A computationally efficient algorithm to estimate the parameters of a two-dimensional chirp model with the product term

Abstract

Chirp signal models and their generalizations have been used to model many natural and man-made phenomena in signal processing and time series literature. In recent times, several methods have been proposed for parameter estimation of these models. However, these methods are either statistically sub-optimal or computationally burdensome, especially for two dimensional chirp models. In this paper, we consider the problem of parameter estimation of two dimensional chirp models and propose a computationally efficient estimator and establish asymptotic theoretical properties of the proposed estimators. Moreover, the proposed estimators are observed to have the same rates of convergence as the least squares estimators. Further, the proposed estimators of chirp rate parameters are shown to be asymptotically optimal. Extensive and detailed numerical simulations are conducted, which support the theoretical results of the proposed estimators.

Appendices

Appendix A

Proof of Theorem 1:

Given the data matrix ${\varvec{Y}}$, we compute the LSEs of $\alpha ^0+n_0\mu ^0$ and $\beta ^0$ corresponding to $n_0^{th}$ column vector of ${\varvec{Y}}$. We denote the obtained estimators by ${\widehat{\alpha }}_{n_0}$ and ${\widehat{\beta }}_{n_0}$ to emphasize that these depend on $n_0$. Similarly, for fixed $m_0^{th}$ row of ${\varvec{Y}}$, we have denoted the LSEs of $\gamma ^0+m_0\mu ^0$ and $\delta ^0$ by ${\widehat{\gamma }}_{m_0}$ and ${\widehat{\delta }}_{m_0}$. Under assumptions that $X(m_0,n_0)$ are stationary (11) and (12), see Nandi and Kundu (2004), we have

$$\begin{aligned} {\widehat{\beta }}_{n_0}&= \beta ^0+o\bigg (\frac{1}{M^2}\bigg ), {\widehat{\alpha }}_{n_0}=\alpha ^0+n_0\mu ^0+o\bigg (\frac{1}{M}\bigg ),\end{aligned}$$

(15)

$$\begin{aligned} {\widehat{\gamma }}_{m_0}&= \gamma ^0+m_0\mu ^0+o\bigg (\frac{1}{N}\bigg ), {\widehat{\delta }}_{m_0}=\delta ^0+o\bigg (\frac{1}{N^2}\bigg ). \end{aligned}$$

(16)

The final estimator of $\beta ^0$ given by

$$\begin{aligned} {\widehat{\beta }}= \frac{\displaystyle \sum _{n_0=1}^{N}{\widehat{\beta }}_{n_0}}{N} \end{aligned}$$

is strongly consistent estimate of $\beta ^0$ which is observed by (15) and also that ${\widehat{\beta }}_{n_0}$ is strongly consistent for $\beta ^0$ as $M\xrightarrow {} \infty$. For proof, one may refer to Lahiri et al. (2015).

Similarly ${\widehat{\delta }}=\displaystyle \frac{1}{M}\sum _{m_0=1}^{M}{\widehat{\delta }}_{m_0}$ is strongly consistent estimator of $\delta ^0$.

Denote

We now prove the consistency of the frequency parameter estimators ${\widehat{\alpha }}, {\widehat{\gamma }}$, and that of ${\widehat{\mu }}$, estimator of the interaction term parameter. From (15) and (16), we have thefollowing:

$$\begin{aligned} \begin{bmatrix} {\widehat{\alpha }}\\ l{\widehat{\gamma }}\\ {\widehat{\mu }} \end{bmatrix} =\Big ({\varvec{\Gamma }}^\top {\varvec{\Gamma }}\Big )^{-1}{\varvec{\Gamma }}^\top {\varvec{\Lambda }}= ({\varvec{\Gamma }}^\top {\varvec{\Gamma }})^{-1}{\varvec{\Gamma }}^\top \bigg ({\varvec{\Gamma }} \begin{bmatrix} \alpha ^0\\ \gamma ^0\\ \mu ^0 \end{bmatrix}+{\varvec{\tau }}\bigg ), \end{aligned}$$

(17)

where ${\varvec{\Gamma }}^\top =\begin{bmatrix} 1&{}1&{}\cdots &{}1&{}0&{}0&{}\cdots &{}0\\ 0&{}0&{}\cdots &{}0&{}1&{}1&{}\cdots &{}1\\ 1&{}2&{}\cdots &{}N&{}1&{}2&{}\cdots &{}M \end{bmatrix}_{3\times (M+N)}$, and ${\varvec{\Lambda }}^\top =\begin{bmatrix} {\widehat{\alpha }}_1&{\widehat{\alpha }}_2&\cdots&{\widehat{\alpha }}_N&{\widehat{\gamma }}_1&{\widehat{\gamma }}_2&\cdots&{\widehat{\gamma }}_M \end{bmatrix}$.

