Modified maximum spacings method for generalized extreme value distribution and applications in real data analysis

Abstract

This paper analyzes weekly closing price data of the S&P 500 stock index and electrical insulation element lifetimes data based on generalized extreme value distribution. A new estimation method, modified maximum spacings (MSP) method, is proposed and obtained by using interior penalty function algorithm. The standard error of the proposed method is calculated through Bootstrap method. The asymptotic properties of the modified MSP estimators are discussed. Some simulations are performed, which show that the proposed method is not only available for the whole shape parameter space, but is also of high efficiency. The benchmark risk index, value at risk (VaR), is evaluated according to the proposed method, and the confidence interval of VaR is also calculated through Bootstrap method. Finally, the results are compared with those derived by empirical calculation and some existing methods.

Appendix A: Proofs of Theorems 1 and 2

Now we present some lemmas, which will be applied in proving Theorems 1 and 2. Lemmas 1–4 can be easily proved based on the result of Ranneby (1984). Consequently, here we omit the proof of Lemmas 1–4, and present the proof of Lemma 5, Theorems 1 and 2 in details.

Lemma 1

The random functions $V_n(N,\varvec{\theta })$ converge to zero, uniformly in $n$ and $\varvec{\theta }$, as $N$ tends to infinity (i.e. $\sup _{n\ge 1,\varvec{\theta }\in \varvec{\Theta }}|V_n(N,\varvec{\theta })|\rightarrow 0$, as $N\rightarrow \infty$).

Lemma 2

The functions $V(N,\varvec{\theta })$ converge to zero, uniformly in $\varvec{\theta }$, as $N$ tends to infinity (i.e. $\sup _{\varvec{\theta }\in \varvec{\Theta }}|V(N,\varvec{\theta })|\rightarrow 0$, as $N\rightarrow \infty$).

Lemma 3

For every fixed $\varvec{\theta }\in \varvec{\Theta }$ and every fixed $M\in {\mathbb {R}}^+, T_n(M,\varvec{\theta })$ converges in probability to $T(M,\varvec{\theta })$, as $n\rightarrow \infty$. If, in addition, Assumption A2 is satisfied, then the convergence is uniform in $\varvec{\Theta }$, that is $\sup _{\varvec{\theta }\in \varvec{\Theta }}|T_n(M,\varvec{\theta })-T(M, \varvec{\theta })|\mathop {\longrightarrow }\limits ^{P}0$ as $n\rightarrow \infty$.

Lemma 4

Let $x_1,x_2,\ldots ,x_n$ be a sequence of i.i.d. random variables with density $g(x;\varvec{\theta }_0)$. Then $T_n(\varvec{\theta }_0)$ converges in probability to $T_0(\varvec{\theta }_0)$ as $n\rightarrow \infty$.

Lemma 5

Let $L(x,y):(0,\infty )\times (0,1)\rightarrow {\mathbb {R}}, L_n(x,y):(0,\infty )\times (0,1)\rightarrow {\mathbb {R}}, n=1,2,\ldots$ satisfy the followings:

• i. For each $y, L(x,y),L_n(x,y), n=1,2,\ldots$ are differentiable w.r.t. $x$ on $(0,\infty )$ and the derivative $\frac{\partial }{\partial x}L(x,y), \frac{\partial }{\partial x}L_n(x,y)$ are continuous in $x$;

• ii. For each $x, L(x,y)$ and $\frac{\partial }{\partial x}L(x,y)$ are continuous in $y$;

• iii. For r.v. $z$ and constant $a, |L_n(z,a)-L(z,a)|\mathop {\longrightarrow }\limits ^{P}0$ and $|\frac{\partial }{\partial z}L_n(z,a)-\frac{\partial }{\partial z}L(z,a)|\mathop {\longrightarrow }\limits ^{P}0$, as $n\rightarrow \infty$;

