An efficient methodology to estimate the parameters of a two-dimensional chirp signal model

Abstract

In various capacities of statistical signal processing two-dimensional (2-D) chirp models have been considered significantly, particularly in image processing—to model gray-scale and texture images, magnetic resonance imaging, optical imaging etc. In this paper we address the problem of estimation of the unknown parameters of a 2-D chirp model under the assumption that the errors are independently and identically distributed (i.i.d.). The key attribute of the proposed estimation procedure is that it is computationally more efficient than the least squares estimation method. Moreover, the proposed estimators are observed to have the same asymptotic properties as the least squares estimators, thus providing computational effectiveness without any compromise on the efficiency of the estimators. We extend the propounded estimation method to provide a sequential procedure to estimate the unknown parameters of a 2-D chirp model with multiple components and under the assumption of i.i.d. errors we study the large sample properties of these sequential estimators. Simulation studies and two synthetic data analyses have been performed to show the effectiveness of the proposed estimators.

Appendices

Appendix A

Henceforth, we will denote $\varvec{\theta }(n_0) = (A(n_0), B(n_0), \alpha , \beta )$ as the parameter vector and $\varvec{\theta }^0(n_0) = (A^0(n_0), B^0(n_0), \alpha ^0, \beta ^0)$ as the true parameter vector of the 1-D chirp model (9).

To prove Theorem 1, we need the following lemma:

Lemma 1

Consider the set $S_c = \{(\alpha , \beta ) : |\alpha - \alpha ^0| \geqslant c \text { or } |\beta - \beta ^0| \geqslant c\}$. If for any c > 0,

$$\begin{aligned} \liminf \inf \limits _{(\alpha , \beta ) \in S_c} \frac{1}{M N}\bigg [R^{(1)}_{MN}(\alpha , \beta ) - R^{(1)}_{MN}(\alpha ^0, \beta ^0)\bigg ] > 0 \ a.s. \end{aligned}$$

(21)

then, ${\hat{\alpha }}$ $\rightarrow$ $\alpha ^0$ and ${\hat{\beta }}$ $\rightarrow$ $\beta ^0$ almost surely as $M \rightarrow \infty$. Note that $\liminf$ stands for limit infimum and $\inf$ stands for the infimum.

Proof

This proof follows along the same lines as that of Lemma 1 of Wu (1981). $\square$

Proof of Theorem 1:

Let us consider the following:

$$\begin{aligned}&\liminf \inf \limits _{(\alpha , \beta ) \in S_c} \frac{1}{M N}\bigg [R^{(1)}_{MN}(\alpha , \beta ) - R^{(1)}_{MN}(\alpha ^0, \beta ^0)\bigg ]&\\&\quad = \liminf \inf \limits _{(\alpha , \beta ) \in S_c} \frac{1}{M N}\bigg [\sum _{{n_0} = 1}^{N}R_{M}(\alpha , \beta , n_0) - \sum _{{n_0} = 1}^{N} R_{M}(\alpha ^0, \beta ^0, n_0)\bigg ]&\\&\quad = \liminf \inf \limits _{(\alpha , \beta ) \in S_c} \frac{1}{M N}\bigg [\sum _{{n_0} = 1}^{N}Q_{M}({\hat{A}}(n_0), {\hat{B}}(n_0), \alpha , \beta ) - \sum _{{n_0} = 1}^{N} Q_{M}({\hat{A}}(n_0), {\hat{B}}(n_0), \alpha ^0, \beta ^0)\bigg ]&\\&\quad \geqslant \liminf \inf \limits _{(\alpha , \beta ) \in S_c} \frac{1}{M N}\bigg [\sum _{{n_0} = 1}^{N}Q_{M}({\hat{A}}(n_0), {\hat{B}}(n_0), \alpha , \beta ) - \sum _{{n_0} = 1}^{N} Q_{M}(A^0(n_0) , B^0(n_0), \alpha ^0, \beta ^0)\bigg ]&\\&\quad \geqslant \liminf \inf \limits _{ \varvec{\theta }(n_0) \in M_c^{n_0}} \frac{1}{M N}\bigg [\sum _{{n_0} = 1}^{N}Q_{M}(A(n_0), B(n_0), \alpha , \beta ) - \sum _{{n_0} = 1}^{N} Q_{M}(A^0(n_0) , B^0(n_0), \alpha ^0, \beta ^0)\bigg ]&\\&\quad \geqslant \frac{1}{N}\sum _{n_0 = 1}^{N} \liminf \inf \limits _{\varvec{\theta }(n_0) \in M_c^{n_0}} \frac{1}{M}\bigg [Q_M(\varvec{\theta }(n_0)) - Q_M(\varvec{\theta }^0(n_0))\bigg ] > 0.&\end{aligned}$$

