Additionally, using the first equation of (1) and Itô’s formula applied to $\log X_T$, we obtain that

$$\begin{aligned} 2 \displaystyle \int _0^T{X_t^{1/2} \, \mathrm {d}B_t} = X_T - x_0 -a \, T -b \, T S_T \end{aligned}$$

(26)

and

$$\begin{aligned} 2\displaystyle \int _0^T{X_t^{-1/2} \, \mathrm {d}B_t} = \log X_T - \log x_0 - b \, T +(2-a) \, T \, \Sigma _T \end{aligned}$$

(27)

where $S_T$ and $\Sigma _T$ are given after (4). Thus, we rewrite $\mathscr {E}_{1,T}$ and $\mathscr {E}_{2,T}$ as functions of $X_T$, $S_T$ and $\Sigma _T$:

$$\begin{aligned} \mathscr {E}_{1,T}= \exp \left( \frac{v_{3,T}}{2} \left( \log X_T- \log x_0-bT+ (2-a) T \Sigma _T\right) \right) \end{aligned}$$

(28)

and

$$\begin{aligned} \mathscr {E}_{2,T}= \exp \left( \frac{1}{2} v_{4,T} \left( X_T-x_0-aT-bTS_T\right) \right) . \end{aligned}$$

(29)

Thus, replacing (25), (29) and (28) into (24), and taking out of the expectation all the deterministic terms, we obtain that

$$\begin{aligned} E_T\left( u\right) =\exp \left( \mathscr {A}_T\right) \mathbb {E}_{\gamma _T,\delta _T}^{a,b}\left[ \exp \left( v_{1,T} T \Sigma _T + v_{2,T} T S_T +\frac{1}{2} \left( v_{3,T} X_T + v_{4,T} \log X_T \right) \right) \right] \end{aligned}$$

(30)

where

$$\begin{aligned} v_{1,T}=\frac{\left( c-\gamma _T\right) ^2}{8\left( 1-\rho ^2\right) }+ \frac{2-a}{2}v_{3,T}, \quad v_{2,T}=\frac{\left( d-\delta _T\right) ^2}{8\left( 1-\rho ^2\right) }- \frac{b}{2}v_{4,T} \end{aligned}$$

and

$$\begin{aligned} \mathscr {A}_T =T \left( 1-\rho ^2\right) ^{-1} \left( c-\gamma _T\right) \left( d-\delta _T\right) /4 - v_{4,T} \left( aT+x_0\right) /2- v_{3,T} \left( \log x_0 + bT\right) /2. \end{aligned}$$

We now make a new change of parameters for a and b, in order to kill the terms involving $S_T$ and $\Sigma _T$. The new time-depending parameters are given, for T large enough, by

$$\begin{aligned} \alpha _T =2+\left( a-2\right) \sqrt{1 - 8\, v_{1,T}/\left( a-2\right) ^2} \quad \text { and } \quad \beta _T=b\sqrt{1-8 \, v_{2,T}/b^2}. \end{aligned}$$

Thus, (30) becomes

$$\begin{aligned} E_T\left( u\right) =\exp \left( -w_{1,T} x_0 - w_{2,T} \log x_0 + T \mathscr {C}_T\right) \mathbb {E}_{\gamma _T, \delta _T}^{\alpha _T,\beta _T}\left( \exp \left( w_{1,T} X_T\right) X_T ^{w_{2,T}}\right) \end{aligned}$$

(31)

where $w_{1,T} =\left( b-\beta _T + 2 v_{4,T}\right) /4$, $w_{2,T} =\left( a-\alpha _T + 2 v_{3,T}\right) /4$ and

$$\begin{aligned} 4 \mathscr {C}_T=- 2 a v_{4,T} - 2 b v_{3,T}+ \beta _T \alpha _T - ab - (c-\gamma _T) (d-\delta _T)/(1-\rho ^2). \end{aligned}$$

Therefore, taking the logarithm of (31), we obtain that

$$\begin{aligned} \log E_T\left( u\right) = -w_{1,T} x_0 - w_{2,T} \log x_0 + T \mathscr {C}_T+ \log \mathbb {E}_{\gamma _T, \delta _T}^{\alpha _T,\beta _T}\left( \exp \left( w_{1,T} X_T\right) X_T ^{w_{2,T}}\right) . \end{aligned}$$

