If $P\left\{\lVert Q_{\boldsymbol{\omega}}\rVert>\eta\right\}$ falls sufficiently fast for $\eta\rightarrow\infty$ (for example with $e^{-\eta}$), then using an appropriate $\eta$, an arbitrarily good agreement with the value $\bar{\boldsymbol{\omega}}$ with $\E\left\{Q_{\boldsymbol{\omega}}\right\}(\boldsymbol{\omega})=\mathbf{0}$ can be achieved. With the bounding through $\eta$, the expected value of the estimate is steered. If the correction vector is a linear function of $\boldsymbol{\omega}$, it is shown in [27] that such a saturation for an appropriate $\eta$ provides a robust estimator.

To 4:

As in 1, 2, it is questionable whether $\E\left\{\mathbf{G}\right\}$ exists. For a constant input, the differential equation (5.1-5) does not have a global solution, because the right side of the differential equation is not continuous. For an arbitrary starting point, a local solution that can be continued until a point $\bar{\boldsymbol{\omega}}$ with $\grad Q(\boldsymbol{\omega})=\mathbf{0}$ exists. A convergence with probability $1$ cannot necessarily be expected. However, there can be a convergence in distribution (weak convergence) of the $\boldsymbol{\omega}_n$.

These four methods are now compared for the particularly critical Signal Model 4, whereby $\mu$ is chosen according to (5.1-10). The separation of the targets is $0.45BW$ with ELAN 11 L, the average SNR on an individual array element is $12\,\text{dB}$.  The convergence for the original algorithm and Versions $1$ and $2$ is shown in Fig. <ref>. With Version 1, dampening of the fluctuations in the first two iteration steps leads to better convergence. With Version 2, the probability for a large correction near the minimum is still high. The dampening of the $Q$-function for deterministic inputs through the $\ln$ or $\arctan$ functions thus does not transfer to stochastic inputs. The same cases as Fig. <ref> are shown in <ref>,  but all correction vectors have a saturation bound $\eta$ like Version 3 ($\eta=0.8\text{\ae}$, $\text{\ae}=\frac{1}{3}\sum_{i=1}^3\left\lVert\grad Q(\boldsymbol{\omega}_0,\mathbf{z}_i)\right\rVert$ as in (5.1-10)). The introduction of such a saturation is the decisive method of dampening the fluctuations.

Using Version $3$ with the saturation bound $\eta$ in Fig. <ref> shows a compromise exists between dampening fluctuations and estimator bias with a particular number of iterations exists (Array and target configurations are the same as in Figs. <ref> and <ref>).

The behavior given a poor initial estimate when using Versions $3$ and $1$ with a poor initial estimate and bounding $(\eta=0.8\text{\ae})$ and also of Version $4$ is shown in Fig. <ref>. With a good starting value, Version $4$ has a higher residual noise than the other versions, but with a bad initial estimate, it converges particularly fast to a bias-free process. This is also shown in Fig. <ref>. There, the estimate after $17$ iterations (residual noise) is shown as a function of the initial direction estimate $\mathbf{u}_0=\left(\mathbf{u}_g-0.9BW,\mathbf{u}_g+0.9BW\right)$.

Given target positions $\mathbf{u}^{\text{ex}}=(-0.45BW,0.45BW)$, both targets are also in the search region with $u_g=-0.45BW$. The curves (a) and (b) show that for Algorithm 1 (with bounding $\eta=0.8\text{\ae}$) the convergence with a poor initialization can be improved using a larger value of $\delta$ in (<ref>). The convergence to the minimum takes place very cautiously, in time increments $t_n=\sum_{k=1}^n\frac{1}{k}$, where $n$ is the number of iterations [6].

§.§ Application to Planar Antenna Arrays

Though in the mathematical formulation, no difference needs to be made between linear and planar antenna arrays, significant differences become evident through simulation.

§.§.§ Limiting the Search Area

The necessary restriction with planar arrays to limit the search areas to a symmetric region with $\epsilon$-spacing from surface $u_i=u_k$, proves to be superfluous in practice for planar arrays because the case $u_i\approx u_k$ and $v_i\approx v_k$ $(i\neq k)$ so rarely occurs. (In $\mathbb{R}^M$ one has $\dim\left\{\left.\mathbf{u}\in\mathbb{R}^M\right|u_i=u_k,i\neq k\right\}\leq M-1$, but in $\mathbb{R}^{2M}$ it is $\dim\left\{\left.(\mathbf{u},\mathbf{v})\right|u_i=u_k\wedge v_i=v_k,i\neq k\right\}\leq2M-2!$). This simplifies the cumbersome determination of the defined region of iteration, because by simultaneous azimuth and elevation estimation, there is no natural numbering of the targets.

§.§.§ The Choice of the Initial Estimate and Poor Convergence

Convergence of the stochastic approximation with two targets.

Convergence of the stochastic approximation with two targets.

The starting value $\boldsymbol{\omega}_0$ should generally be chosen such that $\lVert\boldsymbol{\omega}_0-\boldsymbol{\omega}_{\text{ex}}\rVert$ is small and the starting directions are as far apart as possible. With two targets, prior knowledge about the orientation in the $u-v$-plane may accelerate the convergence. The poor convergence for Signal Model 1 in Fig. <ref> with $\Delta\varphi=180^\circ$ arises with other phase relations with additional dependence on the directions. There are more instances in which the strong $M$-regularity of the ordering is poorly fulfilled. For example if $\Delta\varphi=0^\circ$, the order $\boldsymbol{\omega}_1=(-u,u,0,0)$ can be barely distinguished from the ordering $\boldsymbol{\omega}_1=(0,0,-v,v)$. That means that $\left\lVert\grad |\E\left\{Q\right\}\right\rVert(\boldsymbol{\omega}_2)$ is small for $\lVert\grad \E\left\{Q\right\}\rVert(\boldsymbol{\omega}_1)=0$. This is shown in Fig. <ref> in the $(u-v)$ plane for ELAN 6. The starting value of the iteration is $\boldsymbol{\omega}_2=\left(0,0,-0.9\frac{BW}{2},0.9\frac{BW}{2}\right)$, the actual target position is $\boldsymbol{\omega}_1=\left(-\frac{BW}{4},\frac{BW}{4},0,0\right)$ with $\Delta\varphi=0^\circ$ and SNR=$12\,\text{dB}$. Although in the region of $\boldsymbol{\omega}_2$, the value $\left\lVert\grad\E Q\right\rVert\neq0$, the slope is so small that the iteration does not move.  The same configuration is shown in Fig. <ref>, but for Signal Model 2 with $\Delta\varphi\sim\mathcal{N}\left(0,\left(\frac{2\pi}{8}\right)^2\right)$. This supports the demand for higher bandwidths and more frequent revisit periods.

§.§.§ Attenuating the Fluctuations of $\grad Q$

Versions $3$ and $4$ in Chapter 5.2 (limiting $\lVert\grad Q\rVert$) are also suitable iteration procedures with planar arrays. Versions $1$ and $2$ (normalization $\frac{\grad Q}{Q}$ and $\frac{\grad Q}{1+Q^2}$), however, behave differently. The convergence for the $u$-direction for ELAN 6 is shown in Fig. <ref>. Version $1$ provides an estimate that is uniformly jumpy, as Version $2$ was in the linear case (See Fig. <ref>).

With regard to Version $2$, the choice of $\mu$ is so small that no convergence can be seen after $30$ iterations. By bounding the magnitude of the correction vector as in Version 3, one can again attain convergence. This is shown in Figure <ref> ($\eta=0.8\text{\ae}$).

In any case, a limitation of the correction vector should be applied. Whether Version $1$, $2$, or $4$ should be used depends on the desired properties of the estimate. These three procedures provide an unbiased estimate after a small number of iterations.

Convergence of the stochastic approximation with three targets.

An example of the estimation of three targets is shown in Fig. <ref> for Algorithm $3$ with $\eta=0.8\text{\ae}$. Initial values of the iteration are

\begin{align*}
\omega_1^{(0)}=&(-0.6 BW/2,-0.6BW/2)\\
\omega_2^{(0)}=&(0,0)\\
\omega_3^{(0)}=&(0.6BW/2,0.6BW/2)
\end{align*}

The average SNR at an individual element is $16.8\,\text{dB}$. One recognizes that the algorithm tries a different numbering of targets with different Monte Carlo runs.

§.§ The Influence of Perturbations

For simplicity, in the following subsection, only the original algorithm with bounding $(\eta=0.8\text{\ae})$ is analyzed.