• iv. $\int _0^1E\{L(W,u)\}du<\infty$ and $\int _0^1E\{W\frac{\partial }{\partial W}L(W,u)\}du<\infty$, where $W\sim E(1)$; Then,

  $$\begin{aligned} \frac{1}{n}\sum _{k=1}^nL_n(nT_k,\xi _k)\mathop {\longrightarrow }\limits ^{P} \int \limits _0^1E\{L(W,u)\}du, \end{aligned}$$

  (14)

  where $\xi _k\triangleq (k-\frac{1}{2})/n$ and $T_k\triangleq U_{(k)}-U_{(k-1)}, k=1,\ldots ,n$ with $U_{(k)}$ be the $k^{th}$ order statistic for $U(0,1)$.

Proof

From Pyke (1965), $\{nT_k\}_{k=1}^n$ have the same distribution as independent exponential random variables divided by their mean. Thus, we can derive that $\{W_k/\overline{W}\}_{k=1}^n\mathop {=}\limits ^{d}\{nT_k\}_{k=1}^n$, where $\{W_k\}_{k=1}^n$ are i.i.d. and $W_k\sim {\hbox {Exp}}(1),~k=1,\ldots ,n$. The symbol $a\mathop {=}\limits ^{d}b$ means that $a$ is equivalent to $b$ in distribution. As the $L_n(x,y)$ is continuous at $x$ for every $n$, the following statement holds:

$$\begin{aligned} \frac{1}{n}\sum _{k=1}^nL_n(nT_k,\xi _k)\mathop {=}\limits ^{d}\frac{1}{n}\sum _{k=1}^nL_n \left( \frac{W_k}{\overline{W}},\xi _k\right) . \end{aligned}$$

(15)

By a Taylor expansion of $\overline{W}$ around its mean $1$, (15) can be written as

$$\begin{aligned} \frac{1}{n}\sum _{k=1}^nL_n\left( \frac{W_k}{\overline{W}},\xi _k\right) =\frac{1}{n}\sum _{k=1}^n\bigg \{L_n(W_k,\xi _k)- \frac{(\overline{W}-1)W_k}{\varpi _{kn}^2}L_n^{1,0}\left( \frac{W_k}{\varpi _{kn}},\xi _k\right) \bigg \}, \end{aligned}$$

(16)

where $\varpi _{kn}=1-\alpha _{kn}+\alpha _{kn}\overline{W},~0\le \alpha _{kn}\le 1$, and $L_n^{1,0}(x,y)$ denotes $\displaystyle \frac{\partial }{\partial x}L(x,y)$. Now we will proof that the first term on the right side of (16) converges to the desired limit while the second term goes to zero in probability.

$$\begin{aligned}&\bigg |\frac{1}{n}\sum _{k=1}^nL_n(W_k,\xi _k)-\int \limits _0^1E\{L(W,u)\}du\bigg | \le \bigg |\frac{1}{n}\sum _{k=1}^nL_n(W_k,\xi _k)-\frac{1}{n}\sum _{k=1}^nL(W_k,\xi _k)\bigg |\nonumber \\&\quad +\bigg |\frac{1}{n}\sum _{k=1}^n[L(W_k,\xi _k)-E\{L(W,\xi _k)\}]\bigg |+\bigg |\frac{1}{n}\sum _{k=1}^nE\{L(W,\xi _k)\}-\int \limits _0^1E\{L(W,u)\}du\bigg |.\qquad \qquad \end{aligned}$$

(17)

From the assumption (iii) in Lemma 1, it can be easily found that the first term on the right side of the inequality (17)

$$\begin{aligned} \bigg |\frac{1}{n}\sum _{k=1}^nL_n(W_k,\xi _k)-\frac{1}{n}\sum _{k=1}^nL(W_k,\xi _k)\bigg |\mathop {\longrightarrow }\limits ^{P}0, \end{aligned}$$

as $n\rightarrow \infty$. Since $W_k$ are i.i.d., by Kolomogorov’s SLLN (see Shiryayev 1984), the second term on the right side of (17) converge to zero in probability. Due to the continuity of $L(x,y)$ in $y$, the third term also converges to zero. Thus, the first term of (16) converges to the desired limit. Next the second term will be proved to go to zero in probability.