(32)

We now have to divide (32) by $\lambda _T$ and investigate the limit for T going to infinity. We consider each term of the right-hand side of (32) separately. First of all, as $w_{1,T}$ and $w_{2,T}$ tend to zero for T going to infinity, we immediately deduce that

$$\begin{aligned} \lim _{T \rightarrow +\infty } \lambda _T^{-1} \left( -w_{1,T} x_0 - w_{2,T} \log x_0 \right) =0. \end{aligned}$$

(33)

We now consider the term $T\mathscr {C}_T/ \lambda _T$. On the one hand, replacing $\gamma _T$ and $\delta _T$ by their respective definitions, we easily obtain that

$$\begin{aligned} \frac{1}{4 \left( 1-\rho ^2\right) } \left( c-\gamma _T\right) \left( d-\delta _T\right) \frac{T}{\lambda _T}= \left( 1-\rho ^2\right) u_3 u_4. \end{aligned}$$

(34)

On the other hand, as $\lambda _T/T$ goes to zero for T tending to infinity, we expand $\beta _T \alpha _T$ up to order two in $\sqrt{\lambda _T/T}$ and obtain that:

$$\begin{aligned} \begin{array}{ll} \frac{T}{4\lambda _T}\left( -2a v_{4,T} - 2 b v_{3,T}+ \beta _T \alpha _T - ab \right) =&{} \displaystyle -\frac{b}{2\left( a-2\right) } \left( \left( u_1+\rho u_3\right) ^2 + \left( 1- \rho ^2\right) u_3^2\right) \\ &{}\displaystyle -\frac{a}{2b}\left( \left( u_2+\rho u_4\right) ^2+ \left( 1-\rho ^2\right) u_4^2\right) \\ &{}\displaystyle + \left( u_1+\rho u_3\right) \left( u_2 +\rho u_4\right) + o\left( 1\right) . \end{array} \end{aligned}$$

(35)

Thus, for T going to infinity, the limit value of $T \mathscr {C}_T / \lambda _T$ is the sum of the limits of (34) and (35). Before concluding, we will now show that

$$\begin{aligned} \lim _{T \rightarrow +\infty } \lambda _T^{-1} \log \mathbb {E}_{\gamma _T, \delta _T}^{\alpha _T,\beta _T}\left( \exp \left( w_{1,T} X_T\right) X_T ^{w_{2,T}}\right) = 0 \end{aligned}$$

(36)

The density function of the solution $X_T$ associated with parameters $\alpha _T$ and $\beta _T$ and initial point $x_0$ is given, for any positive real y, by

$$\begin{aligned} f(y)= K_T \exp \left( -y/2\right) y^{(\alpha _T-2)/4} I_{(\alpha _{T}-2)/2}\left( \sqrt{y \xi _T}\right) \end{aligned}$$

where $I_{\nu }$ is the modified Bessel function of the first kind and $\xi _T$ and $K_T$ are two constants respectively given by

$$\begin{aligned} \xi _T= -\frac{x_0 \beta _T}{e^{-\beta _T T}-1} \quad \text {and}\quad K_T= \frac{e^{-\xi _T/2}}{2 \xi _T^{(\alpha _T-2)/4}}, \end{aligned}$$

see for instance (Lamberton and Lapeyre 1997). Thus, using formulas 6.643(2) and 9.220(2) of Gradshteyn and Ryzhik (1980), we compute the expectation in the last term of (32) as follows

$$\begin{aligned} \begin{array}{lcl} \mathbb {E}_{\gamma _T, \delta _T}^{\alpha _T,\beta _T}\left( \exp \left( w_{1,T} X_T\right) X_T ^{w_{2,T}}\right) &{} = &{} \int _0^{+\infty } \exp \left( w_{1,T} y\right) y ^{w_{2,T}} f(y) \, \mathrm {d}y\\ &{}=&{}\displaystyle \frac{\Gamma (w_{2,T} + \alpha _T/2)}{\Gamma (\alpha _T/2)} \, 2^{(\alpha _T+2)/4} \left( 1/2- w_{1,T}\right) ^{-(2 w_{2,T}+\alpha _T-1)/2}\\ &{} &{} \times \, \displaystyle e^{-\xi _T/2}\, {}_1F_1\left( w_{2,T} + \alpha _T/2, \alpha _T/2, \xi _T/(2-4 w_{1,T}) \right) \end{array} \end{aligned}$$

