Maximum likelihood estimation of autoregressive models with a near unit root and Cauchy errors

Abstract

This paper studies maximum likelihood estimation of autoregressive models of order 1 with a near unit root and Cauchy errors. Autoregressive models with an intercept and with an intercept and a linear time trend are also considered. The maximum likelihood estimator (MLE) for the autoregressive coefficient is $n^{3/2}$-consistent with n denoting the sample size and has a mixture-normal distribution in the limit. The MLE for the scale parameter of Cauchy distribution is $n^{1/2}$-consistent, and its limiting distribution is normal. The MLEs of the intercept and the linear time trend are $n^{1/2}$- and $n^{3/2}$-consistent, respectively. It is also shown that the t statistic for the null hypothesis of a unit root based on the MLE has a standard normal distribution in the limit. In addition, finite-sample properties of the MLE are compared with those of the least square estimator (LSE). It is found that the MLE is more efficient than the LSE when the errors have a Cauchy distribution or a distribution which is a mixture of Cauchy and normal distributions. It is also shown that empirical power of the MLE-based t test for a unit root is much higher than that of the Dickey–Fuller t test.

Appendices

Appendix A: Proofs

The following lemma will be used to prove Lemmas 2, 3, 5 which are used for the proofs of all the theorems of this paper.

Lemma 1

Define $Z_{n} (r) =n^{ -1} \sum _{t =1}^{[n r]}\varepsilon _{t} ,\;B_{n} (r) =n^{ -1/2} \sum _{t =1}^{[n r]}\frac{ \partial \ln f (\varepsilon _{t} ,\sigma ^{o})}{ \partial \varepsilon _{t}}\;$and $Q_{n} (r) =n^{ -1/2} \sum _{t =1}^{[n r]}\frac{ \partial \ln f (\varepsilon _{t} ,\sigma )}{ \partial \sigma }_{ \mid \sigma =\sigma ^{o}}$ for $0 \le r \le 1.$ Then, $(Z_{n} (r_{1})\text {,}$$B_{n} (r_{2})\text {,}$$Q_{n} (r_{3}))\overset{d}{ \rightarrow }(Z (r_{1})\text {,}$$B (r_{2})\text {,}$$Q (r_{3}))$ in $D [0 ,1]^{3}\text {,}$ where$\;Z (r_{1})\text {,}$$B (r_{2})\,$and $Q (r_{3})$ are defined in Theorem 1.

Proof

This lemma is similar to Lemma 1 of Zhang and Chan (2012) except that the limit results for $B_{n} (r_{2})$ and $Q_{n} (r_{3})$ under the assumption of Cauchy errors are not dealt with there. Thus, we only need to show marginal weak convergence results for $B_{n} (r_{2})$ and $Q_{n} (r_{3})$ under the assumption of Cauchy errors. Because $E \genfrac(){}{}{ \partial \ln f (\varepsilon _{t} ,\sigma ^{o})}{ \partial \varepsilon _{t}} =0$ by standard theory and $E \genfrac(){}{}{ \partial \ln f (\varepsilon _{t} ,\sigma ^{o})}{ \partial \varepsilon _{t}}^{2} =\frac{1}{2 \sigma ^{o 2}} <\infty$ by Lemma 7, we can apply the classical functional central limit theorem for the sequence of i.i.d. random variables obtaining

$$\begin{aligned} B_{n} (r)\overset{d}{ \rightarrow }B (r)\text { in }D [0 ,1]\text {.} \end{aligned}$$

Likewise, we have

$$\begin{aligned} Q_{n} (r)\overset{d}{ \rightarrow }Q (r)\text { in }D [0 ,1]\text {.} \end{aligned}$$

Since $E \genfrac(){}{}{ \partial \ln f (\varepsilon _{i} ,\sigma ^{o})}{ \partial \varepsilon _{i}} \left( \frac{ \partial \ln f (\varepsilon _{j} ,\sigma )}{ \partial \sigma }_{ \mid \sigma =\sigma ^{o}}\right) =0$ for every i and $j ,\;B (r)$ and$\;Q (s)$ are independent for $0 \le r \le 1$ and $0 \le s \le 1.$$\square$

The following lemma will be used to prove Theorem 1.

Lemma 2

• (i) $\;n^{ -3/2} \sum _{t =1}^{n}\frac{ \partial \ln f (\varepsilon _{t} ,\sigma ^{o})}{ \partial \varepsilon _{t}} Y_{t -1}\overset{d}{ \rightarrow } \smallint _{0}^{1}S (r) \mathrm{d} B (r) ;$

• (ii) $n^{ -2} \sum _{t =1}^{n}Y_{t -1}\overset{d}{ \rightarrow } \smallint _{0}^{1}S (r) \mathrm{d} r ;$

• (iii) $n^{ -3} \sum _{t =1}^{n}Y_{t -1}^{2}\overset{d}{ \rightarrow } \smallint _{0}^{1}S^{2} (r) \mathrm{d} r ;$

• (iv) $n^{ -3} \sum _{t =1}^{n}Y_{t -1}^{2} \left[ \frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma ^{o})}{ \partial \varepsilon _{t}^{2}} -E \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma ^{o})}{ \partial \varepsilon _{t}^{2}}\right] \overset{p}{ \longrightarrow }0 ;$

• (v) $n^{ -2} \sum _{t =1}^{n}Y_{t -1} \left[ \frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t} \partial \sigma }_{ \mid \sigma =\sigma ^{o}} -E \left( \frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t} \partial \sigma }_{ \mid \sigma =\sigma ^{o}}\right) \right] \overset{p}{ \longrightarrow }0 ;$

• (vi) $n^{ -1} \sum _{t =1}^{n}\left[ \frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \sigma ^{2}}_{ \mid \sigma =\sigma ^{o}} -E \left( \frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \sigma ^{2}}_{ \mid \sigma =\sigma ^{o}}\right) \right] \overset{p}{ \longrightarrow }0.$

Proof

Because $S_{n} (r)\overset{d}{ \rightarrow }S (r)$ in $D [0 ,1]\text {,}$ where $S_{n} (r) =n^{ -1} Y_{[n r]}$ and S(r) is defined in Theorem 1, as shown in Chan et al. (2006), parts (i), (ii) and (iii) follow as in Lemma 4 of Zhang and Chan (2012). Because the variances of and $\left\{ \frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t} \partial \sigma }_{ \mid \sigma =\sigma ^{o}}\right\}$ are finite due to Lemma 10, we obtain (iv) and (v) by using the same method as for equation (16) of Zhang and Chan (2012). Part (vi) is trivial. $\square$

Proof of Theorem 1

• (i) We need to check the four conditions of Newey and McFadden (1994) Theorem 2.5. Conditions (i), (ii) and (iii) are trivially satisfied. To check condition (iv), write

  $$\begin{aligned} f_{t} (\rho _{n} ,\sigma )= & {} \frac{\sigma }{\pi \{\sigma ^{2} +(Y_{t} -\rho _{n} Y_{t -1})^{2}\}} \\= & {} \frac{\sigma }{\pi \{\sigma ^{2} +\varepsilon _{t}^{2} +2 (\rho _{n}^{o} -\rho _{n}) \varepsilon _{t} Y_{t -1} +(\rho _{n}^{o} -\rho _{n})^{2} Y_{t -1}^{2}\}}\text {.} \end{aligned}$$

  Suppose that $Y_{t -1}$ is given. Then, as in Newey and McFadden (1994; p.2125),

  $$\begin{aligned} \left| \ln f_{t} (\rho _{n} ,\sigma )\right|\le & {} \ln \sigma -\ln \pi -\ln \{\sigma ^{2} +\varepsilon _{t}^{2} +2 (\rho _{n}^{o} -\rho _{n}) \varepsilon _{t} Y_{t -1} +(\rho _{n}^{o} -\rho _{n})^{2} Y_{t -1}^{2}\} \\\le & {} C_{1} +\ln (C_{2} +C_{3} \varepsilon _{t}^{2})\end{aligned}$$

  for some positive constants $C_{1} ,\;C_{2}$ and $C_{3} .\;$Since $E \left[ \ln (C_{2} +C_{3} \varepsilon _{t}^{2})\right] <\infty \;$for every $t ,\;$we obtain

  $$\begin{aligned}E \left( \underset{\rho _{n} ,\sigma }{\sup }\left| \ln f_{t} (\rho _{n} ,\sigma )\right| \right) =E E \left( \underset{\rho _{n} ,\sigma }{\sup }\left| \ln f_{t} (\rho _{n} ,\sigma )\right| \mid Y_{t -1}\right) <\infty \text {,} \end{aligned}$$

  as desired.