As the $L_n^{1,0}(x,y)$ is continuous in $x$ and $\overline{W}-1={\mathcal {O}}_p(n^{-1/2})$, it can be derived that

$$\begin{aligned} \frac{1}{n}\sum _{k=1}^n\frac{W_k}{\varpi _{kn}^2}L_n^{1,0}\left( \frac{W_k}{\varpi _{kn}},\xi _k\right) = \frac{1}{n}\sum _{k=1}^n\left[ W_kL_n^{1,0}(W_k,\xi _k)\right] +o_p(1). \end{aligned}$$

(18)

By the same argument as before and the assumption (iv) in Lemma 1, it can be shown that

$$\begin{aligned} \frac{1}{n}\sum _{k=1}^n[W_kL_n^{1,0}(W_k,\xi _k)]\mathop {\longrightarrow }\limits ^{P}\int \limits _0^1E\{WL^{1,0}(W,u)\}du<\infty . \end{aligned}$$

(19)

Since $\overline{W}-1={\mathcal {O}}_p(n^{-1/2})$, then

$$\begin{aligned} (\overline{W}-1)\frac{1}{n}\sum _{k=1}^n[W_kL_n^{1,0}(W_k,\xi _k)]\mathop {\longrightarrow }\limits ^{P}0, \end{aligned}$$

(20)

as $n\rightarrow \infty$, in other word, the second term on the right side of (16) goes to zero in probability. Thus, Lemma 5 has been proved. $\square$

1.1 A.1 Proof of Theorem 1

Proof

Let $\delta$ be arbitrary and denote

$$\begin{aligned} \inf _{\varvec{\theta }\in \varvec{\Theta }_\delta }T(M,\varvec{\theta })-T_0(\varvec{\theta }_0)=a(M,\delta ). \end{aligned}$$

(21)

By Assumption A1, there exists an integer $M_1$ such that $a(M,\delta )>0$ for all $M\ge M_1$. Put $\epsilon =a(M_1,\delta )/2$. We have

$$\begin{aligned} T_n(\varvec{\theta })\ge T_n(M,\varvec{\theta })+\frac{1}{n+1}h_n(n+1), \end{aligned}$$

(22)

so if $n$ is chosen such that

$$\begin{aligned} \frac{1}{n+1}h_n(n+1)>-\frac{\epsilon }{2} \quad \hbox {and} \quad -\frac{h_n(c_n)}{n+1}>\frac{\epsilon }{4}, \end{aligned}$$

we get

$$\begin{aligned} T_n(M,\widehat{\varvec{\theta }}_n)-\frac{\epsilon }{2}<T_n(M,\widehat{\varvec{\theta }}_n)+\frac{1}{n+1}h_n(n+1) <T_n(\varvec{\theta }_0)-\frac{\epsilon }{4}. \end{aligned}$$

(23)

It follows from Lemma 3 and 4 that

$$\begin{aligned} T(M,\widehat{\varvec{\theta }}_n)<T_n(M,\widehat{\varvec{\theta }}_n)-\epsilon \quad \hbox {and} \quad T_n(\varvec{\theta }_0)<T_0(\varvec{\theta }_0)-\frac{\epsilon }{4}, \end{aligned}$$

both hold with a probability tending to 1 as $n\rightarrow \infty$. Hence, for all $M\ge M_1$,

$$\begin{aligned} T(M,\widehat{\varvec{\theta }}_n)<T_0(\varvec{\theta }_0)-2\epsilon =T_0(\varvec{\theta }_0)-a(M,\delta )=\inf _{\varvec{\theta }\in \varvec{\Theta }_\delta }T(M,\varvec{\theta }), \end{aligned}$$