This implies that

$$\begin{aligned}&\begin{bmatrix} {\widehat{\alpha }}\\ {\widehat{\gamma }}\\ {\widehat{\mu }} \end{bmatrix}= \begin{bmatrix} \alpha ^0\\ \gamma ^0\\ \mu ^0 \end{bmatrix}+({\varvec{\Gamma }}^\top {\varvec{\Gamma }})^{-1}\begin{bmatrix} N\times o\bigg (\frac{1}{M}\bigg )\\ M\times o\bigg (\frac{1}{N}\bigg )\\ \frac{N(N+1)}{2}\times o\bigg (\frac{1}{M}\bigg )+\frac{M(M+1)}{2}\times o\bigg (\frac{1}{N}\bigg ) \end{bmatrix}. \end{aligned}$$

(18)

Now we look at the first element $a_1+a_2-a_3$ of the following matrix

$$\begin{aligned}({\varvec{\Gamma }}^\top {\varvec{\Gamma }})^{-1}\begin{bmatrix} N\times o\bigg (\frac{1}{M}\bigg )\\ M\times o\bigg (\frac{1}{N}\bigg )\\ \frac{N(N+1)}{2}\times o\bigg (\frac{1}{M}\bigg )+\frac{M(M+1)}{2}\times o\bigg (\frac{1}{N}\bigg ) \end{bmatrix}.\end{aligned}$$

where,

$$\begin{aligned} a_1&=\bigg (MK-\frac{M^2(M+1)^2}{4}\bigg ) N\times o\bigg (\frac{1}{M}\bigg )/\** ,&\\ a_2&= \frac{MN(M+1)(N+1)}{4} M\times o\bigg (\frac{1}{N}\bigg )/\** ,&\\ a_3&= \frac{MN(N+1)}{2}\Bigg (\frac{N(N+1)}{2}\times o\bigg (\frac{1}{M}\bigg )+\frac{M(M+1)}{2}\times o\bigg (\frac{1}{N}\bigg )\Bigg )/\** ,&\end{aligned}$$

and

$$\begin{aligned} \** = \frac{MN}{12} \bigg (N(N^2-1)+M(M^2-1)\bigg ),K=\frac{N(N+1)(2N+1)}{6}+\frac{M(M+1)(2M+1)}{6}. \end{aligned}$$

Now we look at $a_1, a_2$ and $a_3$ individually and compute their limits.

$$\begin{aligned} a_1&=\bigg (MK-\frac{M^2(M+1)^2}{4}\bigg ) N\times o\bigg (\frac{1}{M}\bigg )/\**&\\~\\&= N\times \frac{o(1)}{\** }\times \bigg (\frac{N(N+1)(2N+1)}{6}+\frac{M(M+1)(M-1)}{12}\bigg )&\\~\\ {}&= \frac{o(1)}{M}\times \frac{\bigg (2N(N+1)(2N+1)+M(M^2-1)\bigg )}{\bigg (N(N^2-1)+M(M^2-1)\bigg )}.&\end{aligned}$$

$\hbox { This implies that } a_1\xrightarrow {a.s.} 0\hbox { as } \min \{M,N\}\rightarrow \infty .$ Here,

$$\begin{aligned} a_2&= \frac{MN(M+1)(N+1)}{4} M\times o\bigg (\frac{1}{N}\bigg )/\** = o(1)\times \frac{M(M+1)}{\bigg (N(N^2-1)+M(M^2-1)\bigg )}.&\end{aligned}$$

$\hbox { This implies that } a_2 \xrightarrow {a.s.} 0\hbox { as } \min \{M,N\}\rightarrow \infty .$

$$\begin{aligned} a_3&= \frac{MN(N+1)}{2}\Bigg (\frac{N(N+1)}{2}\times o\bigg (\frac{1}{M}\bigg )+\frac{M(M+1)}{2}\times o\bigg (\frac{1}{N}\bigg )\Bigg )/\**&\\~\\ {}&= \frac{(N+1)}{2\** }\Bigg (N^2(N+1)\times o(1)+M^2(M+1)\times o(1)\Bigg ).&\end{aligned}$$

$\hbox { This implies that } a_3 \xrightarrow {a.s.} 0\hbox { as } \min \{M,N\}\rightarrow \infty .$ Hence, ${\widehat{\alpha }}$ is strongly consistent estimate of $\alpha ^0$. Similarly strong consistency of ${\widehat{\gamma }}$ for $\gamma ^0$ can be derived.