This follows from the proof of Theorem 1 of Kundu and Nandi (2008). Here, $Q_{M}(A(n_0), B(n_0), \alpha , \beta ) = {{\varvec{Y}}}^{\top }_{n_0}({{\varvec{\textit{I}}}} - {{\varvec{Z}}}_M(\alpha , \beta )({{\varvec{Z}}}_M(\alpha , \beta )^{\top }{{\varvec{Z}}}_M(\alpha , \beta ))^{-1}{{\varvec{Z}}}_M(\alpha , \beta )^{\top }){{\varvec{Y}}}_{n_0}$. Also note that the set $M_c^{n_0}$ = $\{\varvec{\theta }(n_0) : |A(n_0) - A^0(n_0)| \geqslant c \text { or } |B(n_0) - B^0(n_0)| \geqslant c \text { or } |\alpha - \alpha ^0| \geqslant c \text { or } |\beta - \beta ^0| \geqslant c\}$ which implies $S_c \subset M_c^{n_0}$, for all $n_0 \in \{1, \ldots , N\}$. Thus, using Lemma 1, ${\hat{\alpha }} \xrightarrow {a.s.} \alpha ^0$ and ${\hat{\beta }} \xrightarrow {a.s.} \beta ^0$. $\square$

Proof of Theorem 3:

Let us denote $\varvec{\xi } = (\alpha , \beta )$ and $\hat{\varvec{\xi }} = ({\hat{\alpha }}, {\hat{\beta }})$, the estimator of $\varvec{\xi }^0 = (\alpha ^0, \beta ^0)$ obtained by minimising the function $R_{MN}^{(1)}(\varvec{\xi }) = R_{MN}^{(1)}(\alpha , \beta )$ defined in (10).

Using multivariate Taylor series, we expand the $1 \times 2$ gradient vector ${{\varvec{R}}}_{MN}^{(1)'}(\hat{\varvec{\xi }})$ of the function $R_{MN}^{(1)}(\varvec{\xi })$, around the point $\varvec{\xi }^0$ as follows:

$$\begin{aligned} {{\varvec{R}}}_{MN}^{(1)'}(\hat{\varvec{\xi }}) - {{\varvec{R}}}_{MN}^{(1)'}(\xi ^0) = (\hat{\varvec{\xi }} - \varvec{\xi }^0){{\varvec{R}}}_{MN}^{(1)''}(\bar{\varvec{\xi }}), \end{aligned}$$

where $\bar{\varvec{\xi }}$ is a point between $\hat{\varvec{\xi }}$ and $\varvec{\xi }^0$ and ${{\varvec{R}}}_{MN}^{(1)''}(\bar{\varvec{\xi }})$ is the $2 \times 2$ Hessian matrix of the function $R_{MN}^{(1)}(\varvec{\xi })$ at the point $\bar{\varvec{\xi }}$. Since $\hat{\varvec{\xi }}$ minimises the function $R_{MN}^{(1)}(\varvec{\xi })$, ${{\varvec{R}}}_{MN}^{(1)'}(\hat{\varvec{\xi }}) = 0$. Thus, we have

$$\begin{aligned} (\hat{\varvec{\xi }} - \varvec{\xi }^0) = - {{\varvec{R}}}_{MN}^{(1)'}(\varvec{\xi }^0)[{{\varvec{R}}}_{MN}^{(1)''}(\bar{\varvec{\xi }})]^{-1}. \end{aligned}$$