§.§.§ Coupling and Subarrays

The influence of the coupling can be interpreted as a change in the characteristic responses of the individual elements. According to (4.1-8), if all of the individual elements have the same characteristic response, then this has no effect on the form of the $Q$ function. Thus, coupling has no influence as long as its effect on every element is the same, for example, by a large array on a regular grid. Additionally, preprocessing of the element outputs by summing them to form subarrays does not change the $Q$ function as long as the subarrays all have the same response (As new element coordinates $\mathbf{x},\mathbf{y}$, one must use the phase centers of the subarrays). With different element characteristic responses, if the number of elements is sufficiently large the distortion of the $Q$ function and thus the worsening of the estimate is not very large, because the antenna response differences are mitigated through averaging. This is shown in Fig. <ref> for the antenna ELAN 192 with the subarray ordering that is used in the ELRA system of the FFM with $24$ subarrays of $8$ elements. The convergence for Signal Model 4 is shown. The position of the targets and $\mathbf{u}=\left(-0.3\frac{BW}{2},0.3\frac{BW}{2}\right)$, $\mathbf{v}=\left(-0.3\frac{BW}{2},0.3\frac{BW}{2}\right)$; the starting value for the iteration is $\mathbf{u}_0=(0,0)$, $\mathbf{v}_0=\left(-0.9\frac{BW}{2},0.9\frac{BW}{2}\right)$. The influence of coupling on ELAN 39 is shown in Fig. <ref>. The coupling is simulated using the coupling matrix $\mathbf{C}$, (signal $\mathbf{C}\mathbf{s}$ instead of $\mathbf{s}$) that was generated using a program developed at the FFM for the V-shaped dipoles in the ELRA array (in [28] this program is used). In such a model, each antenna element is represented using the equivalent circuit

so that $u_e=\frac{z_i}{z_s+z_i}u_a$ and $\mathbf{C}=z_i\left(\mathbf{Z}_s+z_i\mathbf{I}\right)^{-1}$.

In both instances, the procedure appears to not be sensitive to deviations from isotropic element responses as long as the number of elements used is sufficiently large.

§.§.§ Extended Targets and Bandwidth

According to Signal Model 4, extended targets can be approximately simulated using normal numbers with an appropriate covariance matrix according to (2.2-2) and (2.2-3). The bandwidth of the receiver can be similarly simulated. If one considers the target to be a pure noise source, then an antenna element receives a signal with frequency $\bar{f}$ and bandwidth $2B$ with the following frequency mixture that is delayed with respect to the element at $(0,0)$:

\begin{equation*}
\end{equation*}

The frequency-component weightings $b(f)$ are independent for $f\neq f'$ and are zero on average (for example, with uniformly distributed initial phases). At element $(0,0)$, one has the power density $\wp^2$. Then $b$ can be formulated as a stochastic process with $\E\left\{b(f)\right\}=0$ and $\E\left\{b(f)b^*(f')\right\}=\frac{\wp^2}{2B}\delta(f-f')$ so that

\begin{align*}
\E\left\{E{z_{x,y}  z*_{x',y'}}\right\}=&\frac{\wp^2}{2B}\int_{\bar{f}-B}^{\bar{f}+B}e^{-j\frac{2\pi}{c}f\left((x-x')u+(y-y')v\right)}df\\
\end{align*}

One obtains the same expressions for the correlation of a bandwidth-limited impulse $s(t)=\frac{\sin2\pi Bt}{2\pi Bt}e^{j2\pi ft}$.

Thus, the bandwidth of the receiver with noise sources according to (2.2-2) functions like an extended quadratic target with the direction-dependent bandwidth $2s=\frac{2Bu}{\bar{f}}$. For small $s$, the covariance matrix of $\mathbf{z}$ only changes a little so that $\E\{Q\}=\trace \boldsymbol{\Gamma}\mathbf{R}$ is also only slightly perturbed. With ELAN 7 L, for example, with two extended targets of width $s=0.2BW$, there was no noticeable difference in the plot of $\E\{Q\}$ corresponding to Fig. <ref>; the minimum of $\E\{Q\}$ was displaced by less than $0.05BW$. Simulations show a larger difference with planar arrays. In Fig. <ref>, curve (a) shows this in comparison to Curve (a) in Fig. <ref> for ELAN 39. In Fig. <ref> (a), two extended targets with $r=0.2BW$ are assumed.

In general,  these effects can be neglected for targets extending only a few percent of the beamwidth and less than $1\%$ of the bandwidth.

§.§.§ Correlated Noise (External Interference) and Amplifier Fluctuations

If the noise vector $\mathbf{n}$ has the properties

\begin{align*}
\E\left\{\mathbf{n}\mathbf{n}^*\right\}=\mathbf{W}\neq \mathbf{I}\quad\text{but}\quad\E\left\{\mathbf{n}\right\}=\mathbf{0},\quad\E\left\{\mathbf{n}^*\mathbf{s}\right\}=0,
\end{align*}

then one gets a displacement of the expected value of the estimate so that

\begin{equation*}
\E\left\{Q\right\}=\trace\boldsymbol{\Gamma}\mathbf{A}\mathbf{B}\mathbf{A}^*+\trace\boldsymbol{\Gamma}\mathbf{W}\quad\text{and}\quad\trace\boldsymbol{\Gamma}\mathbf{W}\neq\text{constant}.
\end{equation*}

If one has external jammers that are sufficiently far from the directions to be estimated, then the jammers just go into the $Q$ function dampened by the sidelobes, because $\boldsymbol{\Gamma}$ represents the construction of simultaneous sum beams. If, for example, the jammer power is on the order of the target power, then the displacement is small. For example, with ELAN 11 L with two targets at $\mathbf{u}=\left(0,\frac{BW}{2}\right)$ and an extended target according to (2.2-2) with $\bar{u}=0.5$, $s=\frac{BW}{2}$, the displacement is less than $0.02BW$ ($\text{SNR}=3\,\text{dB}$). In this case, the form of the minimum remains almost unchanged. However, the values of $\E\left\{Q_{\text{min}}\right\}$ with jamming are much larger than $N-M=\E\left\{Q_{\text{min}}\right\}$ without jamming, which can become noticeable when a test uses $Q_{\text{min}}$ (see Chapter <ref>). If one has fluctuations in the receiver noise power, then $\mathbf{W}=\diag\sigma_i^2$. For the simulation, noise-power fluctuations of $1\,\text{dB}$ are assumed and $\sigma_i^2\sim\mathcal{R}\left(\left[\sigma^2(1-1/4),\sigma^2(1+1/4)\right]\right)$ is chosen. The convergence in the $u$ direction for the same target configuration and antenna array as in Fig <ref> (a) are shown in Fig. <ref> for the the following scenarios:

* (a) Two extended targets according to (2.2-3) with $r=0.2BW$.

* (b) Two point targets with $1$-$\text{dB}$ noise-power fluctuations at the receiver.

* (c) Two point targets and also an extended jammer of the same power according to (2.2-3) with $(u,v)=(-0.5,0.1)$ and $r=0.9BW$.

If the external jammers become stronger, then additional measures for interference suppression must be taken, such as a pre-filtering of the simultaneous sum-and-difference beam outputs or through the use of an appropriate projection matrix as in (3.2-3).

§.§.§ Quantization and Clipping

If the input data $\mathbf{z}$ is quantized into real and imaginary parts having the same quantization levels $k\Delta$ $(k=\ldots-1,0,1,2,\ldots)$, then if $\Delta$ is small, this acts as an increased noise, because for $\mathbf{z}_q=\mathbf{z}+\boldsymbol{\epsilon}$,

\begin{equation*}
\E\left\{\mathbf{z}_q\mathbf{z}_q^*\right\}=\E\left\{\mathbf{z}\mathbf{z}^*\right\}+\E\left\{\boldsymbol{\epsilon}\boldsymbol{\epsilon}^*\right\}
\end{equation*}

and if $\text{Re }\epsilon_i\sim\mathcal{R}(\Delta)$, $\text{Im }\epsilon_i\sim\mathbb{R}(\Delta)$ can be assumed to be independent, then $\E\left\{\boldsymbol{\epsilon}\boldsymbol{\epsilon}^*\right\}=\diag\left(\frac{\Delta^3}{6}\right)$.

With rougher quantization, the assumption that the real and imaginary parts of the error are independent is no longer correct, and one gets correlations. That means that $\E\left\{\boldsymbol{\epsilon}\boldsymbol{\epsilon}^*\right\}$ is not diagonal and thus the function $\E\left\{Q\right\}$ is distorted. Such a distortion arises notably through clipping of the input data. The $Q$ function is, however, not particularly sensitive to such distortions. With ELAN 11 L, simulations of $\E\left\{\mathbf{z}_q\mathbf{z}_q^*\right\}$ showed that the quantization and clipping only caused a very small displacement of the minimum of $\E\left\{Q\right\}$. For example, with $3$-bit quantization and clipping to $\left|z_{q_i}\right|\leq\sqrt{\E\left\{z_iz_i^*\right\}}$, the displacement is less than $0.012BW$ (Separation of the targets $BW/2$, $\text{SNR}=12.5\,\text{dB}$). However, the minimum became notably flatter. This reduction in the curvature of the minimum is particularly strong with sharper clipping. This is shown in Fig. <ref> for the same target and antenna-array situation as in Fig. <ref> (a) for different quantizations and clippings $B_q$ when $\left|z_{q_i}\right|\leq B_q\sqrt{\E\left\{z_qz_q^*\right\}}$.

In all, the simulations show that the procedure is quite robust with low levels of interference and distortion.

§ AVERAGED GRID SEARCH

Summary of Chapter 6

This chapter studies whether the suggested minimization procedure utilizing a grid search can be improved when the results are subsequently averaged. For linear antennas, the computational load of such an averaged grid search is higher than for the stochastic approximation when maintaining the same accuracy. For planar antennas,  the complexity is so much higher that stochastic approximation is generally preferred.