Now, the third element of $({\varvec{\Gamma }}^\top {\varvec{\Gamma }})^{-1}\begin{bmatrix} N\times o\bigg (\frac{1}{M}\bigg )\\ M\times o\bigg (\frac{1}{N}\bigg )\\ \frac{N(N+1)}{2}\times o\bigg (\frac{1}{M}\bigg )+\frac{M(M+1)}{2}\times o\bigg (\frac{1}{N}\bigg ) \end{bmatrix}$ can be written as $-b_1-b_2+b_3$, where

$$\begin{aligned} b_1&=\frac{MN(N+1)}{2}\times N\times o\bigg (\frac{1}{M}\bigg )/\** theorem. $\square$

Appendix B

Proof of Theorem 2:

Suppose ${\varvec{\kappa }}^\top =(A,B,\alpha ,\beta )$ and

${\varvec{\kappa }}^{0^\top }= (A^0(n_0),B^0(n_0),\alpha ^0+n_0\mu ^0,\beta ^0)$. We define sum of squares as follows:

$$\begin{aligned}Q_{n_0}({\varvec{\kappa }})= \displaystyle \sum _{m_0=1}^M\bigg (y(m_0,n_0)-A\cos (\alpha m_0+\beta m_0^2)-B\sin (\alpha m_0+\beta m_0^2)\bigg )^2.\end{aligned}$$

Let $\widehat{{\varvec{\kappa }}}$ be the minimizer of $Q_{n_0}({\varvec{\kappa }})$, then using Taylor Series expansion on the first derivative vector $Q_{n_0}^{\prime }({\varvec{\kappa }})$ around the point ${\varvec{\kappa }}^{0}$, we get:

$$\begin{aligned} {\varvec{Q}}_{n_0}^{\prime }(\widehat{{\varvec{\kappa }}})-{\varvec{Q}}_{n_0}^{\prime }({\varvec{\kappa }}^0) = {\varvec{Q}}_{n_0}^{\prime \prime }(\breve{{\varvec{\kappa }}}) (\widehat{{\varvec{\kappa }}}-{\varvec{\kappa }}^0), \end{aligned}$$

(19)

where $\breve{{\varvec{\kappa }}}$ is a point between $\widehat{{\varvec{\kappa }}}$ and ${\varvec{\kappa }}^0$. Also ${\varvec{Q}}_{n_0}^{\prime }(\widehat{{\varvec{\kappa }}})=0$ as $\widehat{{\varvec{\kappa }}}$ is LSE of ${\varvec{\kappa }}^0$.

Let us denote ${\varvec{D}}_1^{-1}=diag(M^{1/2},M^{1/2},M^{3/2},M^{5/2})$. Then on multiplying ${\varvec{D}}_1^{-1}$ both sides of Eq. (19), it gives:

$$\begin{aligned}&-\big [{\varvec{D_1}}{\varvec{Q}}_{n_0}^{\prime \prime }(\breve{{\varvec{\kappa }}}){\varvec{D_1}}\big ]^{-1}{\varvec{\Sigma }}_{n_0} \displaystyle {\varvec{\Sigma }}_{n_0}^{-1}{\varvec{D_1}}{\varvec{Q}}_{n_0}^{\prime }({\varvec{\kappa }}^0), = {\varvec{D_1}}^{-1}(\widehat{{\varvec{\kappa }}}-{\varvec{\kappa }}^0)\nonumber \\&\quad \hbox {where } \lim \limits _{M\rightarrow \infty }\big [{\varvec{D_1}}{\varvec{Q}}_{n_0}^{\prime \prime }(\breve{{\varvec{\kappa }}}){\varvec{D_1}}\big ]^{-1}{\varvec{\Sigma }}_{n_0} = I_{4\times 4}, \end{aligned}$$

