Confidence intervals centred on bootstrap smoothed estimators: an impossibility result

Abstract

Frequentist confidence intervals that include some element of data-based model selection or model averaging is an active area of research. Assessments of the performance, in terms of coverage and expected length, of such intervals yield few positive results. Efron, JASA 2014, proposed a confidence interval centred on a bootstrap smoothed estimator, with width proportional to an estimator of Efron’s delta method approximation to the standard deviation of this estimator. Recently, Kabaila and Wijethunga assessed the performance of this confidence interval using a testbed consisting of two nested linear regression models, with error variance assumed known. This interval was shown to have far better coverage properties than the corresponding post-model-selection confidence interval. However, its expected length properties were not as good as had been hoped for. For this testbed, we ask the following question. Does there exist a formula for the data-based width of a confidence interval centred on the bootstrap smoothed estimator so that it has good performance in terms of both coverage and expected length? Using a decision-theoretic performance bound we answer this question in the negative.

Appendix

We will express all quantities of interest in terms of the random vector $(\widehat{\theta }, \widehat{\gamma })$, which has the following bivariate normal distribution:

$$\begin{aligned} \left[ {\begin{array}{c} \widehat{\theta } \\ \widehat{\gamma }\end{array} } \right] \sim N \left( \left[ {\begin{array}{c} \theta \\ \gamma \end{array} } \right] , \left[ {\begin{array}{cc} \sigma ^2 \, v_\theta &{} \rho \, \sigma \, {v_\theta }^{1/2} \\ \rho \, \sigma \, {v_\theta }^{1/2} &{} 1 \end{array} } \right] \right) . \end{aligned}$$

(6)

1.1 A.1 Proof of Theorem 1

The following proof is based, in part, on the derivations described in Section 4.3 of Giri (2008). The coverage probability of the confidence interval $\text {CI}(s)$ is

$$\begin{aligned} \nonumber&P(\theta \in \text {CI}(s)) \\&= P\left( \widehat{\theta } - \sigma \, v_\theta ^{1/2}\, b(\widehat{\gamma }) - \sigma \, v_\theta ^{1/2}\, s(\widehat{\gamma }) \le \theta \le \widehat{\theta } - \sigma \, v_\theta ^{1/2}\, b(\widehat{\gamma }) + \sigma \, v_\theta ^{1/2}\, s(\widehat{\gamma }) \right) \\ \nonumber&= P\left( \sigma \, v_\theta ^{1/2} \big (b(\widehat{\gamma }) - s(\widehat{\gamma })\big ) \le \widehat{\theta } - \theta \le \sigma \, v_\theta ^{1/2} \big (b(\widehat{\gamma }) + s(\widehat{\gamma })\big ) \right) \\ \nonumber&= P\left( b(\widehat{\gamma }) - s(\widehat{\gamma }) \le \frac{\widehat{\theta } - \theta }{\sigma \, v_\theta ^{1/2}} \le b(\widehat{\gamma }) + s(\widehat{\gamma }) \right) \\&= P \big ( b(\widehat{\gamma }) - s(\widehat{\gamma }) \le G \le b(\widehat{\gamma }) + s(\widehat{\gamma }) \big ), \end{aligned}$$

where $G = (\widehat{\theta } - \theta )/\big (\sigma \,{v_\theta ^{1/2}} \big )$. It follows from (6) that

$$\begin{aligned} \left[ {\begin{array}{c} G \\ \widehat{\gamma }\end{array} } \right] \sim N \left( \left[ {\begin{array}{c} 0 \\ \gamma \end{array} } \right] , \left[ {\begin{array}{cc} 1 &{} \rho \\ \rho &{} 1 \end{array} } \right] \right) . \end{aligned}$$

(7)

For given $\rho$ and function s, the coverage probability of $\text {CI}(s)$ is a function of $\gamma$. We denote this coverage probability by $c(\gamma ; s, \rho )$.