• (ii) Let $\theta _{n} =(\rho _{n} ,\sigma )\text {,}$$\hat{\theta }_{n} =(\hat{\rho }_{n} ,\hat{\sigma }_{n})\text {,}$$\theta _{n}^{o} =(\rho _{n}^{o} ,\sigma ^{o})$ and $\theta _{n}^{ *}$ be on the line joining $\theta _{n}^{o}$ and $\hat{\theta }_{n} .\;$Because $\sum _{t =1}^{n}\frac{ \partial \ln f}{ \partial \theta _{n}}_{ \mid _{\theta _{n} =\widehat{\theta }_{n}}} =0\text {,}$ we obtain by the mean value theorem

  $$\begin{aligned}0 =\sum _{t =1}^{n}\frac{ \partial \ln f}{ \partial \theta _{n}}_{ \mid _{\theta _{n} =\theta _{n}^{o}}} +\sum _{t =1}^{n}\frac{ \partial ^{2}\ln f}{ \partial \theta _{n} \partial \theta _{n}^{ \prime }}_{ \mid _{\theta _{n} =\theta _{n}^{ *}}} \left( \begin{array}{l}\hat{\rho }_{n} -\rho _{n}^{o} \\ \hat{\sigma }_{n} -\sigma ^{o}\end{array}\right) \text {,} \end{aligned}$$

which gives

$$\begin{aligned}&\left( \begin{array}{l}n^{3/2} \left( \hat{\rho }_{n} -\rho _{n}^{o}\right) \\ n^{1/2} \left( \hat{\sigma }_{n} -\sigma ^{o}\right) \end{array}\right) \\&\quad = -\left[ J_{n}^{ -1} \sum _{t =1}^{n}\frac{ \partial ^{2}\ln f}{ \partial \theta _{n} \partial \theta _{n}^{ \prime }}_{ \mid _{\theta _{n} =\theta _{n}^{ *}}} J_{n}^{ -1}\right] ^{ -1} J_{n}^{ -1} \sum _{t =1}^{n}\frac{ \partial \ln f}{ \partial \theta _{n}}_{ \mid _{\theta _{n} =\theta _{n}^{o}}} \\&\quad = -\left( A_{n \mid _{\theta _{n} =\theta _{n}^{ *}}}\right) ^{ -1} J_{n}^{ -1} \sum _{t =1}^{n}\frac{ \partial \ln f}{ \partial \theta _{n}}_{ \mid _{\theta _{n} =\theta _{n}^{o}}} ,\text {say,} \end{aligned}$$

where $J_{n} =d i a g (n^{3/2} ,n^{1/2}) .\;$Because $\theta _{n}^{ *} -\theta _{n}^{o}\overset{p}{ \longrightarrow }0$ as $n \rightarrow \infty ,\;$it follows that

$$\begin{aligned} A_{n \mid _{\theta _{n} =\theta _{n}^{ *}}} -A_{n \mid _{\theta _{n} =\theta _{n}^{o}}}\overset{p}{ \longrightarrow }0. \end{aligned}$$

(8)

Since

$$\begin{aligned} A_{n \mid _{\theta _{n} =\theta _{n}^{o}}} =\left[ \begin{array}{ll}n^{ -3} \sum \nolimits _{t =1}^{n}Y_{t -1}^{2} \frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t}^{2}} &{} -n^{ -2} \sum \nolimits _{t =1}^{n}Y_{t -1} \frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t} \partial \sigma }\\ -n^{ -2} \sum \nolimits _{t =1}^{n}Y_{t -1} \frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t} \partial \sigma } &{} n^{ -1} \sum \nolimits _{t =1}^{n}\frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \sigma ^{2}}\end{array}\right] _{ \mid \theta _{n} =\theta _{n}^{o}}\text {,} \end{aligned}$$

parts (iv), (v) and (vi) of Lemma 2 show that $A_{n \mid _{\theta _{n} =\theta _{n}^{o}}}$ has the same limiting distribution as

$$\begin{aligned} \left[ \begin{array}{ll}n^{ -3} \sum \nolimits _{t =1}^{n}Y_{t -1}^{2} E \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t}^{2}} &{} -n^{ -2} \sum \nolimits _{t =1}^{n}Y_{t -1} E \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t} \partial \sigma } \\ -n^{ -2} \sum \nolimits _{t =1}^{n}Y_{t -1} E \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t} \partial \sigma } &{} n^{ -1} \sum \nolimits _{t =1}^{n}E \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \sigma ^{2}}\end{array}\right] _{ \mid \theta _{n} =\theta _{n}^{o}}\text {,} \end{aligned}$$

which is equal to, due to Lemmas 8 and 9,

$$\begin{aligned} -\frac{1}{2 \sigma ^{o 2}} \left[ \begin{array}{ll}n^{ -3} \sum \nolimits _{t =1}^{n}Y_{t -1}^{2} &{}\quad 0 \\ 0 &{}\quad 1\end{array}\right] \text {.} \end{aligned}$$

Thus, using relation (8) and part (iii) of Lemma 2, we obtain

$$\begin{aligned} A_{n \mid _{\theta _{n} =\theta _{n}^{ *}}} \overset{d}{ \longrightarrow } -\frac{1}{2 \sigma ^{o 2}} \left[ \begin{array}{ll} \smallint _{0}^{1}S^{2} (r) \mathrm{d} r &{}\quad 0 \\ 0 &{}\quad 1\end{array}\right] . \end{aligned}$$

(9)

In addition, Lemmas 1 and 2 yield

$$\begin{aligned} J_{n}^{ -1} \sum _{t =1}^{n}\frac{ \partial \ln f}{ \partial \theta _{n}}_{ \mid _{\theta _{n} =\theta _{n}^{o}}}&=\left( \begin{array}{l} -n^{ -3/2} \sum \nolimits _{t =1}^{n}\frac{ \partial \ln f (\varepsilon _{t} ,\sigma ^{o})}{ \partial \varepsilon _{t}} Y_{t -1} \\ n^{ -1/2} \sum \nolimits _{t =1}^{n}\frac{ \partial \ln f (\varepsilon _{t} ,\sigma )}{ \partial \sigma }_{ \mid \sigma =\sigma ^{o}}\end{array}\right) \nonumber \\&\overset{d}{ \longrightarrow }\left( \begin{array}{l} - \smallint _{0}^{1}S (r) \mathrm{d} B (r) \\ Q (1)\end{array}\right) . \end{aligned}$$

(10)

The stated result follows, once the continuous mapping theorem is applied to relations (9) and (10). $\square$

It is straightforward to show that $S_{n} (r) =n^{ -1} Y_{[n r]}\overset{d}{ \rightarrow }S (r)$ for model (3). Thus, we continue to use Lemma 2 to prove Theorem 2. In addition to Lemma 2, we need the following lemma to prove Theorem 2.