(20)

and

$$\begin{aligned} \displaystyle \displaystyle {\varvec{\Sigma }}_{n_0}^{-1}= \frac{2}{A^{0^2}+B^{0^2}} \begin{bmatrix} \frac{A^{0^2}(n_0)+9B^{0^{2}}(n_0)}{2}&{}-4A^0(n_0)B^0(n_0)&{}-18B^0(n_0)&{}15B^0(n_0)\\ -4A^0(n_0)B^0(n_0)&{}\frac{9A^{0^2}(n_0)+B^{0^{2}}(n_0)}{2}&{}18A^0(n_0)&{}-15A^0(n_0)\\ -18B^0(n_0)&{}18A^0(n_0)&{}96&{}-90\\ 15B^0(n_0)&{}-15A^0(n_0)&{}-90&{}90 \end{bmatrix}. \end{aligned}$$

The expression of ${\varvec{\Sigma }}_{n_0}^{-1}$ can be obtained by proof of Theorem 2 shown in Lahiri et al. (2015). By using Eq. (20) and above expression of ${\varvec{\Sigma }}_{n_0}^{-1}$, we get following asymptotically equivalent (a.e.) expression of the third element of vector ${\varvec{D_1}}^{-1}(\widehat{{\varvec{\kappa }}}-{\varvec{\kappa }}^0)$ as:

(21)

where, $\eta (m_0,n_0)=X(m_0,n_0)\big (A^0\sin \phi (m_0,n_0,{\varvec{\xi }}^0)-B^0\cos \phi (m_0,n_0,{\varvec{\xi }}^0)\big )$,

$\phi (m_0,n_0,{\varvec{\xi }}^0)= \alpha ^0m_0+\beta ^0m_0^2+\gamma ^0n_0+\delta ^0n_0^2+\mu ^0 m_0n_0$. We have used these notations for brevity. Similarly expressions of fourth, fifth and sixth element of ${\varvec{D_1}}^{-1}(\widehat{{\varvec{\kappa }}}-{\varvec{\kappa }}^0)$ can be written as in Eqs. (22), (23) and (24) respectively:

(22)

(23)

(24)

From Eqs. (22) and (24) above, we can see that estimators ${\widehat{\beta }}$ and ${\widehat{\delta }}$ of $\beta ^0$ and $\delta ^0$ are asymptotically equivalent to the LSEs as they have same asymptotic variances by applying central limit theorem for stationary linear processes, see Fuller (1996). So now, remaining is to show asymptotic properties of estimators $( {\widehat{\alpha }}, {\widehat{\gamma }}, {\widehat{\mu }})^\top$.

The expression of proposed estimators of $(\alpha ^0,\gamma ^0, \mu ^0)^\top$ obtained is:

$$\begin{aligned}{} & {} \begin{bmatrix} {\widehat{\alpha }}\\ {\widehat{\gamma }}\\ {\widehat{\mu }} \end{bmatrix} = \begin{bmatrix} \bigg (MK-\frac{M^2\big (M+1\big )^2}{4}\bigg )c_{\alpha }+ \frac{MN(M+1)(N+1)}{4}c_{\gamma }-\frac{MN(N+1)}{2}c_{\alpha \gamma }\\ ~\\ \frac{MN(M+1)(N+1)}{4}c_{\alpha }+\bigg (KN-\frac{N^2(N+1)^2}{4}\bigg )c_{\gamma }-\frac{NM(M+1)}{2}c_{\alpha \gamma }\\ ~\\ -\frac{MN(N+1)}{2}c_{\alpha }-\frac{NM(M+1)}{2}c_{\gamma }+MNc_{\alpha \gamma } \end{bmatrix}\nonumber \\{} & {} \quad /| {\varvec{\Gamma }}^\top {\varvec{\Gamma }}|,\nonumber \\{} & {} \quad c_{\alpha }=\displaystyle \sum _{n_0=1}^{N}{\widehat{\alpha }}_{n_0}, c_{\gamma }=\displaystyle \sum _{m_0=1}^M{\widehat{\gamma }}_{m_0},\nonumber \\{} & {} \quad c_{\alpha \gamma } = \displaystyle \sum _{n_0=1}^{N}n_0{\widehat{\alpha }}_{n_0}+\displaystyle \sum _{m_0=1}^Mm_0{\widehat{\gamma }}_{m_0},\nonumber \\{} & {} \quad K=\frac{N(N+1)(2N+1)}{6}+\frac{M(M+1)(2M+1)}{6}, \big |{\varvec{\Gamma }}^\top {\varvec{\Gamma }}\big |\nonumber \\{} & {} \quad = \frac{MN}{12} \bigg (N(N^2-1)+M(M^2-1)\bigg ). \end{aligned}$$