Multiplying both sides by the diagonal matrix ${{\varvec{D}}}_1^{-1} = \text {diag}(M^{\frac{-3}{2}}N^{\frac{-1}{2}}, M^{\frac{-5}{2}}N^{\frac{-1}{2}})$, we get:

$$\begin{aligned} (\hat{\varvec{\xi }} - \varvec{\xi }^0){{\varvec{D}}}_1^{-1} = - {{\varvec{R}}}_{MN}^{(1)'}(\varvec{\xi }^0){{\varvec{D}}}_1[{{\varvec{D}}}_1R_{MN}^{(1)''}(\bar{\varvec{\xi }}){{\varvec{D}}}_1]^{-1}. \end{aligned}$$

(22)

Consider the vector,

$$\begin{aligned} {{\varvec{R}}}_{MN}^{(1)'}(\varvec{\xi }^0){{\varvec{D}}}_1 = \begin{bmatrix} \frac{1}{M^{3/2}N^{1/2}}\frac{\partial R_{MN}^{(1)}(\varvec{\xi }^0)}{\partial \alpha },&\frac{1}{M^{5/2}N^{1/2}}\frac{\partial R_{MN}^{(1)}(\varvec{\xi }^0)}{\partial \beta } \end{bmatrix}. \end{aligned}$$

On computing the elements of this vector and using preliminary result (5) (see Sect. 2.1) and the definition of the function:

$$\begin{aligned} R_{MN}^{(1)}(\alpha , \beta ) = \sum \limits _{{n_0} = 1}^{N}R_M(\alpha , \beta , n_0) \end{aligned}$$

we obtain the following result:

$$\begin{aligned} -{{\varvec{R}}}_{MN}^{(1)'}(\varvec{\xi }^0){{\varvec{D}}}_1 \xrightarrow {d} \varvec{{\mathcal {N}}}_2(\mathbf{0 }, 2\sigma ^2\varvec{\Sigma }) \text { as } M \rightarrow \infty . \end{aligned}$$

(23)

Since $\hat{\varvec{\xi }} \xrightarrow {a.s.} \varvec{\xi }^0$, and as each element of the matrix ${{\varvec{R}}}_{MN}^{(1)''}(\varvec{\xi })$ is a continuous function of $\varvec{\xi }$, we have

$$\begin{aligned} \lim _{M \rightarrow \infty }{{\varvec{D}}}_1{{\varvec{R}}}_{MN}^{(1)''}(\bar{\varvec{\xi }}){{\varvec{D}}}_1 = \lim _{M \rightarrow \infty }{{\varvec{D}}}_1{{\varvec{R}}}_{MN}^{(1)''}(\varvec{\xi }^0){{\varvec{D}}}_1. \end{aligned}$$

Now using preliminary result (6) (see Sect. 2.1), it can be seen that:

$$\begin{aligned} \lim _{M \rightarrow \infty }{{\varvec{D}}}_1{{\varvec{R}}}_{MN}^{(1)''}(\varvec{\xi }^0){{\varvec{D}}}_1 \rightarrow \varvec{\Sigma }^{-1}. \end{aligned}$$

(24)

On combining (22), (23) and (24), we have the desired result. $\square$

Appendix B

To prove Theorem 5, we need the following lemmas:

Lemma 2

Consider the set $S_c^{1}$ = $\{(\alpha , \beta ) : |\alpha - \alpha _1^0| \geqslant c \text { or } |\beta - \beta _1^0| \geqslant c\}$.If for any c > 0,

$$\begin{aligned} \liminf \inf \limits _{(\alpha , \beta ) \in S_c^{1}} \frac{1}{M N}[R^{(1)}_{1,MN}(\alpha , \beta ) - R^{(1)}_{1,MN}(\alpha _1^0, \beta _1^0)] > 0 \ a.s. \end{aligned}$$

(25)

then, ${\hat{\alpha }}_1$ $\rightarrow$ $\alpha _1^0$ and ${\hat{\beta }}_1$ $\rightarrow$ $\beta _1^0$ almost surely as $M \rightarrow \infty$.