Since b and s are odd and even functions, respectively,

$$\begin{aligned} c(\gamma ; s, \rho )&= P \big ( -b(-\widehat{\gamma }) - s(-\widehat{\gamma }) \le G \le -b(-\widehat{\gamma }) + s(-\widehat{\gamma }) \big ) \\&= P \big ( b(-\widehat{\gamma }) - s(-\widehat{\gamma }) \le -G \le b(-\widehat{\gamma }) + s(-\widehat{\gamma }) \big ) \\&= P \big ( b(\widehat{\gamma }^{\prime }) - s(\widehat{\gamma }^{\prime }) \le G^{\prime } \le b(\widehat{\gamma }^{\prime }) + s(\widehat{\gamma }^{\prime }) \big ), \end{aligned}$$

where $G^{\prime } = -G$ and $\widehat{\gamma }^{\prime } = - \widehat{\gamma }$. It follows from (7) that

$$\begin{aligned} \left[ {\begin{array}{c} G^{\prime } \\ \widehat{\gamma }^{\prime } \end{array} } \right] \sim N \left( \left[ {\begin{array}{c} 0 \\ -\gamma \end{array} } \right] , \left[ {\begin{array}{cc} 1 &{} \rho \\ \rho &{} 1 \end{array} } \right] \right) . \end{aligned}$$

Hence $c(\gamma ; s, \rho ) = c(-\gamma ; s, \rho )$.

Let $\bar{k}(x) = k(x)$ for $|x| < c \,$; otherwise $\bar{k}(x) = 0$. Since $b(x) = \rho \, \bar{k}(x)$,

$$\begin{aligned} c(\gamma ; s, \rho )&= P \big ( \rho \, \bar{k}(\widehat{\gamma }) - s(\widehat{\gamma }) \le G \le \rho \, \bar{k}(\widehat{\gamma }) + s(\widehat{\gamma }) \big )\\&= P \big ( - \rho \, \bar{k}(\widehat{\gamma }) - s(\widehat{\gamma }) \le - G \le - \rho \, \bar{k}(\widehat{\gamma }) + s(\widehat{\gamma }) \big )\\&= P \big ( (- \rho ) \, \bar{k}(\widehat{\gamma }) - s(\widehat{\gamma }) \le G^{\prime } \le (- \rho ) \, \bar{k}(\widehat{\gamma }) + s(\widehat{\gamma }) \big ), \end{aligned}$$

where $G^{\prime } = -G$. It follows from (7) that

$$\begin{aligned} \left[ {\begin{array}{c} G^{\prime } \\ \widehat{\gamma } \end{array} } \right] \sim N \left( \left[ {\begin{array}{c} 0 \\ \gamma \end{array} } \right] , \left[ {\begin{array}{cc} 1 &{} - \rho \\ - \rho &{} 1 \end{array} } \right] \right) . \end{aligned}$$

Hence $c(\gamma ; s, \rho ) = c(\gamma ; s, -\rho )$.

It follows from (7) that the probability distribution of G, conditional on $\widehat{\gamma }=h$, is $N\big (\rho (h-\gamma ), 1-\rho ^2 \big )$. Note that

$$\begin{aligned}&P \big ( b(\widehat{\gamma }) - s(\widehat{\gamma }) \le G \le b(\widehat{\gamma }) + s(\widehat{\gamma }) \big )\\&= \int _{-\infty }^{\infty } P \big ( b(h) - s(h) \le G \le b(h) + s(h) \big | \widehat{\gamma }=h \big ) \, \phi (h-\gamma )\, dh\\&= \int _{-\infty }^{\infty } P \big ( b(h) - s(h) \le \widetilde{G} \le b(h) + s(h) \big ) \, \phi (h-\gamma )\, dh, \end{aligned}$$