(25)

From (25), we get:

$$\begin{aligned} M^{3/2}N^{1/2}\big ({\widehat{\alpha }}-\alpha ^0\big )&= \bigg (\frac{2N(N+1)(2N+1)+M(M^2-1)}{N(N^2-1)+M(M^2-1)}\bigg )\\&\quad \; \frac{1}{\sqrt{N}}\displaystyle \sum _{n_0=1}^{N}M^{3/2}\big ({\widehat{\alpha }}_{n_0}-(\alpha ^0+n_0\mu ^0)\big )\\&\quad \;+ \bigg (\frac{3M^2(M+1)}{N(N^2-1)+M(M^2-1)}\bigg )\\&\quad \; \frac{1}{\sqrt{M}}\displaystyle \sum _{m_0=1}^MN^{3/2}\big ({\widehat{\gamma }}_{m_0}-(\gamma ^0+m_0\mu ^0)\big )\\&\quad \;- \bigg (\frac{6M^{3/2}N^{1/2}(N+1)}{N(N^2-1)+M(M^2-1)}\bigg )\bigg [\displaystyle \sum _{n_0=1}^{N}n_0\big ({\widehat{\alpha }}_{n_0}-(\alpha ^0+n_0\mu ^0)\big )\\ {}&\quad \;+\displaystyle \sum _{m_0=1}^Mm_0\big ({\widehat{\gamma }}_{m_0}-(\gamma ^0+m_0\mu ^0)\big )\bigg ]. \end{aligned}$$

For sufficiently large M and N, we have:

$$\begin{aligned} M^{3/2}N^{1/2}\big ({\widehat{\alpha }}-\alpha ^0\big )&=\bigg (\frac{4N^3+M^3}{N^3+M^3}\bigg ) \frac{1}{\sqrt{N}}\displaystyle \sum _{n_0=1}^{N}M^{3/2}\big ({\widehat{\alpha }}_{n_0}-(\alpha ^0+n_0\mu ^0)\big )\nonumber \\ {}&\quad \;+\bigg (\frac{3M^3}{N^3+M^3}\bigg )\frac{1}{\sqrt{M}}\displaystyle \sum _{m_0=1}^MN^{3/2}\big ({\widehat{\gamma }}_{m_0}-(\gamma ^0+m_0\mu ^0)\big )\nonumber \\ {}&\quad \;-\bigg (\frac{6N^3}{N^3+M^3}\bigg )\frac{1}{N^{3/2}}\displaystyle \sum _{n_0=1}^{N}n_0M^{3/2}\big ({\widehat{\alpha }}_{n_0}-(\alpha ^0+n_0\mu ^0)\big )\nonumber \\ {}&\quad \;- \bigg (\frac{6M^3}{N^3+M^3}\bigg )\frac{1}{M^{3/2}}\displaystyle \sum _{m_0=1}^Mm_0N^{3/2}\big ({\widehat{\gamma }}_{m_0}-(\gamma ^0+m_0\mu ^0)\big ). \end{aligned}$$

(26)

Asymptotic normality of the estimators for $M=N\rightarrow \infty$ with given rates of convergence follows by applying central limit theorem for stationary processes (Fuller, 1996).

We now present some important results which will be used to find the asymptotic variance-covariance matrix of the proposed estimators of non-linear parameters. Using Eqs. (21) and (23) in the paper, we have the following observations:

• 1. $Asy Var \bigg (\frac{1}{2\sqrt{N}}\displaystyle \sum _{n_0=1}^{N}M^{3/2}\big ({\widehat{\alpha }}_{n_0}-(\alpha ^0+n_0\mu ^0)\big )\bigg )=\frac{c\sigma ^2 96}{A^{0^2}+B^{0^2}}$,

• 2. $Asy Var \bigg (\frac{1}{2\sqrt{M}}\displaystyle \sum _{m_0=1}^MN^{3/2}\big ({\widehat{\gamma }}_{m_0}-(\gamma ^0+m_0\mu ^0)\big )\bigg )=\frac{c\sigma ^296}{A^{0^2}+B^{0^2}}$,