Lemma 3

• (i) $n^{ -2} \sum _{t =1}^{n}Y_{t -1} \left[ \frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma ^{o})}{ \partial \varepsilon _{t}^{2}} -E \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma ^{o})}{ \partial \varepsilon _{t}^{2}}\right] \overset{p}{ \longrightarrow }0 ;$

• (ii) $n^{ -1} \sum _{t =1}^{n}\left[ \frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t} \partial \sigma }_{ \mid \sigma =\sigma ^{o}} -E \left( \frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t} \partial \sigma }_{ \mid \sigma =\sigma ^{o}}\right) \right] \overset{p}{ \longrightarrow }0 ;$

• (iii) $n^{ -1} \sum _{t =1}^{n}\left[ \frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma ^{o})}{ \partial \varepsilon _{t}^{2}} -E \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma ^{o})}{ \partial \varepsilon _{t}^{2}}\right] \overset{p}{ \longrightarrow }0.$

Proof

Using the same method as for Eq. (16) of Zhang and Chan (2012), we obtain (i). Parts (ii) and (iii) are trivial. $\square$

Proof of Theorem 2

• (i) Use the same method as for the proof of Theorem 1 (i).

• (ii) Let $\theta _{n} =(\rho _{n} ,\mu ,\sigma )\text {,}$$\hat{\theta }_{n} =(\hat{\rho }_{n} ,\hat{\mu }_{n} ,\hat{\sigma }_{n})\text {,}$$\theta _{n}^{o} =(\rho _{n}^{o} ,\mu ^{o} ,\sigma ^{o})$ and $\theta _{n}^{ *}$ be on the line joining $\theta _{n}^{o}$ and $\hat{\theta }_{n}\text {.}$ Because $\sum _{t =1}^{n}\frac{ \partial \ln f}{ \partial \theta _{n}}_{ \mid _{\theta _{n} =\widehat{\theta }_{n}}} =0\text {,}$ the mean value theorem yields

  $$\begin{aligned}0 =\sum _{t =1}^{n}\frac{ \partial \ln f}{ \partial \theta _{n}}_{ \mid _{\theta _{n} =\theta _{n}^{o}}} +\sum _{t =1}^{n}\frac{ \partial ^{2}\ln f}{ \partial \theta _{n} \partial \theta _{n}^{ \prime }}_{ \mid _{\theta _{n} =\theta _{n}^{ *}}} \left( \begin{array}{l}\hat{\rho }_{n} -\rho _{n}^{o} \\ \hat{\mu }_{n} -\mu ^{o} \\ \hat{\sigma }_{n} -\sigma ^{o}\end{array}\right) \text {,} \end{aligned}$$

which gives

$$\begin{aligned}&\left( \begin{array}{l}n^{3/2} \left( \hat{\rho }_{n} -\rho _{n}^{o}\right) \\ n^{1/2} \left( \hat{\mu }_{n} -\mu ^{o}\right) \\ n^{1/2} \left( \hat{\sigma }_{n} -\sigma ^{o}\right) \end{array}\right) \\&\quad = -\left[ K_{n}^{ -1} \sum _{t =1}^{n}\frac{ \partial ^{2}\ln f}{ \partial \theta _{n} \partial \theta _{n}^{ \prime }}_{ \mid _{\theta _{n} =\theta _{n}^{ *}}} K_{n}^{ -1}\right] ^{ -1} K_{n}^{ -1} \sum _{t =1}^{n}\frac{ \partial \ln f}{ \partial \theta _{n}}_{ \mid _{\theta _{n} =\theta _{n}^{o}}} \\&\quad = -\left( E_{n \mid _{\theta _{n} =\theta _{n}^{ *}}}\right) ^{ -1} K_{n}^{ -1} \sum _{t =1}^{n}\frac{ \partial \ln f}{ \partial \theta _{n}}_{ \mid _{\theta _{n} =\theta _{n}^{o}}} ,\text {say,}\end{aligned}$$

where $K_{n} =d i a g (n^{3/2} ,n^{1/2} ,n^{1/2}) .\;$Because $\theta _{n}^{ *} -\theta _{n}^{o}\overset{p}{ \longrightarrow }0$ as $n \rightarrow \infty ,\;$it follows that

$$\begin{aligned} E_{n \mid _{\theta _{n} =\theta _{n}^{ *}}} -E_{n \mid _{\theta _{n} =\theta _{n}^{o}}}\overset{p}{ \longrightarrow }0. \end{aligned}$$

(11)

Since

$$\begin{aligned}&E_{n \mid _{\theta _{n} =\theta _{n}^{o}}} \\&\quad =\left[ \begin{array}{lll}n^{ -3} \sum \limits _{t =1}^{n}Y_{t -1}^{2} \frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t}^{2}} &{} n^{ -2} \sum \limits _{t =1}^{n}Y_{t -1} \frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t}^{2}} &{} -n^{ -2} \sum \limits _{t =1}^{n}Y_{t -1} \frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t} \partial \sigma } \\ n^{ -2} \sum \limits _{t =1}^{n}Y_{t -1} \frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t}^{2}} &{} n^{ -1} \sum \limits _{t =1}^{n}\frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t}^{2}} &{} -n^{ -1} \sum \limits _{t =1}^{n}\frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t} \partial \sigma } \\ -n^{ -2} \sum \limits _{t =1}^{n}Y_{t -1} \frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t} \partial \sigma } &{} -n^{ -1} \sum \limits _{t =1}^{n}\frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t} \partial \sigma } &{} n^{ -1} \sum \limits _{t =1}^{n}\frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \sigma ^{2}}\end{array}\right] _{ \mid \theta _{n} =\theta _{n}^{o}}\text {,} \end{aligned}$$

parts (iv), (v) and (vi) of Lemma 2 and parts (i), (ii) and (iii) of Lemma 3 show that $E_{n \mid _{\theta _{n} =\theta _{n}^{o}}}$ has the same limiting distribution as

$$\begin{aligned}\left[ \begin{array}{lll}n^{ -3} \sum \limits _{t =1}^{n}Y_{t -1}^{2} E \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t}^{2}} &{} n^{ -2} \sum \limits _{t =1}^{n}Y_{t -1} E \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t}^{2}} &{} -n^{ -2} \sum \limits _{t =1}^{n}Y_{t -1} E \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t} \partial \sigma } \\ n^{ -2} \sum \limits _{t =1}^{n}Y_{t -1} E \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t}^{2}} &{} n^{ -1} \sum \limits _{t =1}^{n}E \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t}^{2}} &{} -n^{ -1} \sum \limits _{t =1}^{n}E \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t} \partial \sigma } \\ -n^{ -2} \sum \limits _{t =1}^{n}Y_{t -1} E \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t} \partial \sigma } &{} -n^{ -1} \sum \limits _{t =1}^{n}E \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t} \partial \sigma } &{} n^{ -1} \sum \limits _{t =1}^{n}E \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \sigma ^{2}}\end{array}\right] _{ \mid \theta _{n} =\theta _{n}^{o}}\text {,} \end{aligned}$$

which is equal to, due to Lemmas 8 and 9,

$$\begin{aligned} -\frac{1}{2 \sigma ^{o 2}} \left[ \begin{array}{lll}n^{ -3} \sum \nolimits _{t =1}^{n}Y_{t -1}^{2} &{}\quad n^{ -2} \sum \nolimits _{t =1}^{n}Y_{t -1} &{}\quad 0 \\ n^{ -2} \sum \nolimits _{t =1}^{n}Y_{t -1} &{}\quad 1 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 1\end{array}\right] \text {.} \end{aligned}$$