where $\widetilde{G} \sim N\big (\rho (h-\gamma ), 1-\rho ^2 \big )$. Thus

$$\begin{aligned} c(\gamma ; s, \rho )&= \int _{-\infty }^{\infty } \ell (h, \gamma ; s(h)) \, \phi (h-\gamma )\, dh\nonumber \\&= \int _{-\infty }^{-c} \ell (h, \gamma ; s(h)) \, \phi (h-\gamma )\, dh + \int _{-c}^{c} \ell (h, \gamma ; s(h)) \, \phi (h-\gamma )\, dh\nonumber \\&\quad + \int _{c}^{\infty } \ell (h, \gamma ; s(h)) \, \phi (h-\gamma )\, dh \nonumber \\&= \int _{-\infty }^{-c} \ell ^{\dag }(h, \gamma ) \, \phi (h-\gamma )\, dh + \int _{-c}^{c} \ell (h, \gamma ; s(h)) \, \phi (h-\gamma )\, dh\nonumber \\&\quad + \int _{c}^{\infty } \ell ^{\dag }(h, \gamma ) \, \phi (h-\gamma )\, dh. \end{aligned}$$

(8)

The usual $1-\alpha$ confidence interval based on the full model $\mathcal{M}_2$ has coverage probability $1-\alpha$. Thus

$$\begin{aligned} 1 - \alpha&= P \big ( -z_{1-\alpha /2} \le G \le z_{1-\alpha /2} \big )\\&= \int _{-\infty }^{\infty } P\big ( -z_{1-\alpha /2} \le G \le z_{1-\alpha /2} \, \big | \, \widehat{\gamma }=h \big ) \, \phi (h-\gamma ) \, dh\\&= \int _{-\infty }^{\infty } \ell ^\dag (h, \gamma )\, \phi (h-\gamma ) \, dh. \end{aligned}$$

Therefore $1-\alpha$ is equal to

$$\begin{aligned} \int _{-\infty }^{-c} \ell ^{\dag }(h, \gamma ) \, \phi (h-\gamma )\, dh + \int _{-c}^{c} \ell ^{\dag }(h, \gamma ) \, \phi (h-\gamma )\, dh + \int _{c}^{\infty } \ell ^{\dag }(h, \gamma ) \, \phi (h-\gamma )\, dh. \end{aligned}$$

It follows from this equality and (8) that

$$\begin{aligned} c(\gamma ; s, \rho )&= 1 - \alpha + \int _{-c}^{c} \big ( \ell (h, \gamma ; s(h)) - \ell ^{\dag }(h, \gamma ) \big ) \, \phi (h-\gamma )\, dh\\&= 1 - \alpha - \Big ( \int _{0}^{c} \left( \ell (h, \gamma ; s(h)) - \ell ^{\dag }(h, \gamma ) \right) \, \phi (h-\gamma )\, dh\\&\quad + \int _{-c}^{0} \left( \ell (h, \gamma ; s(h)) - \ell ^{\dag }(h, \gamma ) \right) \, \phi (h-\gamma )\, dh \Big ). \end{aligned}$$

Change the variable of integration to $y=-h$ in the second integral. The result $c(\gamma ; s, \rho ) = 1 - \alpha - R_1(s, \gamma )$ now follows from the fact that both s and $\phi$ are even functions. $\square$

1.2 A.2 Proof of Theorem 2

The following proof is based, in part, on the derivations described in Section 4.3 of Giri (2008). Note that

$$\begin{aligned} e(\gamma ; s)&= \frac{1}{z_{1-\alpha /2}} \int _{-\infty }^{\infty } s(h) \, \phi (h-\gamma ) \, dh\nonumber \\&= \int _{-\infty }^{-c} \phi (h-\gamma )\, dh + \frac{1}{z_{1 - \alpha /2}} \int _{-c}^{c} s(h) \, \phi (h-\gamma )\, dh + \int _{c}^{\infty } \phi (h-\gamma ) \, dh, \end{aligned}$$

(9)

since $s(x)=z_{1-\alpha /2}$ for all $|x| \ge c$. Obviously,

$$\begin{aligned} 1 = \int _{-\infty }^{-c} \phi (h-\gamma )\, dh + \int _{-c}^{c} \phi (h-\gamma )\, dh + \int _{c}^{\infty } \phi (h-\gamma )\, dh. \end{aligned}$$