• 3. $Asy Var \bigg (\frac{1}{2N^{3/2}}\displaystyle \sum _{n_0=1}^{N}n_0M^{3/2}\big ({\widehat{\alpha }}_{n_0}-(\alpha ^0+n_0\mu ^0)\big )\bigg )=\frac{c\sigma ^232}{A^{0^2}+B^{0^2}}$,

• 4. $Asy Var \bigg (\frac{1}{2M^{3/2}}\displaystyle \sum _{m_0=1}^Mm_0N^{3/2}\big ({\widehat{\gamma }}_{m_0}-(\gamma ^0+m_0\mu ^0)\big )\bigg )=\frac{c\sigma ^232}{A^{0^2}+B^{0^2}}$,

• 5. $Asy Covar \bigg (\frac{1}{2\sqrt{N}}\displaystyle \sum _{n_0=1}^{N}M^{3/2}\big ({\widehat{\alpha }}_{n_0}-(\alpha ^0+n_0\mu ^0)\big ),\frac{1}{2\sqrt{M}}\displaystyle \sum _{m_0=1}^MN^{3/2}\big ({\widehat{\gamma }}_{m_0}-(\gamma ^0+m_0\mu ^0)\big )\bigg )=0$,

• 6. $Asy Covar \bigg (\frac{1}{2\sqrt{N}}\displaystyle \sum _{n_0=1}^{N}M^{3/2}\big ({\widehat{\alpha }}_{n_0}-(\alpha ^0+n_0\mu ^0)\big ),\frac{1}{2N^{3/2}}\displaystyle \sum _{n_0=1}^{N}n_0M^{3/2}\big ({\widehat{\alpha }}_{n_0}-(\alpha ^0+n_0\mu ^0)\big )\bigg )=\frac{c\sigma ^248}{A^{0^2}+B^{0^2}}$,

• 7. $Asy Covar \bigg (\frac{1}{2\sqrt{N}}\displaystyle \sum _{n_0=1}^{N}M^{3/2}\big ({\widehat{\alpha }}_{n_0}-(\alpha ^0+n_0\mu ^0)\big ),\frac{1}{2M^{3/2}}\displaystyle \sum _{m_0=1}^Mm_0N^{3/2}\big ({\widehat{\gamma }}_{m_0}-(\gamma ^0+m_0\mu ^0)\big )\bigg )=0$,

• 8. $Asy Covar \bigg (\frac{1}{2\sqrt{M}}\displaystyle \sum _{m_0=1}^MN^{3/2}\big ({\widehat{\gamma }}_{m_0}-(\gamma ^0+m_0\mu ^0)\big ),\frac{1}{2N^{3/2}}\displaystyle \sum _{n_0=1}^{N}n_0M^{3/2}\big ({\widehat{\alpha }}_{n_0}-(\alpha ^0+n_0\mu ^0)\big )\bigg )=0$,

• 9. $Asy Covar \bigg (\frac{1}{2\sqrt{M}}\displaystyle \sum _{m_0=1}^MN^{3/2}\big ({\widehat{\gamma }}_{m_0}-(\gamma ^0+m_0\mu ^0)\big ),\frac{1}{2M^{3/2}}\displaystyle \sum _{m_0=1}^Mm_0N^{3/2}\big ({\widehat{\gamma }}_{m_0}-(\gamma ^0+m_0\mu ^0)\big )\bigg )=\frac{c\sigma ^248}{A^{0^2}+B^{0^2}}$,

• 10. $Asy Covar \bigg (\frac{1}{2N^{3/2}}\displaystyle \sum _{n_0=1}^{N}n_0M^{3/2}\big ({\widehat{\alpha }}_{n_0}-(\alpha ^0+n_0\mu ^0)\big ),\frac{1}{2M^{3/2}}\displaystyle \sum _{m_0=1}^Mm_0N^{3/2}\big ({\widehat{\gamma }}_{m_0}-(\gamma ^0+m_0\mu ^0)\big )\bigg )=\frac{c\sigma ^2}{2(A^{0^2}+B^{0^2})}$.

where $c=\displaystyle {\sum _{i=-\infty }^{\infty }\sum _{j=-\infty }^{\infty }}a^2(i,j)$.