Thus, using relation (11) and parts (ii) and (iii) of Lemma 2, we obtain

$$\begin{aligned} E_{n \mid _{\theta _{n} =\theta _{n}^{ *}}} \overset{d}{ \longrightarrow } -\frac{1}{2 \sigma ^{o 2}} \left[ \begin{array}{lll} \smallint _{0}^{1}S^{2} (r) \mathrm{d} r &{}\quad \smallint _{0}^{1}S (r) \mathrm{d} r &{}\quad 0 \\ \smallint _{0}^{1}S (r) \mathrm{d} r &{}\quad 1 &{}\quad 0 \\ 0 &{}\quad 0 &{} \quad 1\end{array}\right] . \end{aligned}$$

(12)

In addition, Lemmas 1 and 2 yield

$$\begin{aligned} K_{n}^{ -1} \sum _{t =1}^{n}\frac{ \partial \ln f}{ \partial \theta _{n}}_{ \mid _{\theta _{n} =\theta _{n}^{o}}}&=\left( \begin{array}{l} -n^{ -3/2} \sum \nolimits _{t =1}^{n}\frac{ \partial \ln f (\varepsilon _{t} ,\sigma ^{o})}{ \partial \varepsilon _{t}} Y_{t -1} \\ -n^{ -1/2} \sum \nolimits _{t =1}^{n}\frac{ \partial \ln f (\varepsilon _{t} ,\sigma ^{o})}{ \partial \varepsilon _{t}} \\ n^{ -1/2} \sum \nolimits _{t =1}^{n}\frac{ \partial \ln f (\varepsilon _{t} ,\sigma )}{ \partial \sigma }_{ \mid \sigma =\sigma ^{o}}\end{array}\right) \nonumber \\&\overset{d}{ \rightarrow }\left( \begin{array}{l} - \smallint _{0}^{1}S (r) \mathrm{d} B (r)\\ -B (1) \\ Q (1)\end{array}\right) . \end{aligned}$$

(13)

The stated result is obtained by applying the continuous mapping theorem to relations (12) and (13). $\square$

The following two lemmas will be used to prove Theorem 3.

Lemma 4

For model (5), we have $S_{n} (r) =n^{ -1} Y_{[n r]}\overset{d}{ \rightarrow }U (r) =S (r) +b_{0} r$ in $D [0 ,1]\text {.}$

Proof

Because $X_{t} =\rho _{n} X_{t -1} +\varepsilon _{t}\text {,}$$n^{ -1} X_{[n r]}\overset{d}{ \rightarrow }S (r)$ in D[0, 1], we obtain

$$\begin{aligned} S_{n} (r) =n^{ -1} Y_{[n r]} =n^{ -1} \left( a_{0} +b_{0} [n r] +X_{[n r]}\right) \overset{d}{ \rightarrow }U (r) =b_{0} r +S (r)\text { in }D [0 ,1]\text {,} \end{aligned}$$

as stated. $\square$

Lemma 5

• (i) $\;n^{ -3/2} \sum _{t =1}^{n}\frac{ \partial \ln f (\varepsilon _{t} ,\sigma ^{o})}{ \partial \varepsilon _{t}} Y_{t -1}\overset{d}{ \rightarrow } \smallint _{0}^{1}U (r) \mathrm{d} B (r) ;$

• (ii) $n^{ -2} \sum _{t =1}^{n}Y_{t -1}\overset{d}{ \rightarrow } \smallint _{0}^{1}U (r) \mathrm{d} r ;$

• (iii) $n^{ -3} \sum _{t =1}^{n}Y_{t -1}^{2}\overset{d}{ \rightarrow } \smallint _{0}^{1}U^{2} (r) \mathrm{d} r ;$

• (iv) $n^{ -3} \sum _{t =1}^{n}t Y_{t -1}\overset{d}{ \rightarrow } \smallint _{0}^{1}r U (r) \mathrm{d} r ;$

• (v) $n^{ -3/2} \sum _{t =1}^{n}t \frac{ \partial \ln f (\varepsilon _{t} ,\sigma ^{o})}{ \partial \varepsilon _{t}}\overset{d}{ \rightarrow } \smallint _{0}^{1}r \mathrm{d} B (r) ;$

• (vi) $n^{ -3} \sum _{t =1}^{n}t Y_{t -1} \left[ \frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma ^{o})}{ \partial \varepsilon _{t}^{2}} -E \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma ^{o})}{ \partial \varepsilon _{t}^{2}}\right] \overset{p}{ \longrightarrow }0 ;$

• (vii) $n^{ -2} \sum _{t =1}^{n}t \left[ \frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t} \partial \sigma }_{ \mid \sigma =\sigma ^{o}} -E \left( \frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma )}{ \partial \varepsilon _{t} \partial \sigma }_{ \mid \sigma =\sigma ^{o}}\right) \right] \overset{p}{ \longrightarrow }0 ;$

• (viii) $n^{ -2} \sum _{t =1}^{n}t \left[ \frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma ^{o})}{ \partial \varepsilon _{t}^{2}} -E \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma ^{o})}{ \partial \varepsilon _{t}^{2}}\right] \overset{p}{ \longrightarrow }0 ;$

• (ix) $n^{ -3} \sum _{t =1}^{n}t^{2} \left[ \frac{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma ^{o})}{ \partial \varepsilon _{t}^{2}} -E \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{t} ,\sigma ^{o})}{ \partial \varepsilon _{t}^{2}}\right] \overset{p}{ \longrightarrow }0.$

Proof

Because $S_{n} (r)\overset{d}{ \rightarrow }U (r)$ in D[0, 1] by Lemma 4, parts (i), (ii), (iii) and (iv) follow as in Lemma 4 of Zhang and Chan (2012). Part (v) follows from the central limit theorem. Using the same method as for Eq. (16) of Zhang and Chan (2012), we obtain (vi), (vii), (viii) and (ix). $\square$

Proof of Theorem 3

• (i) Using the same method as for the proof of Theorem 1 (i), it is straightforward to prove this.

• (ii) Let $\theta _{n} =(\rho _{n} ,\mu ,\beta ,\sigma )\text {,}$$\hat{\theta }_{n} =(\hat{\rho }_{n} ,\hat{\mu }_{n} ,\hat{\beta }_{n\text {,}} \hat{\sigma }_{n})\text {,}$$\theta _{n}^{o} =(\rho _{n}^{o} ,\mu ^{o} ,\beta ^{o} ,\sigma ^{o})$ and $\theta _{n}^{ *}$ be on the line joining $\theta _{n}^{o}$ and $\hat{\theta }_{n}\text {.}$ Because $\sum _{t =1}^{n}\frac{ \partial \ln f}{ \partial \theta _{n}}_{ \mid _{\theta _{n} =\widehat{\theta }_{n}}} =0\text {,}$ we obtain by the mean value theorem

  $$\begin{aligned}0 =\sum _{t =1}^{n}\frac{ \partial \ln f}{ \partial \theta _{n}}_{ \mid _{\theta _{n} =\theta _{n}^{o}}} +\sum _{t =1}^{n}\frac{ \partial ^{2}\ln f}{ \partial \theta _{n} \partial \theta _{n}^{ \prime }}_{ \mid _{\theta _{n} =\theta _{n}^{ *}}} \left( \begin{array}{l}\hat{\rho }_{n} -\rho _{n}^{o} \\ \hat{\mu }_{n} -\mu ^{o} \\ \hat{\beta }_{n} -\beta ^{o} \\ \hat{\sigma }_{n} -\sigma ^{o}\end{array}\right) \text {.} \end{aligned}$$