It follows from this equality and (9) that

$$\begin{aligned} e(\gamma ; s)&= 1 + \int _{-c}^{c} \left( \frac{s(h)}{z_{1-\alpha /2}} - 1 \right) \phi (h-\gamma )\, dh\\&= 1 + \int _{-c}^{0} \left( \frac{s(h)}{z_{1-\alpha /2}} - 1 \right) \phi (h-\gamma )\, dh + \int _{0}^{c} \left( \frac{s(h)}{z_{1-\alpha /2}} - 1 \right) \phi (h-\gamma )\, dh. \end{aligned}$$

Change the variable of integration to $y=-h$ in the first integral on the right-hand side. The fact that both s and $\phi$ are even functions implies that (1) is true. $\square$

1.3 A.3 Computation of the function $\varvec{s_{\gamma , \nu }}$ for given $\varvec{(\gamma ,\nu )}$

Throughout this section we suppose that $(\varvec{\gamma },\varvec{\nu })$ is given. We describe the computation of $s_{\gamma , \nu }$, a value of $s \in \mathcal{D}$ that minimizes $\widetilde{g}(s, \varvec{\gamma }, \varvec{\nu })$. Straightforward manipulations show that

$$\begin{aligned} \widetilde{g}\big (s, \varvec{\gamma }, \varvec{\nu }\big ) = \int _{0}^{c} q \big (s(h); h, \varvec{\gamma }, \varvec{\nu }\big ) \, dh, \end{aligned}$$

(10)

where $q(x; h, \varvec{\gamma }, \varvec{\nu })$ is defined to be

$$\begin{aligned}&\left( \frac{x}{z_{1-\alpha /2}} - 1 \right) \, \Big ( 2\phi (h) + \sum _{j=1}^{m_2} \nu _2(j)\, \big ( \phi (h-\gamma _2(j)) + \phi (h+\gamma _2(j)) \big ) \Big )\\&\quad + \sum _{j=1}^{m_1} \nu _1(j) \Big ( \left( \ell ^{\dag }(h, \gamma _1(j)) - \ell (h, \gamma _1(j); x )\right) \, \phi (h-\gamma _1(j)) \\&\quad + \left( \ell ^{\dag }(-h, \gamma _1(j)) - \ell (-h, \gamma _1(j); x ) \right) \, \phi (h+\gamma _1(j))\, \Big ). \end{aligned}$$

Recall that the functions $\ell$ and $\ell ^{\dag }$ are defined in the statement of Theorem 1. It follows from (10) that a function $s_{\varvec{\gamma }, \varvec{\nu }}$, defined as a minimizer of $\widetilde{g}(s, \varvec{\gamma }, \varvec{\nu })$ over $s \in \mathcal{D}$, may be found as follows. We set $s_{\varvec{\gamma }, \varvec{\nu }}(h)$, for any $h \in [0, c]$, to be a minimizer over $x\in [0, \infty )$ of $q(x; h, \varvec{\gamma }, \varvec{\nu })$.

Now $q(x; h, \varvec{\gamma }, \varvec{\nu })$ is a continuous function of $x \in [0, \infty )$ for all $h \in [0,c]$ and every given $(\varvec{\gamma }, \varvec{\nu })$. An examination of some examples of this function of $x \in [0, \infty )$ show that this function may have several local minima, including the possibility of a local minimum at $x = 0$. Consequently, the value of $x \in [0, \infty )$ that minimizes $q(x; h, \varvec{\gamma }, \varvec{\nu })$ may change discontinuously, as h increases. In other words, the function $s_{\varvec{\gamma }, \varvec{\nu }}(h)$ may have discontinuities. Figure 5 of the Supplementary Information provides some illustrations of functions $q(x; h, \varvec{\gamma }, \varvec{\nu })$ of $x\in [0, \infty )$ that have two local minima. Figure 4 of the Supplementary Information provides an illustration of a function $s_{\varvec{\gamma }, \varvec{\nu }}(h)$ with discontinuities.