Proof

This proof follows along the same lines as proof of Lemma 1. $\square$

Lemma 3

If Assumptions 1, 3 and P4 are satisfied then:

$$\begin{aligned}\begin{aligned}&M({\hat{\alpha }}_1 - \alpha _1^0) \xrightarrow {a.s.} 0,\\&M^2({\hat{\beta }}_1 - \beta _1^0) \xrightarrow {a.s.} 0. \end{aligned} \end{aligned}$$

Proof

Let us denote ${{\varvec{R}}}_{1, MN}^{(1)'}(\varvec{\xi })$ as the $1 \times 2$ gradient vector and ${{\varvec{R}}}_{1, MN}^{(1)''}(\varvec{\xi })$ as the $2 \times 2$ Hessian matrix of the function $R_{1, MN}^{(1)}(\varvec{\xi })$. Using multivariate Taylor series expansion, we expand the function ${{\varvec{R}}}_{1, MN}^{(1)'}(\hat{\varvec{\xi }}_1)$ around the point $\varvec{\xi }_1^0$ as follows:

$$\begin{aligned} {{\varvec{R}}}_{1, MN}^{(1)'}(\hat{\varvec{\xi }}_1) - {{\varvec{R}}}_{1, MN}^{(1)'}(\varvec{\xi }_1^0) = (\hat{\varvec{\xi }}_1 - \varvec{\xi }_1^0){{\varvec{R}}}_{1, MN}^{(1)''}(\bar{\varvec{\xi }}_1) \end{aligned}$$

where $\bar{\varvec{\xi }}_1$ is a point between $\hat{\varvec{\xi }}_1$ and $\varvec{\xi }_1^0$. Note that ${{\varvec{R}}}_{1, MN}^{(1)'}(\hat{\varvec{\xi }}_1) = 0$. Thus, we have:

$$\begin{aligned} (\hat{\varvec{\xi }}_1 - \varvec{\xi }_1^0) = - {{\varvec{R}}}_{1, MN}^{(1)'}(\varvec{\xi }_1^0)[{{\varvec{R}}}_{1, MN}^{(1)''}(\bar{\varvec{\xi }}_1)]^{-1}. \end{aligned}$$

(26)

Multiplying both sides by $\frac{1}{\sqrt{MN}}{{\varvec{D}}}_1^{-1}$, we get:

$$\begin{aligned} (\hat{\varvec{\xi }}_1 - \varvec{\xi }_1^0)(\sqrt{MN}{{\varvec{D}}}_1)^{-1} = - \frac{1}{\sqrt{MN}}{{\varvec{R}}}_{1, MN}^{(1)'}(\varvec{\xi }_1^0){{\varvec{D}}}_1[{{\varvec{D}}}_1{{\varvec{R}}}_{1, MN}^{(1)''}(\bar{\varvec{\xi }}_1){{\varvec{D}}}_1]^{-1}. \end{aligned}$$

(27)

Since each of the elements of the matrix ${{\varvec{R}}}_{1, MN}^{(1)''}(\varvec{\xi })$ is a continuous function of $\varvec{\xi },$

$$\begin{aligned} \lim _{M \rightarrow \infty } {{\varvec{D}}}_1{{\varvec{R}}}_{1, MN}^{(1)''}(\bar{\varvec{\xi }}_1){{\varvec{D}}}_1 = \lim _{M \rightarrow \infty } {{\varvec{D}}}_1{{\varvec{R}}}_{1, MN}^{(1)''}(\varvec{\xi }_1^0){{\varvec{D}}}_1. \end{aligned}$$

By definition,

$$\begin{aligned} R_{1, MN}^{(1)}(\varvec{\xi }) = \sum _{{n_0} = 1}^{N} R_{1,M}(\varvec{\xi }, n_0). \end{aligned}$$

(28)

Using this and the preliminary result (13) and (15) (see Sect. 3.1), it can be seen that:

$$\begin{aligned}&- \frac{1}{\sqrt{MN}}{{\varvec{R}}}_{1, MN}^{(1)'}(\varvec{\xi }_1^0){{\varvec{D}}}_1 \xrightarrow {a.s.} 0 \text { as } M \rightarrow \infty . \end{aligned}$$

(29)

$$\begin{aligned}&{{\varvec{D}}}_1{{\varvec{R}}}_{1, MN}^{(1)''}(\varvec{\xi }_1^0){{\varvec{D}}}_1 \xrightarrow {a.s.} \varvec{\Sigma }_1^{-1} \text { as } M \rightarrow \infty . \end{aligned}$$