Thus, we have

$$\begin{aligned}&\left( \begin{array}{l}n^{3/2} \left( \hat{\rho }_{n} -\rho _{n}^{o}\right) \\ n^{1/2} \left( \hat{\mu }_{n} -\mu ^{o}\right) \\ n^{3/2} \left( \hat{\beta }_{n} -\beta ^{o}\right) \\ n^{1/2} \left( \hat{\sigma }_{n} -\sigma ^{o}\right) \end{array}\right) \\&\quad = -\left[ L_{n}^{ -1} \sum _{t =1}^{n}\frac{ \partial ^{2}\ln f}{ \partial \theta _{n} \partial \theta _{n}^{ \prime }}_{ \mid _{\theta _{n} =\theta _{n}^{ *}}} L_{n}^{ -1}\right] ^{ -1} L_{n}^{ -1} \sum _{t =1}^{n}\frac{ \partial \ln f}{ \partial \theta _{n}}_{ \mid _{\theta _{n} =\theta _{n}^{o}}} \\&\quad = -\left( C_{n \mid _{\theta _{n} =\theta _{n}^{ *}}}\right) ^{ -1} L_{n}^{ -1} \sum _{t =1}^{n}\frac{ \partial \ln f}{ \partial \theta _{n}}_{ \mid _{\theta _{n} =\theta _{n}^{o}}} ,\text {say,}\end{aligned}$$

where $L_{n} =d i a g (n^{3/2} ,n^{1/2} ,n^{3/2} ,n^{1/2}) .\;$Because $\theta _{n}^{ *} -\theta _{n}^{o}\overset{p}{ \longrightarrow }0$ as $n \rightarrow \infty ,\;$it follows that

$$\begin{aligned} C_{n \mid _{\theta _{n} =\theta _{n}^{ *}}} -C_{n \mid _{\theta _{n} =\theta _{n}^{o}}}\overset{p}{ \longrightarrow }0. \end{aligned}$$

(14)

Since

$$\begin{aligned}&C_{n \mid _{\theta _{n} =\theta _{n}^{o}}} \\&\quad = \left[ \begin{array}{llll}n^{ -3} \sum \limits _{t =1}^{n}Y_{t -1}^{2} \frac{ \partial ^{2}\ln f}{ \partial \varepsilon _{t}^{2}} &{} n^{ -2} \sum \limits _{t =1}^{n}Y_{t -1} \frac{ \partial ^{2}\ln f}{ \partial \varepsilon _{t}^{2}} &{} n^{ -3} \sum \limits _{t =1}^{n}t Y_{t -1} \frac{ \partial ^{2}\ln f}{ \partial \varepsilon _{t}^{2}} &{} -n^{ -2} \sum \limits _{t =1}^{n}Y_{t -1} \frac{ \partial ^{2}\ln f}{ \partial \varepsilon _{t} \partial \sigma } \\ n^{ -2} \sum \limits _{t =1}^{n}Y_{t -1} \frac{ \partial ^{2}\ln f}{ \partial \varepsilon _{t}^{2}} &{} n^{ -1}\sum \limits _{t =1}^{n}\frac{ \partial ^{2}\ln f}{ \partial \varepsilon _{t}^{2}} &{} n^{ -2} \sum \limits _{t =1}^{n}t \frac{ \partial ^{2}\ln f}{ \partial \varepsilon _{t}^{2}} &{} -n^{ -1}\sum \limits _{t =1}^{n}\frac{ \partial ^{2}\ln f}{ \partial \varepsilon _{t} \partial \sigma } \\ n^{ -3} \sum \limits _{t =1}^{n}t Y_{t -1} \frac{ \partial ^{2}\ln f}{ \partial \varepsilon _{t}^{2}} &{} n^{ -2} \sum \limits _{t =1}^{n}t \frac{ \partial ^{2}\ln f}{ \partial \varepsilon _{t}^{2}} &{} n^{ -3} \sum \limits _{t =1}^{n}t^{2} \frac{ \partial ^{2}\ln f}{ \partial \varepsilon _{t}^{2}} &{} -n^{ -2} \sum \limits _{t =1}^{n}t \frac{ \partial ^{2}\ln f}{ \partial \varepsilon _{t} \partial \sigma } \\ -n^{ -2} \sum \limits _{t =1}^{n}Y_{t -1} \frac{ \partial ^{2}\ln f}{ \partial \varepsilon _{t} \partial \sigma } &{} -n^{ -1}\sum \limits _{t =1}^{n}\frac{ \partial ^{2}\ln f}{ \partial \varepsilon _{t} \partial \sigma } &{} -n^{ -2} \sum \limits _{t =1}^{n}t \frac{ \partial ^{2}\ln f}{ \partial \varepsilon _{t} \partial \sigma } &{} n^{ -1} \sum \limits _{t =1}^{n}\frac{ \partial ^{2}\ln f}{ \partial \sigma ^{2}}\end{array}\right] _{ \mid \theta _{n} =\theta _{n}^{o}}\text {,} \end{aligned}$$

Lemmas 2, 3 and 5 show that $C_{n \mid _{\theta _{n} =\theta _{n}^{o}}}$ has the same limiting distribution as

$$\begin{aligned} \left[ \begin{array}{llll}n^{ -3} \sum \limits _{t =1}^{n}Y_{t -1}^{2} G &{} n^{ -2} \sum \limits _{t =1}^{n}Y_{t -1} G &{} n^{ -3}\sum \limits _{t =1}^{n}t Y_{t -1} G &{} -n^{ -2} \sum \limits _{t =1}^{n}Y_{t -1} H \\ n^{ -2} \sum \limits _{t =1}^{n}Y_{t -1} G &{} n^{ -1} \sum \limits _{t =1}^{n}G &{} n^{ -2} \sum \limits _{t =1}^{n}t G &{} -n^{ -1} \sum \limits _{t =1}^{n}H \\ n^{ -3} \sum \limits _{t =1}^{n}t Y_{t -1} G &{} n^{ -2} \sum \limits _{t =1}^{n}t G &{} n^{ -3} \sum \limits _{t =1}^{n}t^{2} G &{} -n^{ -2}\sum \limits _{t =1}^{n}t H \\ -n^{ -2} \sum \limits _{t =1}^{n}Y_{t -1} H &{} -n^{ -1}\sum \limits _{t =1}^{n}H &{} -n^{ -2} \sum \limits _{t =1}^{n}t H &{} n^{ -1} \sum \limits _{t =1}^{n}J\end{array}\right] _{ \mid \theta _{n} =\theta _{n}^{o}}\text {,} \end{aligned}$$

where $G =E \genfrac[]{}{}{ \partial ^{2}\ln f}{ \partial \varepsilon _{t}^{2}}\text {,}$$H =E \genfrac[]{}{}{ \partial ^{2}\ln f}{ \partial \varepsilon _{t} \partial \sigma }$ and $J =E \genfrac[]{}{}{ \partial ^{2}\ln f}{ \partial \sigma ^{2}}\text {.}$ Due to Lemmas 8 and 9, the above matrix is equal to

$$\begin{aligned} -\frac{1}{2 \sigma ^{o 2}} \left[ \begin{array}{llll}n^{ -3} \sum \nolimits _{t =1}^{n}Y_{t -1}^{2} &{} n^{ -2} \sum \nolimits _{t =1}^{n}Y_{t -1} &{} n^{ -3} \sum \nolimits _{t =1}^{n}t Y_{t -1} &{} 0 \\ n^{ -2} \sum \nolimits _{t =1}^{n}Y_{t -1} &{} 1 &{} n^{ -2} \sum \nolimits _{t =1}^{n}t &{} 0 \\ n^{ -3} \sum \nolimits _{t =1}^{n}t Y_{t -1} &{} n^{ -2} \sum \nolimits _{t =1}^{n}t &{} n^{ -3} \sum \nolimits _{t =1}^{n}t^{2} &{} 0 \\ 0 &{} 0 &{} 0 &{} 1\end{array}\right] \text {.} \end{aligned}$$