To evaluate the lower bound (3), we need to evaluate

$$\begin{aligned} \widetilde{g}\big (s_{\varvec{\gamma }, \varvec{\nu }}, \varvec{\gamma }, \varvec{\nu }\big ) = \int _{0}^{c} q \big (s_{\varvec{\gamma }, \varvec{\nu }}(h); h, \varvec{\gamma }, \varvec{\nu }\big ) \, dh. \end{aligned}$$

(11)

Although the function $s_{\varvec{\gamma }, \varvec{\nu }}(h)$ may have discontinuities, the integrand of the integral on the right-hand side of (11) is a continuous function of $h \in [0, c]$. An illustration of this, when the function $s_{\varvec{\gamma }, \varvec{\nu }}(h)$ has discontinuities, is provided by Figure 3 of the Supplementary Information.

To carry out the computation of the function $s_{\varvec{\gamma }, \varvec{\nu }}$ accurately and effectively, we use the properties of $dq(x; h, \varvec{\gamma }, \varvec{\nu })/dx$, considered as a function of x, described in Appendix A.4. Suppose that $h \in [0, c]$ is given. Theorem 3 of Appendix A.4 leads to the procedure described at the end of this appendix for finding an interval $\big [0, \widetilde{x}\big ]$ that must contain a value of $x \ge 0$ that minimizes $q(x; h, \varvec{\gamma }, \varvec{\nu })$.

We use the following two step procedure to find the value of $x \in [0, \widetilde{x}]$ that minimizes $q(x; h, \varvec{\gamma }, \varvec{\nu })$. We find all possible local minima in Step 1 and compare them to find the global minimum in Step 2.

Step 1: By considering $dq(x; h, \varvec{\gamma }, \varvec{\nu })/dx$, find all the local minimizers of $q(x; h, \varvec{\gamma }, \varvec{\nu })$ in the interval $[0, \widetilde{x}]$. Define w to be the smallest integer that is greater than or equal to $10 \, \widetilde{x}$. We evaluate $dq(x; h, \varvec{\gamma }, \varvec{\nu })/dx$ on the evenly-spaced grid $x_1=0, x_2=0.1, x_3=0.2, \dots , x_w$ of values of x. To find the values of $x \in [0, x_w]$ that are local minimizers of $q(x; h, \gamma , \nu )$, we need to consider the following two cases.

$Case 1: x = 0$

$x=0$ is a local minimizer of $q(x; h, \varvec{\gamma }, \varvec{\nu })$ if either $dq(0; h, \varvec{\gamma }, \varvec{\nu })/dx > 0$ or $dq(0; h, \varvec{\gamma }, \varvec{\nu })/dx = 0$ and $dq(x_2; h, \varvec{\gamma }, \varvec{\nu })/dx > 0$; otherwise $x=0$ is not a local minimizer.

$Case 2: 0< x < x_w$

If $dq(x_i; h, \varvec{\gamma }, \varvec{\nu })/dx < 0$ and $dq(x_{i+1}; h, \varvec{\gamma }, \varvec{\nu })/dx > 0$, then $dq(x; h, \varvec{\gamma }, \varvec{\nu })/dx$ has a zero in the interval $[x_i, x_{i+1}]$ that is a local minimizer of $q(x; h, \varvec{\gamma }, \varvec{\nu })$. We find this zero using the R function uniroot, to which we provide the interval $[x_i, x_{i+1}]$. Also, if $dq(x_i; h, \varvec{\gamma }, \varvec{\nu })/dx = 0$ and $dq(x_{i-1}; h, \varvec{\gamma }, \varvec{\nu })/dx < 0$ and $dq(x_{i+1}; h, \varvec{\gamma }, \varvec{\nu })/dx > 0$ then $x_i$ is a zero of $dq(x; h, \varvec{\gamma }, \varvec{\nu })/dx$ that is a local minimizer of $q(x; h, \varvec{\gamma }, \varvec{\nu })$.

Step 2 Evaluate $q(x; h, \varvec{\gamma }, \varvec{\nu })$ at the local minimizers of $q(x; h, \varvec{\gamma }, \varvec{\nu })$ found in Step 1. The global minimum of $q(x; h, \varvec{\gamma }, \varvec{\nu })$ is simply the minimum of all of the local minima.