(30)

On combining (27), (29) and (30), we have the desired result. $\square$

Proof of Theorem 5:

Consider the left hand side of (25), that is,

$$\begin{aligned}&\liminf \inf \limits _{(\alpha , \beta ) \in S_c^{1}} \frac{1}{M N}\left[ R^{(1)}_{1,MN}(\alpha , \beta ) - R^{(1)}_{1,MN}(\alpha _1^0, \beta _1^0)\right]&\\&\quad = \liminf \inf \limits _{(\alpha , \beta ) \in S_c^{1}} \frac{1}{M N}\left[ \sum _{{n_0} = 1}^{N}Q_{1,M}({\hat{A}}_1(n_0), {\hat{B}}_1(n_0), \alpha , \beta ) - \sum _{{n_0} = 1}^{N} Q_{1,M}({\hat{A}}_1(n_0), {\hat{B}}_1(n_0), \alpha _1^0, \beta _1^0)\right]&\\&\quad \geqslant \liminf \inf \limits _{(\alpha , \beta ) \in S_c^{1}} \frac{1}{M N}\left[ \sum _{{n_0} = 1}^{N}Q_{1,M}({\hat{A}}_1(n_0), {\hat{B}}_1(n_0), \alpha , \beta ) - \sum _{{n_0} = 1}^{N} Q_{1,M}(A_1^0(n_0) , B_1^0(n_0), \alpha _1^0, \beta _1^0)\right]&\\&\quad \geqslant \liminf \inf \limits _{ \varvec{\theta }_1(n_0) \in M_c^{1,n_0}} \frac{1}{M N}\left[ \sum _{{n_0} = 1}^{N}Q_{1,M}(A_1(n_0), B_1(n_0), \alpha , \beta ) - \sum _{{n_0} = 1}^{N} Q_{1,M}(A_1^0(n_0) , B_1^0(n_0), \alpha _1^0, \beta _1^0)\right]&\\&\quad \geqslant \frac{1}{N}\sum _{n_0 = 1}^{N} \liminf \inf \limits _{\varvec{\theta }_1(n_0) \in M_c^{1,n_0}} \frac{1}{M}\bigg [Q_{1,M}(\varvec{\theta }_1(n_0)) - Q_{1,M}(\varvec{\theta }_1^0(n_0))\bigg ] > 0.&\end{aligned}$$

Here, $Q_{1,M}(A(n_0), B(n_0), \alpha , \beta ) = {{\varvec{Y}}}^{\top }_{n_0}({{\varvec{\textit{I}}}} - {{\varvec{Z}}}_M(\alpha , \beta )({{\varvec{Z}}}_M(\alpha , \beta )^{\top }{{\varvec{Z}}}_M(\alpha , \beta ))^{-1}{{\varvec{Z}}}_M(\alpha , \beta )^{\top }){{\varvec{Y}}}_{n_0}$ and $M_c^{1,n_0}$ can be obtained by replacing $\alpha ^0$ and $\beta ^0$ by $\alpha _1^0$ and $\beta _1^0$ respectively, in the set $M_c^{n_0}$ defined in Lemma 1. The last step follows from the proof of Theorem 2.4.1 of Lahiri et al. (2015). Thus, using Lemma 2, ${\hat{\alpha }}_1 \xrightarrow {a.s.} \alpha _1^0$ and ${\hat{\beta }}_1 \xrightarrow {a.s.} \beta _1^0$ as $M \rightarrow \infty$.