Thus, using relation (14) and parts (ii), (iii) and (iv) of Lemma 5, we obtain

$$\begin{aligned} C_{n \mid _{\theta _{n} =\theta _{n}^{ *}}} \overset{d}{ \longrightarrow } -\frac{1}{2 \sigma ^{o 2}} \left[ \begin{array}{llll} \smallint _{0}^{1}U^{2} (r) \mathrm{d} r &{} \smallint _{0}^{1}U (r) \mathrm{d} r &{} \smallint _{0}^{1}r U (r) \mathrm{d} r &{} 0 \\ \smallint _{0}^{1}U (r) \mathrm{d} r &{} 1 &{} \frac{1}{2} &{} 0 \\ \smallint _{0}^{1}r U (r) \mathrm{d} r &{} \frac{1}{2} &{} \frac{1}{3} &{} 0 \\ 0 &{} 0 &{} 0 &{} 1\end{array}\right] . \end{aligned}$$

(15)

In addition, Lemmas 1 and 5 yield

$$\begin{aligned}&L_{n}^{ -1} \sum _{t =1}^{n}\frac{ \partial \ln f}{ \partial \theta _{n}}_{ \mid _{\theta _{n} =\theta _{n}^{o}}} \nonumber \\&\quad = \left( \begin{array}{l} -n^{ -3/2} \sum \nolimits _{t =1}^{n}\frac{ \partial \ln f (\varepsilon _{t} ,\sigma ^{o})}{ \partial \varepsilon _{t}} Y_{t -1} \\ -n^{ -1/2} \sum \nolimits _{t =1}^{n}\frac{ \partial \ln f (\varepsilon _{t} ,\sigma ^{o})}{ \partial \varepsilon _{t}} \\ -n^{ -3/2} \sum \nolimits _{t =1}^{n}t \frac{ \partial \ln f (\varepsilon _{t} ,\sigma ^{o})}{ \partial \varepsilon _{t}} \\ n^{ -1/2} \sum \nolimits _{t =1}^{n}\frac{ \partial \ln f (\varepsilon _{t} ,\sigma )}{ \partial \sigma }_{ \mid \sigma =\sigma ^{o}}\end{array}\right) \overset{d}{ \longrightarrow }\left( \begin{array}{l} - \smallint _{0}^{1}U (r) \mathrm{d} B (r) \\ -B (1) \\ - \smallint _{0}^{1}r \mathrm{d} B (r) \\ Q (1)\end{array}\right) . \end{aligned}$$

(16)

The stated result follows from relations (15) and (16). $\square$

Proof of Theorem 4

This follows straightforwardly from the mixture normality results (2), (4) and (6), and the block-diagonal structure of the information matrices in the limit. $\square$

Appendix B: Auxiliary lemmas

Lemma 6

$$\begin{aligned} D_{1}= & {} \int _{\mathbb {R}}\frac{1}{\sigma ^{o 2} +\varepsilon _{1}^{2}} \mathrm{d} \varepsilon _{1} =\frac{\pi }{\sigma ^{o}} ,\text { }D_{2} =\int _{\mathbb {R}}\frac{1}{\left( \sigma ^{o 2} +\varepsilon _{1}^{2}\right) ^{2}} \mathrm{d} \varepsilon _{1} =\frac{\pi }{2 \sigma ^{o 3}}\text {,} \\ D_{3}= & {} \int _{\mathbb {R}}\frac{1}{\left( \sigma ^{o 2} +\varepsilon _{1}^{2}\right) ^{3}} \mathrm{d} \varepsilon _{1} =\frac{3 \pi }{8 \sigma ^{o 5}} ,\text { }D_{4} =\int _{\mathbb {R}}\frac{1}{\left( \sigma ^{o 2} +\varepsilon _{1}^{2}\right) ^{4}} \mathrm{d} \varepsilon _{1} =\frac{5 \pi }{16 \sigma ^{o 7}}\text {,} \\ D_{5}= & {} \int _{\mathbb {R}}\frac{1}{\left( \sigma ^{o 2} +\varepsilon _{1}^{2}\right) ^{5}} \mathrm{d} \varepsilon _{1} =\frac{35 \pi }{128 \sigma ^{o 9}}\text {.}\end{aligned}$$

Proof

The first result follows because $\smallint _{\mathbb {R}}\frac{1}{x +\varepsilon _{1}^{2}} \mathrm{d} \varepsilon _{1} =\pi x^{ -1/2} .\;$The rest are obtained by successively differentiating both sides of this equation with respect to x and setting $x =\sigma ^{o 2}\text {.}$ Differentiating within the integral sign is allowed, because for $k \in \mathbb {N}$ and $x \in \mathbb {R}^{ +}\text {,}$$\left| \frac{ \partial \left( x +\varepsilon _{1}^{2}\right) ^{ -k}}{ \partial x} \right| =\left| \frac{k}{\left( x +\varepsilon _{1}^{2}\right) ^{k +1}} \right| \le \left| \frac{k}{\varepsilon _{1}^{2 (k +1)}}\right|$ and $\left| \frac{k}{\varepsilon _{1}^{2 (k +1)}}\right|$ is integrable. $\square$

Lemma 7

• (i) $\;E \left[ \genfrac(){}{}{ \partial \ln f (\varepsilon _{1} ,\sigma ^{o})}{ \partial \varepsilon _{1}}^{2}\right] =\frac{1}{2 \sigma ^{o 2}} ;$

• (ii) $E \left[ \left( \frac{ \partial \ln f (\varepsilon _{1} ,\sigma )}{ \partial \sigma }_{ \mid \sigma =\sigma ^{o}}\right) ^{2}\right] =\frac{1}{2 \sigma ^{o 2}}\text {.}$

Proof

Using Lemma 6, we obtain

$$\begin{aligned}E \left[ \genfrac(){}{}{ \partial \ln f (\varepsilon _{1} ,\sigma ^{o})}{ \partial \varepsilon _{1}}^{2}\right]= & {} \int _{\mathbb {R}}\genfrac(){}{}{2 \varepsilon _{1}}{\varepsilon _{1}^{2} +\sigma ^{o 2}}^{2} \cdot \frac{\sigma ^{o}}{\pi (\varepsilon _{1}^{2} +\sigma ^{o 2})} \mathrm{d} \varepsilon _{1} \\= & {} \frac{4 \sigma ^{o}}{\pi } \int _{\mathbb {R}}\frac{\varepsilon _{1}^{2}}{(\varepsilon _{1}^{2} +\sigma ^{o 2})^{3}} \mathrm{d} \varepsilon _{1} \\= & {} \frac{4 \sigma ^{o}}{\pi } \left[ \int _{\mathbb {R}}\frac{\varepsilon _{1}^{2} +\sigma ^{o 2}}{(\varepsilon _{1}^{2} +\sigma ^{o 2})^{3}} \mathrm{d} \varepsilon _{1} -\int _{\mathbb {R}}\frac{\sigma ^{o 2}}{(\varepsilon _{1}^{2} +\sigma ^{o 2})^{3}} \mathrm{d} \varepsilon _{1}\right] \\= & {} \frac{4 \sigma ^{o}}{\pi } \left[ D_{2} -\sigma ^{o 2} D_{3}\right] =\frac{1}{2 \sigma ^{o 2}}\text {,}\end{aligned}$$