Using the above results in Eq. (26) from the paper, we get asymptotic variance-covariance matrix of

$\begin{bmatrix} M^{3/2}N^{1/2}({\widehat{\alpha }}-\alpha ^0)\\ N^{3/2}M^{1/2}({\widehat{\gamma }}-\gamma ^0)\\ M^{3/2}N^{3/2}({\widehat{\mu }}-\mu ^0) \end{bmatrix}$ as: $\frac{c\sigma ^2}{(A^{0^2}+B^{0^2})}\begin{bmatrix} 996&{}612&{}-1224\\ 612 &{}996&{}-1224\\ -1224 &{}-1224&{}2448 \end{bmatrix}.$

From (21), (22) and (23) equations of the paper, it is further observed that:

• 1. $AsyCovar\bigg ( M^{5/2}N^{1/2}\big ({\widehat{\beta }}-\beta ^0\big ), M^{1/2}N^{5/2}\big ({\widehat{\delta }}-\delta ^0\big )\bigg )=0$,

• 2. $AsyCovar\bigg ( M^{5/2}N^{1/2}\big ({\widehat{\beta }}-\beta ^0\big ),\frac{1}{\sqrt{N}}\displaystyle \sum _{n_0=1}^{N}M^{3/2}\big ({\widehat{\alpha }}_{n_0}-(\alpha ^0+n_0\mu ^0)\big )\bigg )=\frac{-360c\sigma ^2}{A^{0^2}+B^{0^2}}$,

• 3. $AsyCovar\bigg ( M^{5/2}N^{1/2}\big ({\widehat{\beta }}-\beta ^0\big ),\frac{1}{\sqrt{M}}\displaystyle \sum _{m_0=1}^MN^{3/2}\big ({\widehat{\gamma }}_{m_0}-(\gamma ^0+m_0\mu ^0)\big )\bigg )=0$,

• 4. $AsyCovar\bigg ( M^{5/2}N^{1/2}\big ({\widehat{\beta }}-\beta ^0\big ),\frac{1}{N\sqrt{N}}\displaystyle \sum _{n_0=1}^{N}M^{3/2}n_0\big ({\widehat{\alpha }}_{n_0}-(\alpha ^0+n_0\mu ^0)\big )\bigg )=\frac{-180c\sigma ^2}{A^{0^2}+B^{0^2}}$,

• 5. $AsyCovar\bigg ( M^{5/2}N^{1/2}\big ({\widehat{\beta }}-\beta ^0\big ),\frac{1}{M\sqrt{M}}\displaystyle \sum _{m_0=1}^Mm_0N^{3/2}\big ({\widehat{\gamma }}_{m_0}-(\gamma ^0+m_0\mu ^0)\big )\bigg )=0.$

Similar results can be derived for $M^{1/2}N^{5/2}\big ({\widehat{\delta }}-\delta ^0\big )$.

Asymptotic variance-covariance matrix of the proposed estimators of non-linear parameters is given by:

$$\begin{aligned} \frac{c\sigma ^2}{(A^{0^2}+B^{0^2})}\begin{bmatrix} 996&{}-360&{}612&{}0&{}-1224\\ -360&{}360&{}0&{}0&{}0\\ 612&{}0&{}996&{}-360&{}-1224\\ 0&{}0&{}-360&{}360&{}0\\ -1224&{}0&{}-1224&{}0&{}2448 \end{bmatrix}. \end{aligned}$$

(27)

Next, we derive the asymptotics of amplitude estimators, please recall that by using Taylor series expansion of $\cos \phi (m_0,n_0,\widehat{{\varvec{\xi }}})$ around the point ${\varvec{\xi }}^0$, we can write:

$$\begin{aligned}\cos {\widehat{\phi }}-\cos \phi ^0=-\sin \breve{\phi }\big (\widehat{{\varvec{\xi }}}-{\varvec{\xi }}^0\big )^\top \begin{bmatrix} m_0\\ m_0^2\\ n_0\\ n_0^2\\ m_0n_0 \end{bmatrix}.\end{aligned}$$

For brevity, we have denoted $\cos \phi (m_0,n_0,\widehat{{\varvec{\xi }}})$ by $\cos {\widehat{\phi }}$, $\cos \phi (m_0,n_0,\widehat{{\varvec{\xi }}})$ by $\cos \phi ^0$, and $\sin \phi (m_0,n_0,\breve{{\varvec{\xi }}})$ by $\sin \breve{\phi }$, where $\breve{{\varvec{\xi }}}$ is a point lying between $\widehat{{\varvec{\xi }}}$ and ${\varvec{\xi }}^0$.