where ${}_1F_1$ is the degenerate hypergeometric function (see Gradshteyn and Ryzhik 1980). As we want to compute the limit of the logarithm of this expectation, the obtained product becomes a sum and we investigate the limit of each term separately. For T going to infinity, $\alpha _T$ converges to a and $\beta _T$ to b whereas $\xi _T$, $w_{1,T}$ and $w_{2,T}$ vanish. Thus, we obtain the four following limits

$$\begin{aligned} \lim _{T \rightarrow +\infty } \lambda _T^{-1} \log \frac{\Gamma (w_{2,T} + \alpha _T/2)}{\Gamma (\alpha _T/2)} = \lim _{T \rightarrow +\infty } \lambda _T^{-1} \log \frac{\Gamma (a/2)}{\Gamma (a/2)}=0, \end{aligned}$$

$$\begin{aligned} \lim _{T \rightarrow +\infty } \lambda _T^{-1} \frac{\alpha _T+2}{4} \log 2 =0, \end{aligned}$$

$$\begin{aligned} \lim _{T \rightarrow +\infty } \lambda _T^{-1} ( w_{2,T}+\alpha _T/2-1/2) \log \left( 1/2- w_{1,T}\right) = \lim _{T \rightarrow +\infty } \lambda _T^{-1}(a/2-1/2) \log (1/2) =0, \end{aligned}$$

$$\begin{aligned} \lim _{T \rightarrow +\infty } \lambda _T^{-1} \xi _T/2 = 0. \end{aligned}$$

Furthermore,

$$\begin{aligned} \lim _{T \rightarrow +\infty } {}_1F_1\left( w_{2,T} + \alpha _T/2, \alpha _T/2, \xi _T/(2-4 w_{1,T}) \right) = {}_1F_1\left( a/2, a/2, 0 \right) = 1 \end{aligned}$$

which, combined with the previous limits, leads to (36). We conclude from the conjunction of (33), (34), (35) and (36), that

$$\begin{aligned} \begin{array}{lcl} \displaystyle \lim _{T \rightarrow \infty } \lambda _T^{-1}\log E_T\left( u\right) &{}=&{} \displaystyle -\frac{b}{2\left( a-2\right) }\left( u_1^2 + 2 \rho u_1 u_3+ u_3^2\right) -\frac{a}{2b}\left( u_2^2+ 2 \rho u_2 u_4 + u_4^2\right) \\ &{}{}&{} \displaystyle +\, u_1 u_2 + u_3 u_4 + \rho \left( u_1 u_4 + u_2 u_3\right) \\ &{}=&{} \displaystyle \frac{1}{2} u^{\intercal } \Gamma u, \end{array} \end{aligned}$$

(37)

with the matrix $\Gamma$ given by (16). This concludes the proof of (15).

Corollary 1

Let $\Lambda _T$ be the normalized cumulant generating function of $((\lambda _T T)^{-1/2}M_T)$ given, for all $v \in \mathbb {R}^2$, by

$$\begin{aligned} \Lambda _T\left( v\right) = \frac{1}{\lambda _T} \log \mathbb {E}\left[ e^{\sqrt{\lambda _T/T} \, \left\langle v, M_T\right\rangle }\right] . \end{aligned}$$

Its pointwise limit $\Lambda$ satisfies

$$\begin{aligned} \Lambda (v)=\lim _{T \rightarrow +\infty } \Lambda _T \left( v\right) = \frac{1}{2} \, v^{\intercal } \, \Sigma \, v, \end{aligned}$$

(38)

where the matrix $\Sigma$ was previously given in (5).