where $D_{2}$ and $D_{3}$ are defined in Lemma 6. In the same manner, we have

$$\begin{aligned} E \left[ \left( \frac{ \partial \ln f (\varepsilon _{1} ,\sigma )}{ \partial \sigma }_{ \mid \sigma =\sigma ^{o}}\right) ^{2}\right]&= \int _{\mathbb {R}}\frac{(\varepsilon _{1}^{2} -\sigma ^{o 2})^{2}}{\sigma ^{o 2} (\varepsilon _{1}^{2} +\sigma ^{o 2})^{2}} \cdot \frac{\sigma ^{o}}{\pi (\varepsilon _{1}^{2} +\sigma ^{o 2})} \mathrm{d} \varepsilon _{1} \\&= \frac{1}{\sigma ^{o} \pi } \int _{\mathbb {R}}\frac{(\varepsilon _{1}^{2} +\sigma ^{o 2})^{2} -4 \varepsilon _{1}^{2} \sigma ^{o 2}}{(\varepsilon _{1}^{2} +\sigma ^{o 2})^{3}} \mathrm{d} \varepsilon _{1} \\&= \frac{1}{\sigma ^{o} \pi } \left[ \int _{\mathbb {R}}\frac{1}{\varepsilon _{1}^{2} +\sigma ^{o 2}} \mathrm{d} \varepsilon _{1} -4 \sigma ^{o 2} \int _{\mathbb {R}}\frac{\varepsilon _{1}^{2}}{(\varepsilon _{1}^{2} +\sigma ^{o 2})^{3}} \mathrm{d} \varepsilon _{1}\right] \\&= \frac{1}{\sigma ^{o} \pi } \left[ D_{1} -4 \sigma ^{o 2} (D_{2} -\sigma ^{o 2} D_{3})\right] =\frac{1}{2 \sigma ^{o 2}}\text {.}\end{aligned}$$

$\square$

Lemma 8

$E \left[ \frac{ \partial ^{2}\ln f (\varepsilon _{1} ,\sigma )}{ \partial \varepsilon _{1} \partial \sigma }_{ \mid \sigma =\sigma ^{o}}\right] =0.$

Proof

Because

$$\begin{aligned} \frac{ \partial ^{2}\ln f (\varepsilon _{1} ,\sigma )}{ \partial \varepsilon _{1} \partial \sigma }_{ \mid \sigma =\sigma ^{o}} =\frac{4 \varepsilon _{1} \sigma ^{o}}{\left( \varepsilon _{1}^{2} +\sigma ^{o 2}\right) ^{2}}\text {,} \end{aligned}$$

we have

$$\begin{aligned} E \left[ \frac{ \partial ^{2}\ln f (\varepsilon _{1} ,\sigma )}{ \partial \varepsilon _{1} \partial \sigma }_{ \mid \sigma =\sigma ^{o}}\right] =\frac{4 \sigma ^{o 2}}{\pi } \int _{\mathbb {R}}\frac{\varepsilon _{1}}{\left( \varepsilon _{1}^{2} +\sigma ^{o 2}\right) ^{3}} \mathrm{d} \varepsilon _{1}\text {.} \end{aligned}$$

Because the integrand is an odd function, the stated result follows. $\square$

Lemma 9

$E \left[ \frac{ \partial ^{2}\ln f (\varepsilon _{1} ,\sigma )}{ \partial \sigma ^{2}}_{ \mid \sigma =\sigma ^{o}}\right] = -\frac{1}{2 \sigma ^{o 2}}$ and $E \genfrac[]{}{}{ \partial ^{2}\ln f (\varepsilon _{1} ,\sigma ^{o})}{ \partial \varepsilon _{1}^{2}} = -\frac{1}{2 \sigma ^{o 2}}\text {.}$

Proof

We begin with the identity

$$\begin{aligned} 1 =\int _{\mathbb {R}}f (\varepsilon _{1} ,\sigma ) \mathrm{d} \varepsilon _{1}\text {.} \end{aligned}$$

By Leibniz’s rule, taking the derivative of the both sides of the above equation with respect to $\sigma$ results in

$$\begin{aligned} 0= & {} \int _{\mathbb {R}}\frac{ \partial f (\varepsilon _{1} ,\sigma )}{ \partial \sigma } \mathrm{d} \varepsilon _{1} \\= & {} \int _{\mathbb {R}}\frac{ \partial f (\varepsilon _{1} ,\sigma )/ \partial \sigma }{f (\varepsilon _{1} ,\sigma )} f (\varepsilon _{1} ,\sigma ) \mathrm{d} \varepsilon _{1} \\= & {} \int _{\mathbb {R}}\frac{ \partial \ln f (\varepsilon _{1} ,\sigma )}{ \partial \sigma } f (\varepsilon _{1} ,\sigma ) \mathrm{d} \varepsilon _{1} \\= & {} E \genfrac[]{}{}{ \partial \ln f (\varepsilon _{1} ,\sigma )}{ \partial \sigma }\text {.} \end{aligned}$$

Differentiating this again, we obtain by Leibniz’s rule

$$\begin{aligned}&0\overset{}{ =}\int _{\mathbb {R}}\frac{ \partial ^{2}\ln f (\varepsilon _{1} ,\sigma )}{ \partial \sigma ^{2}} f (\varepsilon _{1} ,\sigma ) \mathrm{d} \varepsilon _{1} +\int _{\mathbb {R}}\genfrac(){}{}{ \partial \ln f (\varepsilon _{1} ,\sigma )}{ \partial \sigma }^{2} f (\varepsilon _{1} ,\sigma ) \mathrm{d} \varepsilon _{1}\text {,} \\&\qquad \int _{\mathbb {R}}\frac{ \partial ^{2}\ln f (\varepsilon _{1} ,\sigma )}{ \partial \sigma ^{2}} f (\varepsilon _{1} ,\sigma ) \mathrm{d} \varepsilon _{1} \overset{}{ =} -\int _{\mathbb {R}}\genfrac(){}{}{ \partial \ln f (\varepsilon _{1} ,\sigma )}{ \partial \sigma }^{2} f (\varepsilon _{1} ,\sigma ) \mathrm{d} \varepsilon _{1}\text {,} \\&\qquad E \genfrac[]{}{}{ \partial ^{2}\ln f (\varepsilon _{1} ,\sigma )}{ \partial \sigma ^{2}} \overset{}{ =} -E \genfrac[]{}{}{ \partial \ln f (\varepsilon _{1} ,\sigma )}{ \partial \sigma }^{2}\text {.} \end{aligned}$$

Hence, by Lemma 7,

$$\begin{aligned} E \left[ \frac{ \partial ^{2}\ln f (\varepsilon _{1} ,\sigma )}{ \partial \sigma ^{2}}_{ \mid \sigma =\sigma ^{o}}\right]= & {} -E \left[ \frac{ \partial \ln f (\varepsilon _{1} ,\sigma )}{ \partial \sigma }_{ \mid \sigma =\sigma ^{o}}\right] ^{2} \\= & {} -\frac{1}{2 \sigma ^{o 2}}\text {.} \end{aligned}$$

To prove the second result, consider the relation

$$\begin{aligned} \frac{ \partial ^{2}\ln f (\varepsilon _{1} ,\sigma ^{o})}{ \partial \varepsilon _{1}^{2}} =\frac{2 \varepsilon _{1}^{2} -2 \sigma ^{o 2}}{\left( \sigma ^{o 2} +\varepsilon _{1}^{2}\right) ^{2}}\text {,} \end{aligned}$$