Now consider first element of the following vector,

$$\begin{aligned} \sqrt{MN}\Bigg (\begin{bmatrix} \frac{2}{{MN}}\displaystyle \sum _{m_0=1}^M\sum _{n_0=1}^{N}y(m_0,n_0)\cos {\widehat{\phi }} \\ \frac{2}{{MN}}\displaystyle \sum _{m_0=1}^M\sum _{n_0=1}^{N}y(m_0,n_0)\sin {\widehat{\phi }} \end{bmatrix}-\begin{bmatrix} A^0\\ B^0 \end{bmatrix}\Bigg ), \end{aligned}$$

we get:

$$\begin{aligned} \sqrt{MN}\bigg ( \frac{2}{{MN}}\displaystyle \sum _{m_0=1}^M\sum _{n_0=1}^{N}y(m_0,n_0)\cos {\phi ^0}-\frac{2}{{MN}}\displaystyle \sum _{m_0=1}^M\sum _{n_0=1}^{N}y(m_0,n_0)\sin {\breve{\phi }}R(m_0,n_0)-A^0\bigg ), \end{aligned}$$

(28)

where $R(m_0,n_0)=\big (\widehat{{\varvec{\xi }}}-{\varvec{\xi }}^0\big )^\top \begin{bmatrix} m_0\\ m_0^2\\ n_0\\ n_0^2\\ m_0n_0 \end{bmatrix}$, and second element of the above amplitude vector can be written as:

$$\begin{aligned} \sqrt{MN}\bigg ( \frac{2}{{MN}}\displaystyle \sum _{m_0=1}^M\sum _{n_0=1}^{N}y(m_0,n_0)\sin {\phi ^0}+\frac{2}{{MN}}\displaystyle \sum _{m_0=1}^M\sum _{n_0=1}^{N}y(m_0,n_0)\sin {\breve{\phi }}R(m_0,n_0)-B^0\bigg ). \end{aligned}$$

(29)

Now let us look at the first and the last term of Eq. (28) and putting value of $y(m_0,n_0)$ from the model (1),

(30)

The above result has been obtained from a famous number theory conjecture by Montgomery (1994).

In second term of (28), $R(m_0,n_0)$ is a sum of five terms, so now consider first term of

$\frac{2}{{MN}}\displaystyle \sum _{m_0=1}^M\sum _{n_0=1}^{N}y(m_0,n_0)\sin {\breve{\phi }}R(m_0,n_0)$,

So, finding the asymptotic distribution of (28) boils down to finding asymptotic distribution of

$$\begin{aligned} \frac{2}{\sqrt{MN}}\displaystyle \sum _{m_0=1}^M\sum _{n_0=1}^{N}X(m_0,n_0)\cos \phi ^0-\bigg (\frac{2}{{\sqrt{MN}}}\displaystyle \sum _{m_0=1}^M\sum _{n_0=1}^{N}B^0\sin \phi ^0\sin {\breve{\phi }}\bigg )R(m_0,n_0), \end{aligned}$$

which is further asymptotically equivalent to:

$$\begin{aligned}&\frac{2}{\sqrt{MN}}\displaystyle \sum _{m_0=1}^M\sum _{n_0=1}^{N}X(m_0,n_0)\cos \phi ^0-B^0\bigg (\frac{1}{2}M^{3/2}N^{1/2} ({\widehat{\alpha }}-\alpha ^0)+\frac{1}{3}M^{5/2}N^{1/2}({\widehat{\beta }}-\beta ^0)\nonumber \\&\quad +\frac{1}{2}N^{3/2}M^{1/2}({\widehat{\gamma }}-\gamma ^0) +\frac{1}{3}N^{5/2}M^{1/2}({\widehat{\delta }}-\delta ^0)+\frac{1}{4}M^{3/2}N^{3/2}({\widehat{\mu }}-\mu ^0)\bigg ). \end{aligned}$$

(31)

Asymptotic normality of amplitude estimators is thus proved by Eq. (31). Now we need to derive the expression of their asymptotic variances.

After lengthy calculations, we get the asymptotic variance of ${\widehat{A}}$ as follows:

$$\begin{aligned} \frac{c\sigma ^2}{(A^{0^2}+B^{0^2})}(2A^{0^2}+187B^{0^2}). \end{aligned}$$

Similarly by calculating other terms too, we get complete variance co-variance matrix ${\varvec{\Sigma }}$ as mentioned in Theorem 2.

Hence the result. $\square$