(24)

holds with a probability tending to 1 as $n\rightarrow \infty$. This implies that

$$\begin{aligned} P(\widehat{\varvec{\theta }}_n\in \{\varvec{\theta }:\Vert \varvec{\theta }-\varvec{\theta }_0\Vert <\delta \})\rightarrow 1\quad \hbox {as}~n\rightarrow \infty , \end{aligned}$$

(25)

and since $\delta$ is arbitrary the desired result is obtained. $\square$

1.2 A.2 Proof of Theorem 2

Proof

By a Taylor expansion of the estimate $\widehat{\varvec{\theta }}_n$ around the true parameter $\varvec{\theta }_0$, it can be showed that

$$\begin{aligned} T_{n}'(\widehat{\varvec{\theta }}_n)=T_{n}'(\varvec{\theta }_0) +T_{n}^{(2)}(\varvec{\theta }_0)(\widehat{\varvec{\theta }}_n-\varvec{\theta }_0) +\frac{1}{2}(\widehat{\varvec{\theta }}_n-\varvec{\theta }_0)^TT_{n}^{(3)}(\varvec{\theta }^*)(\widehat{\varvec{\theta }}_n-\varvec{\theta }_0), \end{aligned}$$

(26)

where $\varvec{\theta }^*=\alpha \widehat{\varvec{\theta }}_n+(1-\alpha )\varvec{\theta }_0,~0\le \alpha \le 1$. As $T_{n}'(\widehat{\varvec{\theta }}_n)=0$, (26) can be written as

$$\begin{aligned} \sqrt{n}(\widehat{\varvec{\theta }}_n-\varvec{\theta }_0)= \bigg \{-T_n^{(2)}(\varvec{\theta }_0)-\frac{1}{2}(\widehat{\varvec{\theta }}_n-\varvec{\theta }_0)^TT_{n}^{(3)}(\varvec{\theta }^*)\bigg \}^{-1} \sqrt{n}T_n'(\varvec{\theta }_0). \end{aligned}$$

(27)

Now we will prove that

$$\begin{aligned} \sqrt{n}T_n'(\varvec{\theta }_0)\mathop {\longrightarrow }\limits ^{d}N(\varvec{0},K(\varvec{\theta }_0)), \end{aligned}$$

(28)

$$\begin{aligned} -T_n^{(2)}(\varvec{\theta }_0)\mathop {\longrightarrow }\limits ^{P}R(\varvec{\theta }_0),~~\frac{1}{2} (\widehat{\varvec{\theta }}_n-\varvec{\theta }_0)^TT_{n}^{(3)}(\varvec{\theta }^*)\mathop {\longrightarrow }\limits ^{P}0, \end{aligned}$$

(29)

where $K(\varvec{\theta })$ and $R(\varvec{\theta })$ are defined in Sect. 2.

Step 1. Proof of (28) Writing $\tilde{U}_i\in (G(X_{(i-1)},\varvec{\theta }_0),G(X_{(i)},\varvec{\theta }_0))$ and substituting the expression of $T_n(\varvec{\theta })$ into $\displaystyle \frac{1}{n}T_{n}'(\varvec{\theta }_0)$, it can be obtained that

$$\begin{aligned} \sqrt{n+1}T_n'(\varvec{\theta }_0)&= \frac{1-q_n}{\sqrt{n+1}}\sum _{i=1}^{n+1}\frac{[(n+1)D_i(\varvec{\theta }_0)]^{1-q_n}-1}{1-q_n}h_{\varvec{\theta }_0}(\xi _i) +\frac{1}{\sqrt{n+1}}\sum _{i=1}^{n+1}h_{\varvec{\theta }_0}(\tilde{U}_i)\nonumber \\&+\frac{1-q_n}{\sqrt{n+1}}\sum _{i=1}^{n+1}\frac{[(n+1)D_i(\varvec{\theta }_0)]^{1-q_n}-1}{1-q_n}[h_{\varvec{\theta }_0}(\tilde{U}_i)-h_{\varvec{\theta }_0}(\xi _i)] \triangleq C_1+C_2+C_3.\nonumber \\ \end{aligned}$$