The simplest realization of the grid search described in Chapter <ref> implies that the method can also be used when multiple samples are taken. This can occur by performing a grid minimization for each sample $\mathbf{z}_k$, and then averaging the resulting estimation values $\hat{\boldsymbol{\omega}}_k$ ($k=1,\ldots, K$). One can perform such minimization on a rather coarse grid, because the dispersion of $\hat{\boldsymbol{\omega}}_k$ is quite large. Moreover, one can hope that with phase fluctuations, the worst case scenario (for example, $\Delta\varphi=0^\circ,180^\circ$ when $M=2$) averages out.

The same situations as in Fig. <ref> (ELAN 11 L, $1/6 BW$ grid, distance between targets $0.5BW$, grid spans from $u_g-\frac{BW}{2}$ to $u_g+\frac{BW}{2}$) for Signal Models $1$, $3$, and $4$ averaging four samples are shown in Fig. <ref>.

Whether the averaged grid search or the stochastic approximation is to be preferred depends on the specific application. The grid search requires a number of fixed sum beams that are jointly steerable. The stochastic approximation, on the other hand, requires independent sum-and-difference beams that are steerable during the iteration. Assuming that the hardware for each scenario has the same cost, one can compare both procedures by the number of their basic operations (that means, multiplications and division).

The number of basic operations (complex multiplications) are shown in the following table:

stochastic approximation		    averaged grid search

-2*linear array	   $+i$ real roots			   -2*$m\binom{k}{M}(M^2+M)$

-2*planar array	   $+i$ real roots 			   -2*$m\binom{k}{M}(M^2+M)$

with $i$ being the number of iterations, $k$ the number of beams on the grid, and $m$ the number of samples. A real multiplication counts as $1/4$ of a complex multiplication. It is assumed that the saturation bound $\eta$ (See (5.2-2)) is reached after half of the specified number of iterations. The comparison to determine whether a saturation bound has been passed, as in the grid search, is assumed to have no cost. The number of beams in the grid is with regard to a linear antenna with a $\frac{1}{n}BW$ grid $k=n+1$, with regard to a planar antenna with a triangular grid with $\frac{1}{n}BW$ spacing as follows:

For a linear antenna with, for example two largets and five beams, the average grid search with $60m$ operations is the same complexity as the stochastic approximation with $i12$ operations for $m=3, i=15$ and $m=4,i=20$, and so on. By this number of iterations, one can see from Fig. <ref> that the stochastic approximation is better.

The averaged grid search will only be of interest in special cases. The computational complexity with linear arrays is with the same higher accuracy as that of the stochastic approximation. With planar arrays, the computational complexity is significantly higher.

§ CONSTRUCTION OF THE MULTIHYPOTHESIS TEST

Summary of Chapter 7

With the suboptimal direction estimation algorithm of Chapter 5, which utilizes a stochastic approximation, a likelihood ratio test can no longer be used to determine the correct number of targets.  Thus, this chapter develops an appropriate suboptimal test. The minimal value of the $Q$ function averaged over a number of measurement vectors is suitable and can, moreover, be calculated without considerable computational effort. This test value measures the residual power after the subtraction of the estimated signal.

In order to have a multihypothesis test like that described in Chapter 3, the test must hold to an approximate probability of a Type-$1$ error. This is attained through the specification of a decision threshold by the approximation of the distribution of the test statistic as a $\chi^2$ random variable. Is is demonstrated through simulation that the total multihypothesis test approximately respects this error level, as demanded in Chapter 3.

For a target with amplitude fluctuations according to Signal Model $4$ (“Swerling-II Signal Model”), an approximation of the detection probability of the multihypothesis test is calculated and its validity proven through simulation. The curves for the detection probability can also be used with other signal models, as their detection probabilities proved higher in simulations.

Subsequently, the behavior of the test given deviations in the assumptions in the signal and noise models are tested through simulations, whereby it is observed that deviations from the noise model are the most critical.

In Chapter 3.3 it was shown that the sequential multiple-hypothesis test has the following steps. For $M+1,2,\ldots, N$:

* Test $H_M$ against $K_M=\mathbb{C}^N \backslash H_M$. This means compute $\varphi_M:\mathcal{Z}^K\rightarrow\{0,1\}, (\mathbf{z}_1,\ldots, \mathbf{z}_K)\mapsto\varphi_M(\mathbf{z}_1,\ldots,\mathbf{z}_K)$ for $H_M=\left\{\mathbf{s}\in\mathbb{C}^N\left|\bigvee\limits_{\mathbf{b}\in\mathbb{C}^M}\bigvee\limits_{\boldsymbol{\omega}\in\Omega^M}:\mathbf{s}=\mathbf{A}\mathbf{b}\right.\right\}$.

* If $\phi_M=0$, then decide $\hat{M}:=M$ and stop.

In order to satisfy the total error level $\alpha$ of such a multihypothesis test, according to Chapter 3.3, each $\phi_M$ must satisfy this error bound. The construction of such a test $\phi_M$ shall be considered here. It shall be shown that such a test can be used with the estimated parameters rather than focussing on particularly efficient tests.

§.§ The $\bar{Q}$ Statistic and its Relationship with the Likelihood Ratio

The desired test should be as simple as possible. Using the estimated parameters $\hat{\boldsymbol{\omega}}(\mathbf{z}_1,\ldots,\mathbf{z}_k)$, a test statistic $T(\hat{\boldsymbol{\omega}}(\mathbf{z}_1,\ldots,\mathbf{z}_k))$ should be compared to a threshold $\eta\in\mathbb{R}$. Such a statistic cannot be sufficient for $H_M$, because $H_M$ is an $M$-dimensional partial set of $\mathbb{C}^N$. A sufficient statistic must be at least an $M$-component vector.

The formulation of the hypotheses

\begin{equation*}
H_M=\left\{\mathbf{s}\in\mathbb{C}^N\left|\bigvee\limits_{\mathbf{b}\in\mathbb{C}^M}\bigvee\limits_{\boldsymbol{\omega}\in\Omega}:\mathbf{s}=\mathbf{A}\mathbf{b}\right.\right\},\Omega\in V^M
\end{equation*}

clearly contains parameters on which the decision is dependent. Instead of a sufficient statistic, a statistic that is invariant to a group of transformations is sought, whereby the invariance with respect to the transformations arbitrarily eliminates parameters that depend on the measuring instrument. In this instance, it is demanded that the statistic be independent of the complex amplitudes $\mathbf{b}_1,\ldots,\mathbf{b}_K$. That is, not dependent on the specific point $(\mathbf{A}\mathbf{b}_1,\ldots,\mathbf{A}\mathbf{b}_k)$, rather only the space $S=\bigtimes_{k=1}^K\lin H(\mathbf{A})$. The magnitude of the test statistic should thus be invariant to all affine transformations on $S$; these are matrices

\begin{equation*}
\mathbf{C}=\left(\begin{IEEEeqnarraybox}[\relax][c]{c'c'c'c}
\IEEEstrut
\mathbf{C}_1&0&\ldots&0\\
\vdots&\vdots&\ddots&\vdots\\
\IEEEstrut
\end{IEEEeqnarraybox}\right)
\end{equation*}

with affine transformation $\mathbf{C}_i\in\mathbb{C}^{N\times N}$ that only transforms the components in $\lin H(\mathbf{A})$. That is $\mathbf{C}_i\mathbf{z}=\tilde{\mathbf{z}}_A+\mathbf{z}_A^\bot$ if $\mathbf{z}=\mathbf{z}_A+\mathbf{z}_A^\bot$ for $\mathbb{C}^N=\lin H(\mathbf{A})\oplus\lin H(\mathbf{A})^\bot$.

In the residual space $S^\bot\subset\mathbb{C}^{NK}$, the test statistic should be independent of the rotation of the coordinate system; that is independent of unitary transformations. Let $\mathbf{G}$ be the set of all such transformations. For a given $\mathbf{A}=\mathbf{A}(\boldsymbol{\omega}),\mathbb{C}^{NK}=S\oplus S^\bot$, then

\begin{equation*}
\bigvee\limits_{\mathbf{C}_k\in\mathbb{C}^{N\times N} \atop \mathbf{C}_k| \lin H(\mathbf{A})^\bot=\mathbf{I}}
\bigvee\limits_{\mathbf{U}:S^\bot\rightarrow S^\bot\atop \text{is unitary}}:\mathbf{H}\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\mathbf{z}_1\\
\vdots\\
\mathbf{z}_K
\IEEEstrut
\end{IEEEeqnarraybox}\right)=\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\mathbf{C}_1\mathbf{z}_{1A}\\
\vdots\\
\mathbf{C}_K\mathbf{z}_{KA}
\IEEEstrut
\end{IEEEeqnarraybox}\right)+\mathbf{U}\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\mathbf{z}_{1A}^\bot\\
\vdots\\
\mathbf{z}_{KA}^\bot
\IEEEstrut
\end{IEEEeqnarraybox}\right)
\right.\right\}
\end{equation*}

Then $G$ is a group, because unitary and affine transformations form a group, and such transformations only decompose $\mathbb{C}^{NK}$ into orthogonal parts.

The $Q$-function is thus, with regard to $G$, an invariant reduction of the data $(\mathbf{z}_1,\ldots,\mathbf{z}_k)$.