1.4 A.4    Properties of $\varvec{dq(x; h, \varvec{\gamma }, \varvec{\nu })/dx}$ considered as a function of $\varvec{x}$

It is straightforward to show that

$$\begin{aligned} \ell (h, \gamma ; x) = \Phi \left( \frac{b(h) + x - \rho (h-\gamma )}{(1-\rho ^2)^{1/2}} \right) - \Phi \left( \frac{b(h) - x - \rho (h-\gamma )}{(1-\rho ^2)^{1/2}} \right) \end{aligned}$$

and

$$\begin{aligned} \ell ^{\dag }(h, \gamma ) = \Phi \left( \frac{z_{1 - \alpha /2} - \rho (h-\gamma )}{(1-\rho ^2)^{1/2}} \right) - \Phi \left( \frac{- z_{1 - \alpha /2} - \rho (h-\gamma )}{(1-\rho ^2)^{1/2}} \right) . \end{aligned}$$

It follows that

$$\begin{aligned} \frac{d\, q(x; h, \varvec{\gamma }, \varvec{\nu })}{dx} = t_1(h, \varvec{\gamma }, \varvec{\nu }) - t_2(x; h, \varvec{\gamma }, \varvec{\nu }), \end{aligned}$$

where $t_1(h, \varvec{\gamma }, \varvec{\nu })$ is defined to be

$$\begin{aligned} \frac{1}{z_{1 - \alpha /2}} \Big ( 2\phi (h) + \sum _{j=1}^{m_2} \nu _2(j)\, \big ( \phi (h-\gamma _2(j)) + \phi (h+\gamma _2(j)) \big ) \Big ) \end{aligned}$$

and $t_2(x; h, \varvec{\gamma }, \varvec{\nu })$ is defined to be

$$\begin{aligned} \sum _{j=1}^{m_1} \nu _1(j) \left( \phi (h-\gamma _1(j)) \frac{d\, \ell (h, \gamma _1(j); x)}{dx} + \phi (h+\gamma _1(j)) \frac{d\, \ell (-h, \gamma _1(j); x)}{dx} \right) , \end{aligned}$$

with

$$\begin{aligned} \frac{d\, \ell (h, \gamma ; x)}{dx} = \frac{1}{(1-\rho ^2)^{1/2}}\Bigg ( \phi \left( \frac{b(h) + x - \rho (h-\gamma )}{(1-\rho ^2)^{1/2}} \right) + \phi \left( \frac{b(h) - x - \rho (h-\gamma )}{(1-\rho ^2)^{1/2}} \right) \Bigg ). \end{aligned}$$

Suppose that $h \in [0,c]$ and $(\varvec{\gamma }, \varvec{\nu })$ are given. Then $t_1(h, \varvec{\gamma }, \varvec{\nu })$ is a fixed positive number. Observe that $t_2(x; h, \varvec{\gamma }, \varvec{\nu })$ is a function of $x \in [0, \infty )$ that can only take positive values and $d\ell (h, \gamma ; x)/dx$ approaches 0 as $x \rightarrow \infty$. We will use the following theorem to find $\widetilde{x} < \infty$, such that $dq(x; h, \varvec{\gamma }, \varvec{\nu })/dx > 0$ for all $x \ge \widetilde{x}$. This implies that a value of x that minimizes $q(x; h, \varvec{\gamma }, \varvec{\nu })$ cannot belong to the interval $[\widetilde{x}, \infty )$.

Theorem 3

Let $\mu (h, \rho , \gamma ) = b(h) - \rho (h-\gamma )$. Then $t_2(x; h, \varvec{\gamma }, \varvec{\nu })$ is a decreasing function of $x \in \big [x^*, \infty \big )$, where $x^*$ is equal to

$$\begin{aligned} \max \Big ( \big |\mu (h, \rho , \gamma _1(1)) \big |, \, \big |\mu (-h, \rho , \gamma _1(1))\big |, \dots , \big |\mu (h, \rho , \gamma _1(m_1)) \big |, \, \big |\mu (-h, \rho , \gamma _1(m_1)) \big |\Big ). \end{aligned}$$