(30)

• (1) $C_1$: It can be easily verified that, for r.v. $z, |-h_n(z)-\log {z}|\mathop {\longrightarrow }\limits ^{P}0$ as $n\rightarrow \infty$ and $\{nD_i(\varvec{\theta }_0)\}_{i=1}^n\mathop {=}\limits ^{d}\{nT_k\}_{k=1}^n$, where $\{nT_k\}_{k=1}^n$ are defined in Lemma 5. From Lemma 5, it can be derived that

  $$\begin{aligned} \frac{1}{n+1}\sum _{i=1}^{n+1}\frac{[(n+1)D_i(\varvec{\theta }_0)]^{1-q_n}-1}{1-q_n}h_{\varvec{\theta }_0}(\xi _i)&\mathop {\longrightarrow }\limits ^{P} E\{\log {W}\} \displaystyle \int \limits _0^1h_{\varvec{\theta }_0}(u)du. \end{aligned}$$

  (31)

  It is obviously seen that $E\{\log {W}\}$ is bounded, and $\int _0^1h_{\varvec{\theta }_0}(u)du$, by Assumption A3, is also bounded. As a result, (31) converges to zero in probability. Since $\sqrt{n}(1-q_n)={\mathcal {O}}(1)$, it can be derived that (see Vaart 1998),

  $$\begin{aligned} C_1\mathop {\longrightarrow }\limits ^{P}0, \quad {\hbox {as}}~ n\rightarrow \infty . \end{aligned}$$

  (32)

• (2) $C_2$: From the existence of the limiting distribution of the Kolmogorov-Smirnov statistic, we have $|\tilde{U}_i-\xi _i|\mathop {\longrightarrow }\limits ^{P}0$ as $n\rightarrow \infty$ uniformly in $i$. Then

  $$\begin{aligned} C_2=\frac{1}{\sqrt{n}}\sum _{i=1}^nh_{\varvec{\theta }_0}(\xi _i)+o_p(1). \end{aligned}$$

  (33)

  From the Central Limit Theorem, it can be obtained that

  $$\begin{aligned} C_2\mathop {\longrightarrow }\limits ^{d}N(\varvec{0},K(\varvec{\theta }_0)). \end{aligned}$$

  (34)

• (3) $C_3$: Since $|\tilde{U}_i-\xi _i|\mathop {\longrightarrow }\limits ^{P}0$ as $n\rightarrow \infty$ uniformly in $i$ and $E\{\log {W}\}$ is bounded, similar to the argument in (31), it can be easily derived that, $C_3$ goes to zero in probability.

Hence, the proposition (28) has been proved. Next we consider the asymptotic result in (29).

Step 2. Proof of (29) First we prove that $-T_{n}^{(2)}(\varvec{\theta }_0)$ converges to $R(\varvec{\theta }_0)$ in probability.

$$\begin{aligned} T_{n}^{(2)}(\varvec{\theta }_0)&= \frac{1-q_n}{n+1}\sum _{i=1}^{n+1}[(n+1)D_i(\varvec{\theta }_0)]^{1-q_n}[h_{\varvec{\theta }_0}(\tilde{U}_i)h^T_{\varvec{\theta }_0} (\tilde{U}_i)-h_{\varvec{\theta }_0} (\xi _i)h^T_{\varvec{\theta }_0} (\xi _i)]\nonumber \\&+\frac{(1-q_n)}{n+1}\sum _{i=1}^{n+1}[(n+1)D_i(\varvec{\theta }_0)]^{1-q_n}h_{\varvec{\theta }_0} (\xi _i)h^T_{\varvec{\theta }_0} (\xi _i)\nonumber \\&+\frac{1}{n+1}\sum _{i=1}^{n+1}[(n+1)D_i(\varvec{\theta }_0)]^{1-q_n}h'_{\varvec{\theta }_0}(\xi _i)\nonumber \\&+\frac{1}{n+1}\sum _{i=1}^{n+1}[(n+1)D_i(\varvec{\theta }_0)]^{1-q_n}[h'_{\varvec{\theta }_0}(\tilde{U}_i)-h'_{\varvec{\theta }_0}(\xi _i)]\nonumber \\&\triangleq D_1+D_2+D_3+D_4. \end{aligned}$$