Theorem (7.1-1):
For all $\boldsymbol{\omega}$,

\begin{equation*}
\bar{Q}:=\frac{1}{K}\sum_{k=1}^K\left\lVert\boldsymbol{\Gamma}(\boldsymbol{\omega})\mathbf{z}_k\right\rVert^2
\end{equation*}

is a maximally invariant statistic with respect to the transformation group $G_{\boldsymbol{\omega}}$.

* Part (i) Invariance:
Let $\mathbf{H}\in G$. Thus,

\begin{equation*}
\mathbf{H}=\left(\begin{IEEEeqnarraybox}[\relax][c]{c'c'c'c}
\IEEEstrut
\mathbf{C}_1&0&\ldots&0\\
\vdots&\vdots&\ddots&\vdots\\
\IEEEstrut
\end{IEEEeqnarraybox}\right)+\mathbf{U}
\end{equation*}

and $\mathbf{z}_k=\mathbf{z}_{kA}+\mathbf{z}_{kA}^\bot$ for all $k=1\ldots K$.

\begin{align*}
\bar{Q}\left(\mathbf{H}\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\mathbf{z}_1\\
\vdots\\
\mathbf{z}_K
\IEEEstrut
\end{IEEEeqnarraybox}\right)\right)=&\frac{1}{K}\left\lVert\left(\begin{IEEEeqnarraybox}[\relax][c]{c'c'c'c}
\IEEEstrut
\boldsymbol{\Gamma}&0&\ldots&0\\
\vdots&\vdots&\ddots&\vdots\\
\IEEEstrut
\end{IEEEeqnarraybox}\right)\left(\left(\begin{IEEEeqnarraybox}[\relax][c]{c'c'c'c}
\IEEEstrut
\mathbf{C}_1&0&\ldots&0\\
\vdots&\vdots&\ddots&\vdots\\
\IEEEstrut
\end{IEEEeqnarraybox}\right)\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\mathbf{z}_{1A}\\
\vdots\\
\mathbf{z}_{KA}
\IEEEstrut
\end{IEEEeqnarraybox}\right)+\mathbf{U}\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\mathbf{z}_{1A}^\bot\\
\vdots\\
\mathbf{z}_{KA}^\bot
\IEEEstrut
\end{IEEEeqnarraybox}\right)\right)\right\rVert^2\\
\IEEEstrut
\boldsymbol{\Gamma}&0&\ldots&0\\
\vdots&\vdots&\ddots&\vdots\\
\IEEEstrut
\end{IEEEeqnarraybox}\right)\mathbf{U}\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\mathbf{z}_{1A}^\bot\\
\vdots\\
\mathbf{z}_{KA}^\bot
\IEEEstrut
\end{IEEEeqnarraybox}\right)\right\rVert^2
\end{align*}

where the simplification is because $\mathbf{C}_i\mathbf{z}_{iA}\in\lin H(\mathbf{A})$ and thus $\boldsymbol{\Gamma}\mathbf{C}_i\mathbf{z}_{iA}=\mathbf{0}$. A further simplification is

\begin{align*}
\bar{Q}\left(\mathbf{H}\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\mathbf{z}_1\\
\vdots\\
\mathbf{z}_K
\IEEEstrut
\end{IEEEeqnarraybox}\right)\right)=\left\lVert
\mathbf{U}\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\mathbf{z}_{1A}^\bot\\
\vdots\\
\mathbf{z}_{KA}^\bot
\IEEEstrut
\end{IEEEeqnarraybox}\right)
\right\rVert^2,
\end{align*}

which is because $\left(\begin{IEEEeqnarraybox}[\relax][c]{c'c'c}
\IEEEstrut
\boldsymbol{\Gamma}&&\\
\IEEEstrut
\end{IEEEeqnarraybox}\right)$ is a projection onto $S^\bot$ and $\mathbf{U}\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\mathbf{z}_{1A}^\bot\\
\vdots\\
\mathbf{z}_{KA}^\bot
\IEEEstrut
\end{IEEEeqnarraybox}\right)\in S^\bot$. Continuing,

\begin{align*}
\bar{Q}\left(\mathbf{H}\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\mathbf{z}_1\\
\vdots\\
\mathbf{z}_K
\IEEEstrut
\end{IEEEeqnarraybox}\right)\right)=&\frac{1}{K}\left\lVert\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\mathbf{z}_{1A}^\bot\\
\vdots\\
\mathbf{z}_{KA}^\bot
\IEEEstrut
\end{IEEEeqnarraybox}\right)\right\rVert^2\quad \text{because $\mathbf{U}$ is unitary}\\
\IEEEstrut
\boldsymbol{\Gamma}&0&\ldots&0\\
\vdots&\vdots&\ddots&\vdots\\
\IEEEstrut
\end{IEEEeqnarraybox}\right)\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\mathbf{z}_1\\
\vdots\\
\mathbf{z}_K
\IEEEstrut
\end{IEEEeqnarraybox}\right)\right\rVert^2\\
\end{align*}

* Part (ii), Maximality:
We have to show:

\begin{equation*}
\bigwedge\limits_{\mathbf{z},\mathbf{y}\in\mathbb{C}^{NK}}:\bar{Q}(\mathbf{z})=\bar{Q}(\mathbf{y})\Longrightarrow\bigvee\limits_{\mathbf{H}\in G}:\mathbf{z}=\mathbf{H}\mathbf{y}.
\end{equation*}

Let $\bar{Q}(\mathbf{z})=\bar{Q}(\mathbf{y})$; thus,

\begin{equation*}
\begin{array}{r@{\,}l}
\left\lVert\left(\begin{IEEEeqnarraybox}[\relax][c]{c'c'c'c}
\IEEEstrut
\boldsymbol{\Gamma}&0&\ldots&0\\
\vdots&\vdots&\ddots&\vdots\\
\IEEEstrut
\end{IEEEeqnarraybox}\right)\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\mathbf{z}_1\\
\vdots\\
\mathbf{z}_K
\IEEEstrut
\end{IEEEeqnarraybox}\right)
\right\rVert^2=\left\lVert\left(\begin{IEEEeqnarraybox}[\relax][c]{c'c'c'c}
\IEEEstrut
\boldsymbol{\Gamma}&0&\ldots&0\\
\vdots&\vdots&\ddots&\vdots\\
\IEEEstrut
\end{IEEEeqnarraybox}\right)\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\mathbf{y}_1\\
\vdots\\
\mathbf{y}_K
\IEEEstrut
\end{IEEEeqnarraybox}\right)
\right\rVert^2
\\
\begin{matrix}
\mspace{0mu}\underbrace{\rule{3.8cm}{0pt}}_{\in S^\bot} &
\quad\quad\underbrace{\rule{3.8cm}{0pt}}_{\in S^\bot}\quad
\end{matrix}
\end{array}
\end{equation*}

and there exists a matrix $\mathbf{U}$ unitary on $S^\bot$ such that

\begin{align*}
\left(\begin{IEEEeqnarraybox}[\relax][c]{c'c'c'c}
\IEEEstrut
\boldsymbol{\Gamma}&0&\ldots&0\\
\vdots&\vdots&\ddots&\vdots\\
\IEEEstrut
\end{IEEEeqnarraybox}\right)\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\mathbf{z}_1\\
\vdots\\
\mathbf{z}_K
\IEEEstrut
\end{IEEEeqnarraybox}\right)=&\left(\begin{IEEEeqnarraybox}[\relax][c]{c'c'c'c}
\IEEEstrut
\boldsymbol{\Gamma}&0&\ldots&0\\
\vdots&\vdots&\ddots&\vdots\\
\IEEEstrut
\end{IEEEeqnarraybox}\right)\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\mathbf{y}_1\\
\vdots\\
\mathbf{y}_K
\IEEEstrut
\end{IEEEeqnarraybox}\right)\\
\IEEEstrut
\boldsymbol{\Gamma}&0&\ldots&0\\
\vdots&\vdots&\ddots&\vdots\\
\IEEEstrut
\end{IEEEeqnarraybox}\right)\mathbf{U}\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\mathbf{y}_1\\
\vdots\\
\mathbf{y}_K
\IEEEstrut
\end{IEEEeqnarraybox}\right)
\end{align*}

because $\boldsymbol{\Gamma}$ projects onto $S^\bot$ and $\mathbf{U}$ only functions on $S^\bot$. Therefore,

\begin{equation*}
\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\mathbf{z}_{1A}^\bot\\
\vdots\\
\mathbf{z}_{KA}^\bot
\IEEEstrut
\end{IEEEeqnarraybox}\right)=\mathbf{U}\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\mathbf{y}_{1A}^\bot\\
\vdots\\
\mathbf{y}_{KA}^\bot
\IEEEstrut
\end{IEEEeqnarraybox}\right).
\end{equation*}

For those components projected onto $\lin H(A)$, there exists  constantly affine transformations $\mathbf{C}_i$ so that $\mathbf{z}_{iA}=\mathbf{C}_i\mathbf{y}_{iA}$. Thus, there exists an $\mathbf{H}\in G$.

The demand that there be independence in the sequence of complex amplitudes $\mathbf{b}_1,\ldots,\mathbf{b}_K$ leads directly to the statistic $\bar{Q}$. All other statistics with this property are directly related to $\bar{Q}$. The statistic $\bar{Q}$ is maximally invariant. This means that all other $G$ invariant statistics $T(\mathbf{z}_1,\ldots,\mathbf{z}_K)$ are related to $\bar{Q}$ through a function $g$ such that $T=g(\bar{Q})$ <cit.>. This simplifies the search for more efficient tests.