which gives

$$\begin{aligned} \;E \genfrac[]{}{}{ \partial ^{2}\ln f (\varepsilon _{1} ,\sigma ^{o})}{ \partial \varepsilon _{1}^{2}}= & {} \int _{\mathbb {R}}\frac{2 \varepsilon _{1}^{2} -2 \sigma ^{o 2}}{(\varepsilon _{1}^{2} +\sigma ^{o 2})^{2}} \cdot \frac{\sigma ^{o}}{\pi (\varepsilon _{1}^{2} +\sigma ^{o 2})} \mathrm{d} \varepsilon _{1} \\= & {} \frac{\sigma ^{o}}{\pi } \int _{\mathbb {R}}\frac{2 \varepsilon _{1}^{2} -2 \sigma ^{o 2}}{(\varepsilon _{1}^{2} +\sigma ^{o 2})^{3}} \mathrm{d} \varepsilon _{1} \\= & {} \frac{\sigma ^{o}}{\pi } \left( 2 \int _{\mathbb {R}}\frac{\varepsilon _{1}^{2}}{(\varepsilon _{1}^{2} +\sigma ^{o 2})^{3}} \mathrm{d} \varepsilon _{1} -2 \sigma ^{o 2} \int _{\mathbb {R}}\frac{1}{(\varepsilon _{1}^{2} +\sigma ^{o 2})^{3}} \mathrm{d} \varepsilon _{1}\right) \\= & {} \frac{\sigma ^{o}}{\pi } (2 (D_{2} -\sigma ^{o 2} D_{3}) -2 \sigma ^{o 2} D_{3}) \\= & {} -\frac{1}{2 \sigma ^{o 2}}\text {,}\end{aligned}$$

as stated. $\square$

Lemma 10

$E \left[ \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{1} ,\sigma ^{o})}{ \partial \varepsilon _{1}^{2}}^{2}\right] \text {,}$$E \left[ \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{1} ,\sigma )}{ \partial \sigma ^{2}}_{ \mid \sigma =\sigma ^{o}}^{2}\right]$ and $E \left[ \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{1} ,\sigma )}{ \partial \varepsilon _{1} \partial \sigma }_{ \mid \sigma =\sigma ^{o}}^{2}\right]$ are finite.

Proof

First, we show that $E \left[ \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{1} ,\sigma ^{o})}{ \partial \varepsilon _{1} \partial \sigma }_{ \mid \sigma =\sigma ^{o}}^{2}\right] =\frac{5}{8} \sigma ^{o -4}\text {.}$ By Lemma 6,

$$\begin{aligned}E \left[ \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{1} ,\sigma ^{o})}{ \partial \varepsilon _{1} \partial \sigma }_{ \mid \sigma =\sigma ^{o}}^{2}\right]= & {} \int _{\mathbb {R}}\genfrac(){}{}{4 \varepsilon _{1} \sigma ^{o}}{\left( \varepsilon _{1}^{2} +\sigma ^{o 2}\right) ^{2}}^{2} \cdot \frac{\sigma ^{o}}{\pi (\varepsilon _{1}^{2} +\sigma ^{o 2})} \mathrm{d} \varepsilon _{1} \\= & {} \frac{16 \sigma ^{o 3}}{\pi } \left( D_{4} -\sigma ^{o 2} D_{5}\right) \\= & {} \frac{5}{8} \sigma ^{o -4}\text {.}\end{aligned}$$

Second, $E \left[ \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{1} ,\sigma ^{o})}{ \partial \varepsilon _{1}^{2}}^{2}\right] =\frac{7}{8} \sigma ^{o -4}\text {.}$ By Lemma 6,

$$\begin{aligned}E \left[ \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{1} ,\sigma ^{o})}{ \partial \varepsilon _{1}^{2}}^{2}\right]= & {} \int _{\mathbb {R}}\genfrac(){}{}{2 \varepsilon _{1}^{2} -2 \sigma ^{o 2}}{(\varepsilon _{1}^{2} +\sigma ^{o 2})^{2}}^{2} \cdot \frac{\sigma ^{o}}{\pi (\varepsilon _{1}^{2} +\sigma ^{o 2})} \mathrm{d} \varepsilon _{1} \\= & {} \frac{4 \sigma ^{o}}{\pi } \left( \int _{\mathbb {R}}\frac{\left( \varepsilon _{1}^{2} +\sigma ^{o 2}\right) ^{2}}{(\varepsilon _{1}^{2} +\sigma ^{o 2})^{5}} \mathrm{d} \varepsilon _{1} -4 \sigma ^{o 2} \int _{\mathbb {R}}\frac{\varepsilon _{1}^{2}}{(\varepsilon _{1}^{2} +\sigma ^{o 2})^{5}} \mathrm{d} \varepsilon _{1}\right) \\= & {} \frac{4 \sigma ^{o}}{\pi } \left( D_{3} -4 \sigma ^{o 2} \left( D_{4} -\sigma ^{o 2} D_{5}\right) \right) \\= & {} \frac{7}{8} \sigma ^{o -4}\text {.}\end{aligned}$$

To prove $E \left[ \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{1} ,\sigma )}{ \partial \sigma ^{2}}_{ \mid \sigma =\sigma ^{o}}^{2}\right] =\frac{7}{8} \sigma ^{o -4}\text {,}$ we should note the following equations

$$\begin{aligned}\int _{\mathbb {R}}\frac{\varepsilon _{1}^{2}}{\left( \varepsilon _{1}^{2} +\sigma ^{o 2}\right) ^{5}} \mathrm{d} \varepsilon _{1}= & {} D_{4} -\sigma ^{o 2} D_{5}\text {,} \\ \int _{\mathbb {R}}\frac{\varepsilon _{1}^{4}}{\left( \varepsilon _{1}^{2} +\sigma ^{o 2}\right) ^{5}} \mathrm{d} \varepsilon _{1}= & {} D_{3} -2 \sigma ^{o 2} D_{4} +\sigma ^{o 4} D_{5}\text {,} \\ \int _{\mathbb {R}}\frac{\varepsilon _{1}^{6}}{\left( \varepsilon _{1}^{2} +\sigma ^{o 2}\right) ^{5}} \mathrm{d} \varepsilon _{1}= & {} D_{2} -3 \sigma ^{o 2} D_{3} +3 \sigma ^{o 4} D_{4} -\sigma ^{o 6} D_{5}\text {,} \\ \int _{\mathbb {R}}\frac{\varepsilon _{1}^{8}}{\left( \varepsilon _{1}^{2} +\sigma ^{o 2}\right) ^{5}} \mathrm{d} \varepsilon _{1}= & {} D_{1} -4 \sigma ^{o 2} D_{2} +6 \sigma ^{o 4} D_{3} -4 \sigma ^{o 6} D_{4} +\sigma ^{o 8} D_{5}\text {.}\end{aligned}$$

By using these equations, we have

$$\begin{aligned}&E \left[ \genfrac(){}{}{ \partial ^{2}\ln f (\varepsilon _{1} ,\sigma )}{ \partial \sigma ^{2}}_{ \mid \sigma =\sigma ^{o}}^{2}\right] = \int _{\mathbb {R}}\genfrac(){}{}{\sigma ^{o 4} -4 \varepsilon _{1}^{2} \sigma ^{o 2} -\varepsilon _{1}^{4}}{\sigma ^{o 2} (\varepsilon _{1}^{2} +\sigma ^{o 2})^{2}}^{2} \cdot \frac{\sigma ^{o}}{\pi (\varepsilon _{1}^{2} +\sigma ^{o 2})} \mathrm{d} \varepsilon _{1} \\&\quad = \frac{1}{\pi \sigma ^{o 3}} \int _{\mathbb {R}}\frac{\varepsilon _{1}^{8} +8 \sigma ^{o 2} \varepsilon _{1}^{6} +14 \sigma ^{o 4} \varepsilon _{1}^{4} -8 \sigma ^{o 6} \varepsilon _{1}^{2} +\sigma ^{o 8}}{\left( \varepsilon _{1}^{2} +\sigma ^{o 2}\right) ^{5}} \mathrm{d} \varepsilon _{1} \\&\quad = \frac{1}{\pi \sigma ^{o 3}} \left( D_{1} +4 \sigma ^{o 2} D_{2} -4 \sigma ^{o 4} D_{3} -16 \sigma ^{o 6} D_{4} +16 \sigma ^{o 8} D_{5}\right) \\&\quad = \frac{7}{8} \sigma ^{o -4}\text {.} \end{aligned}$$

$\square$