(35)

From the existence of the limiting distribution of the Kolmogorov-Smirnov statistic, we have $|\tilde{U}_i-\xi _i|\mathop {\longrightarrow }\limits ^{P}0$ as $n\rightarrow \infty$ uniformly in $i$. Ac $h'_{\varvec{\theta }_0}$ and $h_{\varvec{\theta }_0}h^T_{\varvec{\theta }_0}$ are continuous, then we can obtain that

$$\begin{aligned} h'_{\varvec{\theta }_0}(\tilde{U}_i)=h'_{\varvec{\theta }_0}(\xi _i)+o_p(1),~ h_{\varvec{\theta }_0}(\tilde{U}_i)h^T_{\varvec{\theta }_0}(\tilde{U}_i)=h_{\varvec{\theta }_0}(\xi _i)h^T_{\varvec{\theta }_0}(\xi _i)+o_p(1) \end{aligned}$$

(36)

where $o_p(1)$ is uniform in $i$. Meanwhile, from Lemma 5, it can be proved that

$$\begin{aligned} \frac{1}{n+1}\sum _{i=1}^{n+1}[(n+1)D_i(\varvec{\theta }_0)]^{1-q_n}\mathop {\longrightarrow }\limits ^{P}1<\infty . \end{aligned}$$

(37)

As a result, $D_1$ and $D_4$ both go to zero in probability. Then by Lemma 5,

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^n[nD_i(\varvec{\theta }_0)]^{1-q_n}h_{\varvec{\theta }_0} (\xi _i)h^T_{\varvec{\theta }_0} (\xi _i)&\mathop {\longrightarrow }\limits ^{P}&\int \limits _0^1h_{\varvec{\theta }_0}(u)h^T_{\varvec{\theta }_0}(u)du=K(\varvec{\theta }_0). \end{aligned}$$

(38)

According to Remark 6 and the fact $1-q_n={\mathcal {O}}(n^{-1/2}), D_2$ can be found to converge to zero in probability. Applying Lemma 5,

$$\begin{aligned} D_3&\mathop {\longrightarrow }\limits ^{P}&\int \limits _0^1h'_{\varvec{\theta }_0}(u)du=E_{\varvec{\theta }_0}\bigg [\bigg \{\frac{g_{01}(t,\varvec{\theta })}{g(t,\varvec{\theta })}\bigg \}'\bigg |_{\varvec{\theta }=\varvec{\theta }_0}\bigg ]=-R(\varvec{\theta }_0). \end{aligned}$$

(39)

Thus, $-T_{n}^{(2)}(\varvec{\theta }_0)$ has been proved to converge to $R(\varvec{\theta }_0)$ in probability. Next we will prove that $\frac{1}{2}(\widehat{\varvec{\theta }}_n-\varvec{\theta }_0)^TT_{n}^{(3)} \mathop {\longrightarrow }\limits ^{P}0$ as $n\rightarrow \infty$.