It is clear that $\bar{Q}$ is the average residual energy after signal extraction for a signal assumed to be in direction $\boldsymbol{\omega}$. Only this residual energy is independent of $\mathbf{b}_1,\ldots,\mathbf{b}_K$ if $\boldsymbol{\omega}$ is known. For the choice of $\eta$, this is more desirable than the use of the signal energy itself as in [3].

There remains the question whether the data reduction realized through $\bar{Q}$ suffices to make a decision. For this to be true, the distribution of $\bar{Q}$ under $H_M$ and $K_M$ must at least be different.

Theorem (7.1-2):

for $H_M=\left\{\mathbf{s}\in\mathbb{C}^N\left|
\bigvee\limits_{\mathbf{b}\in\mathbb{C}^M}\bigvee\limits_{\boldsymbol{\omega}\in\Omega}:\mathbf{s}:=\mathbf{A}\mathbf{b}\right.\right\}$ and $K_M=\mathbb{C}^N\backslash H_M$, it is true that

\begin{equation*}
\min\limits_{\boldsymbol{\omega}}\E_n\left\{\bar{Q}\left|\mathbf{s}_1\ldots\mathbf{s}_K\in K_M\right.\right\}>\min\limits_{\boldsymbol{\omega}}\E_n\left\{\bar{Q}\left|\mathbf{s}_1\ldots\mathbf{s}_K\in H_M\right.\right\}.
\end{equation*}

It is clear that under $H_M \bigwedge\limits_{i\in\{1\ldots k\}}:\mathbf{s}^*_i\hat{\boldsymbol{\Gamma}}\mathbf{s}_i=0$, but under $K_M \bigvee\limits_{i\in\{1\ldots k\}}:\mathbf{s}_i^*\hat{\boldsymbol{\Gamma}}\mathbf{s}_i>0$, because all $s_i\notin H_M$.

The statistic $\bar{Q}$ can be computed after estimating $\hat{\boldsymbol{\omega}}$ by means of a stochastic approximating averaging $Q(\hat{\boldsymbol{\omega}})$ over a few few (for example two to four) samples of $\mathbf{z}$. In order to set a bound for a test to level $\alpha$, the distribution of $\bar{Q}$ must be computed.

\begin{equation*}
\end{equation*}

\begin{align*}
\mathbf{z}_i=&\mathbf{A}\mathbf{b}_i+\mathbf{n}_i\\
\mathbf{n}_1\sim&\mathcal{N}(\mathbf{0},\sigma^2\mathbf{I}).
\end{align*}

One can now make the approximation that the parameter estimation is good; that is

\begin{equation}
\hat{\boldsymbol{\Gamma}}\mathbf{A}\mathbf{b}_i\approx\mathbf{0}\label{Eq7-1-3}
\end{equation}

(for the stochastic approximation, this is asymptotically satisfied. Thus, one gets

\begin{align*}
\end{align*}

for $\mathbf{x}_k\sim\mathcal{N}_{\mathbb{C}^{N-M}}(\mathbf{0},\sigma^2\mathbf{I})$, because $\boldsymbol{\Gamma}$ is a projection. This means that $\frac{2}{\sigma^2}K\bar{Q}$ is roughly $\chi^2$-distributed with $K(2N-2M)$ degrees of freedom. With this approximation, for $P\left\{\bar{Q}\leq\eta\right\}=1-\alpha$, one can choose

\begin{equation}
\eta=\frac{\sigma^2}{2K}\chi^2_{K(2N-2M);\alpha}\label{Eq7-1-4}
\end{equation}

for when $H_M$ is valid ($\chi^2_{r;a}$ is defined to be the $\alpha$-fractile of the $\chi^2$ distribution with $r$ degrees of freedom). In the case where $N\gg M$ (for example, it suffices when $N=21$ with $M=2$) or a large $K$, the $\chi^2$ distribution can be approximated using a normal distribution. In such an instance,

\begin{equation*}
\frac{\frac{2}{\sigma^2}K\bar{Q}-\E\left\{\frac{2}{\sigma^2}K\bar{Q}\right\}}{\sqrt{\var\left\{\frac{2}{\sigma^2}K\bar{Q}\right\}}}\sim\mathcal{N}_{\mathbb{R}}(0,1).
\end{equation*}

For a central $\chi^2$-distributed random variable $x$, $\E\{x\}=r$ and $\var\{x\}=2r$ so that

\begin{equation*}
\frac{\frac{2}{\sigma^2}K\bar{Q}-K(2N-2M)}{\sqrt{2K(2N-2M)}}\sim\mathcal{N}(0,1)
\end{equation*}

meaning that

\begin{equation}
\frac{\frac{1}{K}\sum_{k=1}^K\left\lVert\hat{\boldsymbol{\Gamma}}\mathbf{z}_k\right\rVert^2-\sigma^2(N-M)}{\sigma^2\sqrt{(N-M)/K}}\sim\mathcal{N}(0,1).\label{Eq7-1-5}
\end{equation}

For $P\{\bar{Q}\leq\eta\}=1-\alpha$, one should then use

\begin{equation}
\eta=\sigma^2(\sqrt{(N-M)/K}U_\alpha+N-M)\label{Eq7-1-6}
\end{equation}

($U_\alpha$ is the $\alpha$-fractile of $\mathcal{N}(0,1)$).

The test in (7.1-4), or respectively (7.1-6) asymptotically, that is for a large number of iterations by the stochastic approximation and for high SNR, provides a test at a level of $\alpha$.

Given Signal Model $1$ with fixed amplitudes $\mathbf{b}_k=\mathbf{b}$ for all $k$, then, using the theorems regarding the asymptotic distribution of the likelihood ratio, the likelihood ratio test can be viewed as an “optimal test” for comparison. In this instance $\mathbf{z}\sim\mathcal{N}_{\mathbb{C}^N}(\mathbf{s},\sigma^2\mathbf{I})$ with

* $\mathbf{s}=\mathbf{A}\mathbf{b}$ if $H$ is true

* $\mathbf{c}\in\mathbb{C}^N\backslash H$ if $K$ is true.

\begin{equation*}
\end{equation*}

it is true that

* $\max\limits_{\mathbf{s}\in\mathbb{C}^N}p(\mathbf{z}_1\ldots\mathbf{z}_K;\mathbf{s})$ is assumed for $\hat{\mathbf{s}}=\frac{1}{K}\sum_{i=1}^K\mathbf{z}_i:=\bar{\mathbf{z}}$.

* $\max\limits_{\mathbf{s}\in\mathbf{H}}p(\mathbf{z}_1\ldots\mathbf{z}_K;\mathbf{s})$ is assumed for $\hat{\mathbf{b}}=\left.\mathbf{A}\left(\mathbf{A}^*\mathbf{A}\right)^{-1}\mathbf{A}^*\right|_{\hat{\boldsymbol{\omega}}}\bar{\mathbf{z}}$
with $\hat{\boldsymbol{\omega}}$ so that $\sum_{i=1}^K\left\lVert\left.\mathbf{z}_i-\mathbf{A}\left(\mathbf{A}^*\mathbf{A}\right)^{-1}\mathbf{A}^*\right|_{\hat{\boldsymbol{\omega}}}\bar{\mathbf{z}}\right\rVert^2=\min !$.

Thus, the logarithm of the likelihood ratio according to (3.1-8) is

\begin{align}
\ln T=&\frac{1}{\sigma^2}\left(\sum_{i=1}^K\left\lVert\mathbf{z}_i-\mathbf{A}\left(\mathbf{A}^*\mathbf{A}\right)^{-1}\mathbf{A}^*\bar{\mathbf{z}}\right\rVert^2-\sum_{i=1}^K\lVert\mathbf{z}_i-\bar{\mathbf{z}}\rVert^2\right)\notag\\
\end{align}

after multiplying out and combining terms. The quantity $2\ln T$ is asymptotically (meaning for $K\rightarrow\infty$) $\chi^2_r$-distributed (See, for example, <cit.>) with

\begin{align*}
2N-3M&with linear arrays (hypotheses parameterized via $\mathbf{u},\mathbf{b})$\\
2N-4M&with planar arrays (hypotheses parameterized via $\mathbf{u},\mathbf{v},\mathbf{b})$
\end{IEEEeqnarraybox}\right.
\end{align*}

so that under $H_M$ for $P\{2\ln T\leq \eta\}=1-\alpha$,

\begin{equation}
\eta=\left\{\begin{IEEEeqnarraybox}[\relax][c]{c's}
\chi^2_{2N-3M; \alpha}&for a linear array\\
\chi^2_{2N-4M;\alpha}&for a planar array
\end{IEEEeqnarraybox}\right.\label{Eq7-1-8}
\end{equation}

should be chosen.