Proof

One can observe that

$$\begin{aligned} \Lambda _T\left( v\right) = \frac{1}{\lambda _T} E_T(v^{\intercal },0,0), \end{aligned}$$

where $E_T$ is defined and computed in the previous proof. Taking $u_3=u_4=0$ in (37), we obtain the announced result. $\square$

Corollary 2

Let $L_T$ be the normalized cumulant generating function of $((\lambda _T T)^{-1/2}N_T)$ given, for all $v \in \mathbb {R}^2$, by

$$\begin{aligned} L_T\left( v\right) = \frac{1}{\lambda _T} \log \mathbb {E}\left[ e^{\sqrt{\lambda _T/T} \, \left\langle v, N_T\right\rangle }\right] . \end{aligned}$$

Its pointwise limit L satisfies

$$\begin{aligned} L(v)=\lim _{T \rightarrow +\infty } L_T \left( v\right) = \frac{1}{2} \, v^{\intercal } \, \Sigma \, v, \end{aligned}$$

(39)

where the matrix $\Sigma$ was previously given in (5).

Proof

As for the previous corollary, one can observe that

$$\begin{aligned} L_T\left( v\right) = \frac{1}{\lambda _T} E_T(0,0,v^{\intercal }), \end{aligned}$$

Taking $u_1=u_2=0$ in (37), we obtain the announced result. $\square$

Appendix 3: Proof of the exponential convergence for $S_T$ and $\Sigma _T$

We start by showing that $p^{\eta }$ given in (13) is strictly negative. We have established, in Lemma 3.1 of du Roy de Chaumaray (2016), that the couple $(S_T,\Sigma _T)$ satisfies an LDP with speed T and good rate function I given for any $(x,y) \in \mathbb {R}^2$ by

$$\begin{aligned} I(x,y) =\left\{ \begin{array}{ll} \displaystyle \frac{y}{2(xy-1)}+\frac{b^2}{8}x+\frac{(a-2)^2}{8}y + \frac{ab}{4} &{}\quad \text {if }\; x>0, y>0\quad \text { and }\quad xy -1 > 0,\\ \displaystyle +\infty &{}\quad \text {otherwise.} \end{array} \right. \end{aligned}$$

(40)

We denote by $\mathscr {D}_I$ the domain of $\mathbb {R}^2$ where I is finite. Using the contraction principle recalled in Lemma 1, we show that the sequence $(\Sigma _T/V_T, S_T/V_T)$ satisfies an LDP with good rate function J given, for all $(z,t) \in \mathbb {R}^2$, by

$$\begin{aligned} J(z,t)= \inf \left\{ I(x,y) | (x,y) \in \mathscr {D}_I, g(x,y)=(z,t) \right\} \end{aligned}$$

where $g(x,y)=(y(xy-1)^{-1},x(xy-1)^{-1})$, and the infimum over the empty set is equal to infinity. One can observe that, if $z \le 0$ or $t \le 0$, $\left( x,y\right)$ is not inside $\mathscr {D}_I$, which means that the finitude domain $\mathscr {D}_J$ of J is included inside $\mathbb {R}^{+}_{*} \times \mathbb {R}^{+}_{*}$. For any $z>0$ and $t>0$, the condition $(z,t)=g(x,y)$ leads to

$$\begin{aligned} y = z (xy-1) \quad \text {and}\quad x= tz^{-1}y. \end{aligned}$$

(41)

Combining both, we obtain that y satisfies $ty^{2}-y-z=0$. Only one solution is strictly positive, it is given by

$$\begin{aligned} y^{*}= \frac{1+\sqrt{1+4tz}}{2t}. \end{aligned}$$

(42)

We deduce from the second equality in (41) that the only possible value for x is given by

$$\begin{aligned} x^{*}= \frac{1+\sqrt{1+4tz}}{2z}. \end{aligned}$$

(43)

Replacing (42) and (43) into (40), we obtain that for any $(z,t) \in \mathbb {R}^{+}_{*} \times \mathbb {R}^{+}_{*}$,

$$\begin{aligned} J(z,t)= I(x^{*},y^{*})= \frac{z}{2}+ \left( \frac{b^2}{16 z} +\frac{(a-2)^2}{16 t}\right) \left( 1+\sqrt{1+4tz}\right) + \frac{ab}{4}. \end{aligned}$$