$$\begin{aligned} T_{n}^{(3)}(\varvec{\theta }^*)&= \frac{3(1-q_n)}{n+1}\sum _{i=1}^{n+1}[(n+1)D_i(\varvec{\theta }^*)]^{1-q_n}[\nabla _{\varvec{\theta }}\{h_{\varvec{\theta }^*}h^T_{\varvec{\theta }^*}\}(\tilde{U}_i) -\nabla _{\varvec{\theta }}\{h_{\varvec{\theta }^*}h^T_{\varvec{\theta }^*}\} (\xi _i)]\nonumber \\&+\frac{(1-q_n)^2}{n+1}\sum _{i=1}^{n+1}[(n+1)D_i(\varvec{\theta }^*)]^{1-q_n}h^3_{\varvec{\theta }^*} (\xi _i)\nonumber \\&+\frac{1}{n+1}\sum _{i=1}^{n+1}[(n+1)D_i(\varvec{\theta }^*)]^{1-q_n}\nabla _{\varvec{\theta }}^2h_{\varvec{\theta }^*} (\xi _i)\nonumber \\&+\frac{(1-q_n)^2}{n+1}\sum _{i=1}^{n+1}[(n+1)D_i(\varvec{\theta }^*)]^{1-q_n}[h^3_{\varvec{\theta }^*} (\tilde{U}_i)-h^3_{\varvec{\theta }^*} (\xi _i)]\nonumber \\&+\frac{3(1-q_n)}{n+1}\sum _{i=1}^{n+1}[(n+1)D_i(\varvec{\theta }^*)]^{1-q_n}\nabla _{\varvec{\theta }}\{h_{\varvec{\theta }^*}h^T_{\varvec{\theta }^*}\} (\xi _i)\nonumber \\&+\frac{1}{n+1}\sum _{i=1}^{n+1}[(n+1)D_i(\varvec{\theta }^*)]^{1-q_n}[\nabla _{\varvec{\theta }}^2h_{\varvec{\theta }^*} (\tilde{U}_i)-\nabla _{\varvec{\theta }}^2h_{\varvec{\theta }^*} (\xi _i)]\nonumber \\&\triangleq E_1+E_2+E_3+E_4+E_5+E_6. \end{aligned}$$

(40)

From Lemma 5 and the continuity of $h^3_{\varvec{\theta }^*}, \nabla _{\varvec{\theta }}\{h_{\varvec{\theta }^*}h^T_{\varvec{\theta }^*}\}$ and $\nabla _{\varvec{\theta }}^2h_{\varvec{\theta }^*}$, since $|\tilde{U}_i-\xi _i|\mathop {\longrightarrow }\limits ^{P}0$ as $n\rightarrow \infty$ uniformly in $i$, it can be proved that, $E_1, E_4$ and $E_6$ all converge to zero in probability. Meanwhile, similar to the argument before, $E_2$ and $E_5$ can be certified to go to zero in probability and $E_3$ can be certified to be bounded in probability under Assumption A3 and Lemma 5. Thus $T_{n}^{(3)}(\varvec{\theta }^*)$ is bounded in probability. Due to the consistency of $\widehat{\varvec{\theta }}_n$, it can be derived that, $\displaystyle \frac{1}{2}(\widehat{\varvec{\theta }}_n-\varvec{\theta }_0)^TT_{n}^{(3)}(\varvec{\theta }^*) \mathop {\longrightarrow }\limits ^{P}0$ as $n\rightarrow \infty$. Thus, the proposition (29) has been proved.

Substituting (28) and (29) in (27), by applying Slutsky’s lemma, we can obtain the desired result:

$$\begin{aligned} \sqrt{n}(\widehat{\varvec{\theta }}_n-\varvec{\theta }_0)\mathop {\longrightarrow }\limits ^{P}N(\varvec{0},I(\varvec{\theta }_0)), \end{aligned}$$

(41)

where $\varvec{I}(\varvec{\theta }_0)=R^{-1}(\varvec{\theta }_0)K(\varvec{\theta }_0)R^{-1}(\varvec{\theta }_0)$. As a result, the asymptotical normality for the modified MSP estimate has been proved. $\square$