Through the likelihood ratio test, the number of degrees of freedom of the remaining noise after the signal extraction (projection) is further reduced by the minimization with respect to $\boldsymbol{\omega}$ that is simultaneously applied to all $\mathbf{n}_k$. This introduces a small variance in the statistic. Approximating $2\ln T$ as in (7.1-5) with a normal distribution, then one gets, for example with a linear array:

\begin{equation}
\frac{2\ln T-(2N-3M)}{\sqrt{2(2N-3M)}}=\frac{\left\lVert\hat{\boldsymbol{\Gamma}}\bar{\boldsymbol{z}}\right\rVert^2-\frac{\sigma^2}{K}\left(N-\frac{3}{2}M\right)}{\frac{\sigma^2}{K}\sqrt{N-\frac{3}{2}M}}\sim\mathcal{N}(0,1).\label{Eq7-1-9}
\end{equation}

The large variance compared with (7.1-5) is the price that must be paid for separating the test statistic from the minimization process and for the independence of the signal model. The smaller variance in (7.1-9) is an indication that the likelihood ratio test has a higher precision.

The knowledge of the thresholds $\eta_{\alpha;M}$ in (7.1-6) is more important in a multiple-hypothesis test than when simply detecting a target, because closely-spaced targets are generally seldom-occurring phenomena and thus one cannot experimentally set the detection thresholds $\eta_{\alpha;M}$ during radar applications.

§.§ Calculation of the Detection Probability of Stochastic Signals

If the number of targets $M$ is given, then the associated measurement vectors $\mathbf{z}$ are elements of the alternative $K_i$ of the tests $H_i\longleftrightarrow K_i$ for $i<M$. If Signal Model $2,$ $3$, or $4$ is given, then one can compute the distribution of $\bar{Q}$ for such measurement vectors from the distribution of the alternative. Thus, one gets the detection probability because according to (3.3-2) and (3.3-3)

\begin{equation*}
P_E=(1-\alpha)P\{\bar{Q}_{M-1}>\eta\}\ldots P\{\bar{Q}_{1}>\eta\}\quad\text{under $H_M$}.
\end{equation*}

To calculate the detection probability, assume $M^{\text{ex}}$ targets are given with $M<M^{\text{ex}}$, $\mathbf{z}\notin H_M$, $\mathbf{z}=\mathbf{s}+\mathbf{n}$ with $\mathbf{s}=\mathbf{A}\mathbf{b}$ and $\text{rank }\mathbf{A}=M^{\text{ex}}$, $\mathbf{n}\sim\mathcal{N}_{\mathbb{C}^N}(\mathbf{0},\sigma^2\mathbf{I})$. Then, $\E\{\mathbf{s}^*\boldsymbol{\Gamma}\mathbf{s}\}>0$ according to (7.1-2), and $\frac{2}{\sigma^2}Q$ is $\chi^2_{2N-2M}(\delta^2)$-distributed when conditioned on $\mathbf{s}$ with non-centrality parameter $\delta^2=\frac{2}{\sigma^2}\mathbf{s}^*\boldsymbol{\Gamma}\mathbf{s}$.

This is because $\frac{2}{\sigma^2}Q$ can be written as a sum of normally distributed values with equal variance as follows:

\begin{align*}
=&\lVert\mathbf{t}+\tilde{\mathbf{n}}\rVert^2\quad\text{ with $\mathbf{t},\tilde{\mathbf{n}}\in\mathbb{R}^{2N-2M},\tilde{n}_i\sim\mathcal{N}_{\mathbb{R}}\left(0,\frac{\sigma^2}{2}\right)$}\\
=&\sum_{i=1}^{2N-2M}\tilde{\tilde{n}}_i^2\quad\text{with $\tilde{\tilde{n}}_i\sim\mathcal{N}_{\mathbb{R}}\left(t_i,\frac{\sigma^2}{2}\right)$.}
\end{align*}

\begin{equation*}
\frac{2}{\sigma^2}Q=\sum_{i=1}^{2N-2M}x_i^2\quad\text{with $x_i\sim\mathcal{N}_{\mathbb{R}}\left(\frac{t_i\sqrt{2}}{\sigma},1\right)$}
\end{equation*}

\begin{equation*}
\delta=\sum_{i=1}^{2N-2M}\left(\frac{t_i\sqrt{2}}{\sigma}\right)^2.
\end{equation*}

For large $N$, the distribution of $Q$ conditioned on $\delta^2$ can be approximated by a normal distribution and thus $\bar{Q}$ can be approximated with a normal distribution. The distribution of $\bar{Q}$ is thus

\begin{align*}
=&\int\phi\left(\frac{\eta-\E_n\{\bar{Q}\}}{\sqrt{\var_n\{\bar{Q}\}}}\right)dP_{\delta_1^2}\ldots dP_{\delta_K^2}
\end{align*}

where $P_{\delta^2_i}$ is the distribution of $\delta_i^2$ from $K_M$, $\phi$ the cumulative distribution function of the normal distribution, and $\E_n$, $\var_n$ represent the expected value and variance only with respect to $n$.

According to Appendix 4, (A.4-1), (A.4-2), one has

\begin{align*}
\E_n\{\bar{Q}\}=&\sigma^2(N-M)+\frac{\sum_{i=1}^K\mathbf{s}_i^*\boldsymbol{\Gamma}\mathbf{s}_i}{K}\\
\end{align*}

\begin{equation*}
\var_n\{\bar{Q}\}=\frac{\sigma^4}{K}(N-M).
\end{equation*}

The problem to be solved with $\eta=\sigma^2\left(\sqrt{\frac{N-M}{K}}U_\alpha+N-M\right)$ according to (7.1-6) is the integral

\begin{equation}
\int_0^\infty\phi\left(\frac{\sigma^2\left(\sqrt{\frac{N_M}{K}}U_{\alpha}+N-M\right)-\left(\sigma^2(N-M)+q\right)}{\sigma^2\sqrt{\frac{N-M}{K}}}\right)dP_q=
\int_0^\infty\phi\left(U_\alpha-\frac{q}{\sigma^2\sqrt{\frac{N-M}{K}}}\right)dP_q.\label{Eq7-2-1}
\end{equation}

For Signal Models $2$ and $3$, the integral (7.2-1) can probably only be solved numerically (as an integral over the phase $\phi_i$). For Signal Model $4$, however, one can compute the distribution $p_{\delta^2}(x)$ and solve the integral analytically. Consequently, for this unfavorable situation, the detection probability of the entire test can be computed. It is

\begin{align*}
\end{align*}

The matrix $\mathbf{G}=\mathbf{A}^*\hat{\boldsymbol{\Gamma}}\mathbf{A}$ is generally a positive-definite Hermitian matrix of dimensions $(M+I)\times(M+I)$ for the case where $M^\text{ex}=M+I$ targets are present and one is solving for $M$ targets. $\mathbf{G}$ is then positive definite when the vectors $\hat{\boldsymbol{\Gamma}}\mathbf{a}_1,\ldots,\hat{\boldsymbol{\Gamma}}\mathbf{a}_{M+I}$ are linearly independent. This is almost always the case for large $N$. Otherwise, instead of $\mathbf{b}$, $\mathbf{x}=\mathbf{L}^{-1}\mathbf{b}$, if $\mathbf{B}=\mathbf{L}\mathbf{L}^*$, then $\mathbf{x}\sim\mathcal{N}(\mathbf{0},\mathbf{I})$ and if

\begin{equation*}
\mathbf{G}=\mathbf{U}\left(\begin{IEEEeqnarraybox}[\relax][c]{c'c'c'c'c}
\IEEEstrut
\lambda_1&		&			&	&\\
&\ddots	&			&	&\\
&		&\lambda_r	&	&\\
&		&			&0	&\\
&		&			&	&0
\IEEEstrut
\end{IEEEeqnarraybox}\right)\mathbf{U}^*
\end{equation*}

with $\mathbf{U}$ unitary, then $\mathbf{y}=\mathbf{U}^*\mathbf{x}\sim\mathcal{N}(\mathbf{0},\mathbf{I})$ so that instead of considering $\mathbf{b}^*\mathbf{G}\mathbf{b}$, one can consider
\mathbf{y}_1^*&\ldots&\mathbf{y}_r^*
\end{IEEEeqnarraybox}\right)\left(\begin{IEEEeqnarraybox}[\relax][c]{c'c'c}
\IEEEstrut
\lambda_1&		&\\
&\ddots	&\\
&		&\lambda_r
\IEEEstrut
\end{IEEEeqnarraybox}\right)\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\mathbf{y}_1\\
\vdots\\
\mathbf{y}_r
\IEEEstrut
\end{IEEEeqnarraybox}\right)$ and once again a positive-definite, Hermitian form is given.

The probability density of such averaged, positive-definite forms can now be computed:

Theorem (7.2-2): (Probability density of averaged Hermitian forms)

* Preconditions

Let $\mathbf{G},\mathbf{B}$ be positive definite Herminitan $I\times I$ matrices, $\mathbf{b}\sim\mathcal{N}(\mathbf{0},\mathbf{B})$,$\lambda_1\ldots\lambda_I$, the eigenvalues of $\mathbf{G}\mathbf{B}$, $\mu_i=\frac{1}{\lambda_i}$ $(i=1\ldots I)$, and

\begin{equation*}
\end{equation*}

* Assertions

* (i):

\begin{equation*}
p_{I,K}(q;\mu_1\ldots\mu_I)=(-1)^{IK+1}\frac{\prod_{i=1}^I\mu_i^K}{(K-1)!}\left.\sum_{l=1}^I\frac{\partial^{K-1}}{\partial z^{K-1}}\frac{e^{-qz}}{\prod_{t\neq l}(z-\mu_t)^K}\right|_{z=\mu_l}
\end{equation*}

* (ii): Special for $I=1$:

\begin{equation*}
\end{equation*}

* (iii): Special for $K=1$:

\begin{equation*}
\prod_{l\neq t}(\mu_t-\mu_l)}
\end{equation*}

* (iv) Special for $I=2$:

\begin{align*}
\end{align*}

where $l_{ij}=\frac{\mu_i}{\mu_i-\mu_j}$

The distribution (iii) agrees with that in [29], which is derived in a different manner.