Following similar arguments, one can obtain the consistency of ${\hat{\gamma }}_1$ and ${\hat{\delta }}_1$ as $N \rightarrow \infty$. Also,

$$\begin{aligned} \begin{aligned}&N({\hat{\gamma }}_1 - \gamma _1^0) \xrightarrow {a.s.} 0,\\&N^2({\hat{\delta }}_1 - \delta _1^0) \xrightarrow {a.s.} 0. \end{aligned} \end{aligned}$$

The proof of the above equations follows along the same lines as the proof of Lemma 3. From Theorem 7, it follows that as min$\{M, N\} \rightarrow \infty$:

$$\begin{aligned} \begin{aligned}&({\hat{A}}_1 - A_1^0) \xrightarrow {a.s.} 0,\\&({\hat{B}}_1 - B_1^0) \xrightarrow {a.s.} 0. \end{aligned} \end{aligned}$$

Thus, we have the following relationship between the first component of model (1) and its estimate:

$$\begin{aligned} \begin{aligned}&{\hat{A}}_1 \cos ({\hat{\alpha }}_1 m + {\hat{\beta }}_1 m^2 + {\hat{\gamma }}_1 n + {\hat{\delta }}_1 n^2) + {\hat{B}}_1 \sin ({\hat{\alpha }}_1 m + {\hat{\beta }}_1 m^2 + {\hat{\gamma }}_1 n + {\hat{\delta }}_1 n^2) \\&\quad = A_1^0 \cos (\alpha _1^0 m + \beta _1^0 m^2 + \gamma _1^0 n + \delta _1^0 n^2) + B_1^0 \sin (\alpha _1^0 m + \beta _1^0 m^2 + \gamma _1^0 n + \delta _1^0 n^2) + o(1). \end{aligned} \end{aligned}$$

(31)

Here a function g is o(1), if $g \rightarrow 0$ almost surely as min$\{M, N\} \rightarrow \infty$.

Using (31) and following the same arguments as above for the consistency of ${\hat{\alpha }}_1$, ${\hat{\beta }}_1$, ${\hat{\gamma }}_1$ and ${\hat{\delta }}_1$, we can show that, ${\hat{\alpha }}_2$, ${\hat{\beta }}_2$, ${\hat{\gamma }}_2$ and ${\hat{\delta }}_2$ are strongly consistent estimators of $\alpha _2^0$, $\beta _2^0$, $\gamma _2^0$ and $\delta _2^0$ respectively. And the same can be extended for $k \leqslant p.$ Hence, the result. $\square$

Proof of Theorem 7:

We will consider the following two cases that will cover both the scenarios—underestimation as well as overestimation of the number of components:

• Case 1 When $k = 1$:

  $$\begin{aligned} \begin{aligned} \begin{bmatrix} {\hat{A}}_1 \\ {\hat{B}}_1 \end{bmatrix} = [{{\varvec{W}}}({\hat{\alpha }}_1, {\hat{\beta }}_1, {\hat{\gamma }}_1, {\hat{\delta }}_1)^{\top }{{\varvec{W}}}({\hat{\alpha }}_1, {\hat{\beta }}_1, {\hat{\gamma }}_1, {\hat{\delta }}_1)]^{-1}{{\varvec{W}}}({\hat{\alpha }}_1, {\hat{\beta }}_1, {\hat{\gamma }}_1, {\hat{\delta }}_1)^{\top } {{\varvec{Y}}}_{MN \times 1} \end{aligned} \end{aligned}$$

  (32)

  Using Lemma 1 of Lahiri et al. (2015), it can be seen that:

  $$\begin{aligned} \frac{1}{MN}[{{\varvec{W}}}({\hat{\alpha }}_1, {\hat{\beta }}_1, {\hat{\gamma }}_1, {\hat{\delta }}_1)^{\top }{{\varvec{W}}}({\hat{\alpha }}_1, {\hat{\beta }}_1, {\hat{\gamma }}_1, {\hat{\delta }}_1)] \rightarrow \frac{1}{2}{{\varvec{\textit{I}}}}_{2 \times 2} \text { as } \min \{M, N\} \rightarrow \infty . \end{aligned}$$

  Substituting this result in (32), we get:

  $$\begin{aligned}\begin{aligned} \begin{bmatrix} {\hat{A}}_1 \\ {\hat{B}}_1 \end{bmatrix}&= \frac{2}{MN}{{\varvec{W}}}({\hat{\alpha }}_1, {\hat{\beta }}_1, {\hat{\gamma }}_1, {\hat{\delta }}_1)^{\top } {{\varvec{Y}}}_{MN \times 1} + o(1)\\&= \begin{bmatrix} \frac{2}{MN}\sum \limits _{n=1}^{N}\sum \limits _{m=1}^{M}y(m,n)\cos ({\hat{\alpha }}_1 m + {\hat{\beta }}_1 m^2 + {\hat{\gamma }}_1 n + {\hat{\delta }}_1 n^2) + o(1) \\ \frac{2}{MN}\sum \limits _{n=1}^{N}\sum \limits _{m=1}^{M}y(m,n)\sin ({\hat{\alpha }}_1 m + {\hat{\beta }}_1 m^2 + {\hat{\gamma }}_1 n + {\hat{\delta }}_1 n^2) + o(1) \end{bmatrix}. \end{aligned} \end{aligned}$$