Proof

We first prove that, for every $h \in \mathbb {R}$, $d \ell (h, \gamma ; x)/dx$ is a decreasing function of $x \in \big [|\mu (h, \rho , \gamma )|, \infty )$. Observe that, for all $x \ge 0$,

$$\begin{aligned}&\phi \left( \frac{b(h) + x - \rho (h-\gamma )}{(1-\rho ^2)^{1/2}} \right) + \phi \left( \frac{b(h) - x - \rho (h-\gamma )}{(1-\rho ^2)^{1/2}} \right) \\&= \phi \left( \frac{b(h) - \rho (h-\gamma ) + x }{(1-\rho ^2)^{1/2}} \right) + \phi \left( \frac{- \big (b(h) - \rho (h-\gamma ) \big ) + x}{(1-\rho ^2)^{1/2}} \right) ,\\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \text {since } \phi \text { is an even function,}\\&= \phi \left( \frac{\mu (h, \rho , \gamma ) + x }{(1-\rho ^2)^{1/2}} \right) + \phi \left( \frac{- \mu (h, \rho , \gamma ) + x}{(1-\rho ^2)^{1/2}} \right) \\&= \phi \left( \frac{|\mu (h, \rho , \gamma )| + x }{(1-\rho ^2)^{1/2}} \right) + \phi \left( \frac{- |\mu (h, \rho , \gamma )| + x}{(1-\rho ^2)^{1/2}} \right) . \end{aligned}$$

This is a decreasing function of $x \in \big [|\mu (h, \rho , \gamma )|, \infty )$. Consequently, $d \ell (h, \gamma ; x)/dx$ and $d \ell (-h, \gamma ; x)/dx$ are decreasing functions of $x \in \big [|\mu (h, \rho , \gamma )|, \infty )$ and $x \in \big [|\mu (-h, \rho , \gamma )|, \infty )$, respectively.

Thus $d \ell (h, \gamma _1(j); x)/dx$ is a decreasing function of $x \in \big [|\mu (h, \rho , \gamma _1(j))|, \infty \big )$, for $j = 1, \dots , m_1$. Therefore

$$\begin{aligned} \sum _{j=1}^{m_1} \nu _1(j)\, \phi (h-\gamma _1(j)) \frac{d\, \ell (h, \gamma _1(j); x)}{dx} \end{aligned}$$

is a decreasing function of $x \in \Big [\underset{j = 1, \dots , m_1}{\textrm{max}} |\mu (h, \rho , \gamma _1(j)) |, \infty \Big )$. Similarly, $d \ell (-h, \gamma _1(j); s)/ds$ is a decreasing function of $s \in \big [|\mu (-h, \rho , \gamma _1(j))|, \infty \big )$, for $j = 1, \dots , m_1$. Therefore

$$\begin{aligned} \sum _{j=1}^{m_1} \nu _1(j)\, \phi (h+\gamma _1(j)) \frac{d\, \ell (-h, \gamma _1(j); s)}{ds} \end{aligned}$$

is a decreasing function of $s \in \Big [\underset{j = 1, \dots , m_1}{\textrm{max}} |\mu (-h, \rho , \gamma _1(j)) |, \infty \Big )$. Therefore $t_2(x; h, \varvec{\gamma }, \varvec{\nu })$ is a decreasing function of $x \in \big [x^*, \infty \big )$. $\square$

We use Theorem 3 to find $\widetilde{x} < \infty$, such that $dq(x; h, \varvec{\gamma }, \varvec{\nu })/dx > 0$ for all $x \ge \widetilde{x}$ as follows. First evaluate $x^*$ and then $dq(x^*; h, \varvec{\gamma }, \varvec{\nu })/dx$. If $dq(x^*; h, \varvec{\gamma }, \varvec{\nu })/dx > 0$ then set $\widetilde{x} = x^*$ and stop; otherwise use the $\textsf{R}$ function $\textsf{uniroot}$ to find the solution for $x \in [x^*, \infty )$ of $dq(x; h, \varvec{\gamma }, \varvec{\nu })/dx = 0$ and then set $\widetilde{x}$ equal to this solution.