* Proof

See Appendix 5 for the proof.

Thus, the integral in (7.2-1) becomes

\begin{align}
\end{align}

\begin{equation*}
\text{\ae}:=\sigma^2\sqrt{\frac{N-\hat{M}}{K}}
\end{equation*}

for $M$ targets present and $H_{\hat{M}}$ being tested, $\hat{M}<M$.

The probability of a Type-$2$ error can be computed with the above expression.

Corollary 7.2.4:

* Preconditions

Let $\beta_{M,K}:=P_{M,K}\{\bar{Q}\leq\eta\}$ for $\eta=\sigma^2\left(U_\alpha\sqrt{\frac{N-\hat{M}}{K}}+N-\hat{M}\right)$ according to (<ref>) and

\begin{align*}
\text{\ae}:=&\sigma^2\sqrt{\frac{N-\hat{M}}{K}}
\end{align*}

and let $\frac{1}{\mu_1}\ldots\frac{1}{\mu_M}$ be the eigenvalues of $\mathbf{A}^*\hat{\boldsymbol{\Gamma}}\mathbf{A}\mathbf{B}$.

* Assertions

* (i)

\begin{align*}
\beta_{1,1}(\mu\text{\ae})=&1-\alpha-\psi f(\Delta(\mu\text{\ae}))\\
\beta_{1,2}(\mu2\text{\ae})=&\beta_{1,1}(\mu 2\text{\ae})-\mu2\text{ae}\psi\left(\left.\Delta f(\Delta)\right|_{\mu2\text{\ae}}+\frac{1}{\sqrt{2\pi}}\right)\\
\beta_{1,3}(\mu3\text{\ae})=&\beta_{1,2}(\mu3\text{\ae})-\frac{1}{2}(\mu3\text{\ae})^2\left.\psi\left((1+\Delta^2)f(\Delta)+\frac{\Delta}{\sqrt{2\pi}}\right)\right|_{\mu3\text{\ae}}
\end{align*}

* (ii)

\begin{align*}
\beta_{2,1}(\mu_1\text{\ae},\mu_2\text{\ae})=&l_{12}\beta_{1,1}(\mu_2\text{\ae})+l_{21}\beta_{1,1}(\mu_1\text{\ae})\\
\beta_{2,2}(\mu_12\text{\ae},\mu_22\text{\ae})=&l_{12}^2\beta_{1,2}(\mu_22\text{\ae})+l_{21}^2\beta_{1,2}(\mu_12\text{\ae})+2l_{12}l_{21}\beta_{2,1}(\mu_12\text{\ae},\mu_22\text{\ae})\\
\beta_{2,3}(\mu_13\text{\ae},\mu_23\text{\ae})=&l_{12}^3\beta_{1,3}(\mu_23\text{\ae})+l_{21}^3\beta_{1,3}(\mu_13\text{\ae})+3l_{12}l_{21}\beta_{2,2}(\mu_13\text{\ae},\mu_23\text{\ae})
\end{align*}

\begin{align*}
\psi:=&e^{-\frac{U_\alpha^2}{2}}\\
\Delta(\mu K\text{\ae}):=&U_\alpha-\mu K\text{\ae}
\end{align*}

* Proof

See Appendix 6 for the proof.

Because the probabilities only depend on the eigenvalues of $\mathbf{G}\mathbf{B}$ in (7.2-2), these formulae also allow the detection probabilities in the presence of subarrays and coupling (if one knows the coupling-scattering matrix), because only the modified eigenvalues need be used.

With (7.2-4), one can investigate particular two-target configurations with equal amplitudes. The estimate for $\hat{M}=1$ would lie directly between both targets so that the matrix $\mathbf{G}$ in (7.2-2) is known. The eigenvalues of such a $2\times2$ matrix are known so that the probability of exceeding the threshold at level $\hat{M}=1$ can be easily computed from (7.2-4). The detection probability is thus

\begin{align*}
=&(1-\alpha)(1-\beta_{2,K})\quad \text{if the test meets level $\alpha$}
\end{align*}

In order to compute the detection probability for more than two targets or for two targets with different magnitudes, the estimated direction $\hat{\boldsymbol{\omega}}$ with $\E\{Q(\hat{\boldsymbol{\omega}})\}=\min!$ for $M<M^{\text{ex}}$ must first be numerically determined in order to compute the eigenvalues of $\bar{G}$.

The test (7.1-6) and the detection probability (7.2-4) are also valid with general normally distributed noise as long as the parameter estimation is still good in the sense of (7.1-3). $\bar{Q}$ is still asymptotically normally distributed as long as the elements of the covariance matrix of the noise are constrained. In particular, if one had unequal receiver noise in the $N$ channels with $\frac{\sum_{i=1}^N\sigma_i^2}{N}=\sigma^2,$ $0<c_1\leq\sigma_i^2\leq c_2$, then the approximation of (7.1-5) or, respectively, (7.2-1) can be used, though with the quantities

\begin{align*}
\E_n\{\bar{Q}\}=&\sum_{i=1}^{N-M}\tilde{\sigma}_i^2+q\rightarrow(N-M)\sigma^2+q\text{ for $N\rightarrow\infty$}\\
\var_n\{\bar{Q}\}=&\frac{1}{K}\sum_{i=1}^{N-M}\tilde{\sigma}_i^4\quad\text{Does not converge to $\frac{\sigma^4}{K}(N-M)$}
\end{align*}

whereby $\tilde{\sigma}_i^2$ are the eigenvalues of $\boldsymbol{\Gamma}\mathbf{D}$ and $\mathbf{D}=\diag(\sigma_i^2)$. For example, for $\tilde{\sigma}_i^2\sim\mathcal{R}(\sigma^2(1-c),\sigma^2(1+c))$, then $\var_n\{\bar{Q}\}=\frac{\sigma^4}{K}\left(1+\frac{c^2}{3}\right)(N-M)$.

§.§ The Behavior of the Multihypothesis Test

Detection characteristics for two targets using the ELAN 21L array with $du=0.55BW$.

Detection characteristics for two targets using the ELAN 21L array with $du=0.55BW$.

Simulations of the tests developed here with the threshold $\eta_{\alpha;M}$ as in (7.1-6) are rather complex, because many simulations are needed in order to make a statement regarding $\alpha$, particularly when the detection probability is high. Thus, for each specified target scenario, $500$ minimizations of $\bar{Q}$ for $M=1,\ldots, M_{\text{ex}}$ are performed. For each, $17$ iterations of the original algorithm, with bounding, are used. The accuracy of these simulations is shown in Fig. <ref>. The plots show the complete test $PD=P\{M=2\}$ for two targets with an SNR of $3,5,7\,\text{dB}$ and $PF1=P\{M>2\}$, which is set by $\alpha$ (The targets lie $0.55 BW$ apart, ELAN 21 L, Signal Model 4). Thus, given an estimated target number of $(M=1$, the estimated direction is assumed to lie in the middle between both targets, with $M=2$, the exact directions are assumed to be found; that is, precisely the conditions of Chapter 7.2. As helping lines, the curves of the optimal Neyman-Pearson test are shown for $PF2=0$ so that $PF1=\alpha$ and $PD=1-\alpha$. That the computed detection probability is not exactly reached by the simulations, is due to the small number of simulations $(500)$ and also the approximation of the $\chi^2$ distribution by the normal distribution. This is shown in Fig. <ref> for the curves with $SNR=3\,\text{dB}$. These curves were made with $5000$ trials and still show the influence of the approximation via the normal distribution. This seems to already be sufficiently accurate for $N=21$.

Detection characteristics for two targets using the ELAN 21L array with $du=0.55BW$.

For the same situation as Fig. <ref>, Fig. <ref> shows the actual detection probability with estimated parameters. The  test agrees quite well within the bounds of the simulation accuracy with the exact curves in Fig. <ref>. Figure <ref>  also shows that by demanding too small a value of $\alpha$, the detection probability can be dramatically reduced. A value of $\alpha$ between $5\%$ and $10\%$ seems to be a good choice. The standard deviation for the direction estimation is with estimated target number $M=2$, $0.04BW, 0.05BW, 0.06BW$ for the respective SNR values $7$, $5$, and $3\,\text{dB}$. This is within the bounds of the predicted dispersion according to Fig. <ref> (b), Curve 5, because the CRLB for $\Delta\psi=90^\circ$ corresponds to Signal Model $4$ and the variance is well described by the stochastic approximation of Fig. <ref>. Applying Curve 5 for the SNR of $3\,\text{dB}$ means that the stochastic approximation converging with 17 iterations corresponds to ML-estimation with averaging over $9$ samples.

Detection characteristics for two targets using the ELAN 21L array with $du=0.55BW$.