(44)

The function I is positive and only vanishes at point $(-ab^{-1}, -b(a-2)^{-1})$, see du Roy de Chaumaray (2016). Thus, (44) implies that J is positive and vanishes for $(x^{*},y^{*})= (-ab^{-1}, -b(a-2)^{-1})$. As, $(x^{*},y^{*})$ satisfy $(z,t)=g(x^{*},y^{*})$, we obtain that J only vanishes at point

$$\begin{aligned} (z_0,t_0)= \left( \frac{x^{*}}{x^{*}y^{*}-1}, \frac{y^{*}}{x^{*}y^{*}-1}\right) =\left( -\frac{a(a-2)}{2b},-\frac{b}{2}\right) . \end{aligned}$$

Besides, applying the contraction principle again, we show that $p^{\eta }$, given in (13), satisfies the following upper bound

$$\begin{aligned} p^{\eta } \le - \inf \left\{ J(z,t) \mid (z,t)\notin \mathscr {B}\left( (z_0,t_0),\eta ^2\right) \right\} . \end{aligned}$$

(45)

We want to prove that $p^{\eta }$ is strictly negative. One can observe that the function J is coercive. Indeed, let K be the compact subset of $(\mathbb {R}_{+}^{*})^2$ given by $K=[\varepsilon ,A] \times [\xi , B]$ with

$$\begin{aligned} \varepsilon = \frac{b^2}{4(4-ab)},\quad A= \frac{4-ab}{2}, \quad \xi = \frac{(a-2)^2}{4(4-ab)}\quad \text {and} \quad B= \frac{4 (4-ab)^5}{b^6}, \end{aligned}$$

(46)

we easily show that

$$\begin{aligned} \forall (z,t) \notin K, \quad J(z,t) \ge 1. \end{aligned}$$

Consequently, the infimum of J over $(\mathbb {R}_{+}^{*})^2 \setminus \mathscr {B}\left( (z_0,t_0),\eta ^2\right)$ reduces to the infimum over the compact subset $K \setminus \mathscr {B}\left( (z_0,t_0),\eta ^2\right)$. As J is a continuous function, this infimum is reached for some $(z^{*},t^{*})$. As J is positive and only vanishes at point $(z_0,t_0)$ which does not belong to $K \setminus \mathscr {B}\left( (z_0,t_0),\eta ^2\right)$, we can conclude that

$$\begin{aligned} \inf \left\{ J(z,t) \mid (z,t)\notin \mathscr {B}\left( (z_0,t_0),\eta ^2\right) \right\} = J(z^{*},t^{*}) >0, \end{aligned}$$

which is exactly the result that we wanted to prove.

We now consider $q_T^{\eta }$. We have established, in Theorem 3.2 of du Roy de Chaumaray (2016), an LDP for the sequence $(V_T)$ with speed T and good rate function K given for all $x \in \mathbb {R}$ by

$$\begin{aligned} K(x) = \left\{ \begin{array}{ll} \displaystyle -\frac{b}{4}\sqrt{(x+1) \left( (a-2)^2+\frac{4}{x}\right) }+\frac{ab}{4} &{}\quad \text {if }\; x>0\\ \displaystyle +\infty &{}\quad \text {if }\; x \le 0. \end{array} \right. \end{aligned}$$

Thus, by the contraction principle, the sequence $(V_T^{-1})$ satisfies an LDP with speed T and good rate function $\widetilde{K}$ which is infinite over $\mathbb {R}^{-}$ and satisfies for any $x>0$, $\widetilde{K}(x)=K(x^{-1})$. Thus, we easily deduce that

$$\begin{aligned} q^{\eta } \le - \inf \left\{ \widetilde{K}(x); \left| \frac{a-2}{2}-x\right| \ge \frac{\eta }{2} \right\} = - \min \left\{ \widetilde{K}\left( \frac{a-2+\eta }{2}\right) ; \widetilde{K}\left( \frac{a-2-\eta }{2}\right) \right\} \end{aligned}$$

which leads to $q^{\eta }<0$, as $\widetilde{K}$ is strictly positive over $\mathbb {R}\setminus \left\{ \frac{a-2}{2}\right\}$.