  Now consider the estimate ${\hat{A}}_1$. Using multivariate Taylor series, we expand the function $\cos ({\hat{\alpha }}_1 m + {\hat{\beta }}_1 m^2 + {\hat{\gamma }}_1 n + {\hat{\delta }}_1 n^2)$ around the point $(\alpha _1^0, \beta _1^0, \gamma _1^0, \delta _1^0)$ and we obtain:

  $$\begin{aligned} \begin{aligned}&{\hat{A}}_1 \\&\quad = \frac{2}{MN}\sum \limits _{n=1}^{N}\sum \limits _{m=1}^{M}y(m,n)\bigg \{\cos (\alpha _1^0 m + \beta _1^0 m^2 + \gamma _1^0 n + \delta _1^0 n^2) \\&\qquad - m ({\hat{\alpha }}_1 - \alpha _1^0)\sin (\alpha _1^0 m + \beta _1^0 m^2 + \gamma _1^0 n \\&\qquad + \delta _1^0 n^2) - m^2({\hat{\beta }}_1 - \beta _1^0)\sin (\alpha _1^0 m + \beta _1^0 m^2 + \gamma _1^0 n + \delta _1^0 n^2) \\&\qquad - n ({\hat{\gamma }}_1 - \gamma _1^0)\sin (\alpha _1^0 m + \beta _1^0 m^2 + \gamma _1^0 n \\&\qquad + \delta _1^0 n^2) - n^2({\hat{\delta }}_1 - \delta _1^0)\sin (\alpha _1^0 m + \beta _1^0 m^2 + \gamma _1^0 n + \delta _1^0 n^2) \bigg \} \\&\qquad \rightarrow 2 \times \frac{A_1^0}{2} = A_1^0 \text { almost surely as } \min \{M, N\} \rightarrow \infty , \end{aligned} \end{aligned}$$

  using (1) and Lemma 1 and Lemma 2 of Lahiri et al. (2015). Similarly, it can be shown that ${\hat{B}}_1 \rightarrow B_1^0$ almost surely as $\min \{M, N\} \rightarrow \infty$.

  For the second component linear parameter estimates, consider:

  $$\begin{aligned}\begin{aligned} \begin{bmatrix} {\hat{A}}_2 \\ {\hat{B}}_2 \end{bmatrix} = \begin{bmatrix} \frac{2}{MN}\sum \limits _{n=1}^{N}\sum \limits _{m=1}^{M}y_1(m,n)\cos ({\hat{\alpha }}_2 m + {\hat{\beta }}_2 m^2 + {\hat{\gamma }}_2 n + {\hat{\delta }}_2 n^2) + o(1)\\ \frac{2}{MN}\sum \limits _{n=1}^{N}\sum \limits _{m=1}^{M}y_1(m,n)\sin ({\hat{\alpha }}_2 m + {\hat{\beta }}_2 m^2 + {\hat{\gamma }}_2 n + {\hat{\delta }}_2 n^2) + o(1) \end{bmatrix}. \end{aligned} \end{aligned}$$

  Here, $y_1(m,n)$ is the data obtained at the second stage after eliminating the effect of the first component from the original data as defined in (18). Using the relationship (31) and following the same procedure as for the consistency of ${\hat{A}}_1$, it can be seen that:

  $$\begin{aligned} {\hat{A}}_2 \xrightarrow {a.s.} A_2^0 \quad \text {and} \quad {\hat{B}}_2 \xrightarrow {a.s.} B_2^0 \text { as } \min \{M, N\} \rightarrow \infty . \end{aligned}$$

  (33)

  It is evident that the result can be easily extended for any $2 \leqslant k \leqslant p$.

• Case 2 When $k = p+1$:

  $$\begin{aligned} \begin{aligned} \begin{bmatrix} {\hat{A}}_{p+1} \\ {\hat{B}}_{p+1} \end{bmatrix} = \begin{bmatrix} \frac{2}{MN}\sum \limits _{n=1}^{N}\sum \limits _{m=1}^{M}y_p(m,n)\cos ({\hat{\alpha }}_{p+1} m + {\hat{\beta }}_{p+1} m^2 + {\hat{\gamma }}_{p+1} n + {\hat{\delta }}_{p+1} n^2) + o(1)\\ \frac{2}{MN}\sum \limits _{n=1}^{N}\sum \limits _{m=1}^{M}y_p(m,n)\sin ({\hat{\alpha }}_{p+1} m + {\hat{\beta }}_{p+1} m^2 + {\hat{\gamma }}_{p+1} n + {\hat{\delta }}_{p+1} n^2) + o(1) \end{bmatrix}, \end{aligned} \end{aligned}$$

  (34)

  where

  $$\begin{aligned} \begin{aligned}&y_p(m,n)\\&\quad = y(m,n) - \sum \limits _{j=1}^{p}\bigg \{{\hat{A}}_j \cos ({\hat{\alpha }}_j m + {\hat{\beta }}_j m^2 + {\hat{\gamma }}_j n + {\hat{\delta }}_j n^2) \\&\qquad + {\hat{B}}_j \sin ({\hat{\alpha }}_j m + {\hat{\beta }}_j m^2 + {\hat{\gamma }}_j n + {\hat{\delta }}_j n^2) \bigg \} \\&\quad = X(m,n) + o(1), \text { using }(31) \text { and case 1 results.} \end{aligned} \end{aligned}$$

  From here, it is not difficult to see that (34) implies the following result:

  $$\begin{aligned} {\hat{A}}_{p+1} \xrightarrow {a.s.} 0 \quad \text {and} \quad {\hat{B}}_{p+1} \xrightarrow {a.s.} 0 \text { as } \min \{M, N\} \rightarrow \infty . \end{aligned}$$

  This is obtained using Lemma 2 of Lahiri et al. (2015). It is apparent that the result holds true for any $k > p.$

$\square$

Proof of Theorem 8:

Consider (26) and multiply both sides of the equation with the diagonal matrix, ${{\varvec{D}}}_1^{-1}$:

$$\begin{aligned} (\hat{\varvec{\xi }}_1 - \varvec{\xi }_1^0){{\varvec{D}}}_1^{-1} = - {{\varvec{R}}}_{1, MN}^{(1)'}(\varvec{\xi }_1^0){{\varvec{D}}}_1[{{\varvec{D}}}_1{{\varvec{R}}}_{1, MN}^{(1)''}(\bar{\varvec{\xi }}_1){{\varvec{D}}}_1]^{-1}. \end{aligned}$$

(35)

Computing the elements of the vector $- {{\varvec{R}}}_{1, MN}^{(1)'}(\varvec{\xi }_1^0){{\varvec{D}}}_1$ and using definition (28) and the preliminary result (14) (Sect. 3.1), we obtain the following result:

$$\begin{aligned} - {{\varvec{R}}}_{1, MN}^{(1)'}(\varvec{\xi }_1^0){{\varvec{D}}}_1 \xrightarrow {d} \varvec{{\mathcal {N}}}_2(\mathbf{0 }, 2\sigma ^2 \varvec{\Sigma }_1^{-1}) \text { as } M \rightarrow \infty . \end{aligned}$$

(36)

On combining (35), (36) and (30), we have:

$$\begin{aligned} (\hat{\varvec{\xi }}_1 - \varvec{\xi }_1^0){{\varvec{D}}}_1^{-1} \xrightarrow {d} \varvec{{\mathcal {N}}}_2(\mathbf{0 }, 2\sigma ^2 \varvec{\Sigma }_1) \end{aligned}$$

This result can be extended for $k = 2$ using the relation (31) and following the same argument as above. Similarly, we can continue to extend the result for any $k \leqslant p$. $\square$