The influence of different noise powers on the receiver (the same target situation as in Fig. <ref>, SNR=$5\,\text{dB}$) is shown in Fig. <ref>. For the noise power, fluctuations of $1\,\text{dB}$ were assumed and $\sigma_i^2\sim\mathcal{R}\left(\sigma^2\left(1-\frac{1}{4}\right),\sigma^2\left(1+\frac{1}{4}\right)\right)$ was randomly drawn. Curve 1 shows the results for the test where $\sigma^2$ is used without knowledge of the fluctuations. In comparison to the theoretical curves, the detection probability drops and the test no longer attains a level $\alpha$. Therefore, the estimates $\hat{\boldsymbol{\omega}}$ are assumed to be exact. Curve 2 shows the test, also for exact $\hat{\boldsymbol{\omega}}$, but $\E\{\bar{Q}\}$ and $\var\{\bar{Q}\}$ corresponding to (7.2-5) for the threshold $\eta$ are used. Curve 3 shows the case where the parameter $\hat{\boldsymbol{\omega}}$ is estimated. The estimation error due to the fluctuations  in the noise power is negligible. The state of the amplifiers should be taken into account for the test if possible. For large numbers of elements, the difference should be even smaller.

Detection characteristics for two targets using the ELAN 21L array with $du=0.55BW$.

That the test is sensitive to deviations from the assumption that the noise in all of the channels is the same is shown by the behavior in the presence of a jammer. The test with the same target position as in Fig. <ref> is shown in Fig. <ref> for the case where there is a jammer in direction $u=-0.5$ as in (2.2-2) with finite width of $s=0.1$ with the effective value $\wp=\sigma/2$. The effective value of the targets is $\sqrt{\E\left\{|b_i|^2\right\}}=1.6\sigma$ so that the signal-to-internal noise ratio is $7\,\text{dB}$ and the SNR is $6\,\text{dB}$.

White noise having power $\sigma^2+\sigma^2/4$ is assumed for the test. Curve 1 shows the  test for the exact parameter values, Curve 2  for the estimated value $\hat{\boldsymbol{\omega}}$.The influence of the error in the estimator is negligible. The deviation of the given noise background, in which the signal is to be detected, is the primary source of the increased estimation error. If one were to only consider the internal noise $\sigma^2$ when computing the threshold $\eta$ as in (7.1-6), the detection probability would sink by $50\%$.

The worsening of the detection probability in the presence of jamming can only be solved if one computes $\E\{\bar{Q}\}, \var\{\bar{Q}\}$ for the given jammer angle and estimation target direction; that is, by estimating the direction and power of the jammer and assuming noise according to Model $4$ or by estimation of the jamming covariance matrix $\mathbf{R}$ and computation of $\E\{\bar{Q}\}=\trace\boldsymbol{\Gamma}\mathbf{R},\var\{\bar{Q}\}=\frac{1}{K}\trace\boldsymbol{\Gamma}\mathbf{R},\boldsymbol{\Gamma}\mathbf{R}$. Another possibility is that one takes all jammers (at a particular range) into account in the $Q$ function and estimates them as targets.

Detection characteristics for two targets using the ELAN 21L array with $du=0.55BW$.

If instead of using Signal Model $4$, Signal Model $3$ were used, the test (and the estimation according to Chapter 5.3) improves. This is shown in Figure <ref> for the case of exact estimation (target situation and estimate values as in Fig. <ref> for SNR=$3,5\,\text{dB}$). The curves for Signal Model $3$ (S3) are above the theoretical detection probability for Signal Model $4$ (S4). The level $\alpha$, however, is still held. The maintenance of level $\alpha$ is thus (with sufficiently accurate estimates) really independent of the signal model. The higher detection probability comes from the higher probability of $\bar{Q}>\eta$ at the level $M=1$.

One can thus view Signal Model $4$ as being unfavorable in the instances where one excludes the phase positions of $\Delta\varphi=0^\circ,180^\circ$ under Models $1$ and $2$. The curves for the theoretical detection probabilities can be viewed as lower limits on the detection probabilities under these other models. One can thus skip timely simulations and study the influence of other parameters.

The resolution of two targets using the ELAN 21L array with $du=0.5BW$.

The resolution of two targets with $du=0.5BW$.

For ELAN 21 L, Fig. <ref> shows the theoretical detection probability of two targets (separation of the targets is $0.5BW$) as a function of SNR for $\alpha=2,4,6,8,10\,\%$ and the number of samples of $\bar{Q}$ averaged $K=1,2,3$. The same detection probabilities as a function of the distance between targets are shown in Fig. <ref>, whereby the signal power $\E\{|S|^2\}$ is proportional to $\frac{1}{R^4}$ and $R0$ is the $3$-$\text{dB}$ distance of the whole SNR, where

\begin{equation}
\frac{\E\{2N|b|^2\}}{\E\{|n|^2\}}=2.
\end{equation}

Even a small number of samples of $Q$ result in large improvements.

Detection characteristics for two targets using the ELAN 25 array with $du=0.5BW$.

The resolution of two targets with $du=0.5BW$.

The coupling only has a small influence. This is shown in Fig. <ref> for ELAN 25 for two targets with a separation of $BW/2$ and SNR$=5\,\text{dB}$. However, the level $\alpha$ is not quite held anymore, which can be explained by the fact that at the level $M=2$, $\boldsymbol{\Gamma}\mathbf{s}\neq\mathbf{0}$ even when using exact direction estimates. The standard deviation of the directional estimates at the level $M=2$ is, with and without coupling, only about $0.08BW$. For the same situation, Fig. <ref> shows the SNR-loss due to coupling for ELAN 25 and 29. For ELAN 39, with the highest element density, the estimation is better, but the loss due to stronger coupling is larger at about $1\,\text{dB}$.

In all, the theoretical detection probabilities provide the ability to investigate the behavior of the test, because the parameter estimation is generally sufficient and the error level $\alpha$ is independent of the signal model and is thus attained. However, the resolution of target in arbitrary Gaussian noise requires an exact knowledge of the first two moments of this noise.

§ SUMMARY

* The resolution problem was formulated for spatially and temporally sampled, overlapping, monochromatic waves with temporally fluctuating amplitudes and phases in Gaussian noise. A solution is theoretically possible using a sequence of estimates from performing likelihood maximization in likelihood ratio tests.

* To assure the existence and uniqueness of a solution, an appropriate restriction of the angle of incidence of the wave is necessary. When only sampling spatially,
a strong $M$-regular antenna layout is necessary to resolve $M$ targets. When also sampling temporally, only a weak $M$-regular layout is needed. These conditions are particularly important when one wishes to resolve targets using as few elements as possible.

* The condition of the estimate can be analyzed by studying the curvature of the $Q$-function and $\E\{Q\}$ at the minimum point. This essentially corresponds to an application of the CRLB.

* Resolution using only a single impulse, which means only with spatial sampling, demands a very high SNR (over $25\,\text{dB}$). Additionally, there are target constellations (depending primarily upon the phase difference of the targets) that even with high SNRs can only be poorly resolved. Phase fluctuations by multiple temporal samples significantly improve the direction estimation. Thus, an increase in the angular resolution necessitates a worsening of the Doppler estimate, corresponding to a direction-frequency ambiguity relation.

* A simple, sequential estimation algorithm for direction estimation using the stochastic approximation is given. This procedure can still be refined. The procedure uses only the current sum-and-difference beam outputs in each iteration as its inputs. The asymptotic behavior of the algorithm is well approximated by the CRLB.

* The stochastic approximation algorithm is relatively robust to deviations from the assumed signal and noise model. Interference powers on the order of the target amplitudes have little influence on the estimates as long as they are located far from the beam.

* A sequential multihypothesis tests was developed to determine the number of targets present. The test can be used with the stochastic approximation and is computationally efficient. Additionally, the test maintains a firm probability that the number of targets present is overestimated.

* Detection and false-alarm probabilities for the resolution of noise sources (Signal Model $4$) can be calculated for this test. Because this model can be viewed as being pessimistic, such probabilities have meanings for the other models as well.

* The test is not very sensitive against signal distortions, as may result from mutual coupling for thinned arrays. It is however sensitive against deviations from the assumed noise model, because this is the basis of the detection criterion for the different signals. This sensitivity can be mitigated by interference pre-filtering, or by including jammers as additional targets to be estimated, or by estimating the first two moments of the interference. Another promising way could be the development of a non-parametric test procedure.

* The attainable resolution can be shown through two examples:

* For the planar array ELAN 25, two targets located $0.5$ beamwidths apart were correctly detected $76\%$ of the time. In $5\%$ of cases, the number of targets was overestimated (a hard level of $\alpha=0.05$). The target directions were estimated with a standard deviation of 0.08 beamwidths in 17 iterations with an average SNR of $5\,\text{dB}$ per element.

* For the linear antenna ELAN 21 L, two targets located $0.5$ beamwidths apart were correctly detected $95\%$ of the time with $\alpha=0.02$. The standard deviation of the angular estimates was $0.04$ beamwidths after $17$ iterations.

§.§ Proofs for Chapter 4

Proof of (4.1-6)(A.1-1)

Without loss of generality, let $\mathbf{A}=(\mathbf{a}(\omega_1),\ldots,\mathbf{a}(\omega_M),\mathbf{a}(\omega_M),\ldots,\mathbf{a}(\omega_M))$ with $\omega_i\neq\omega_k$ for all $i,k\in\{1,\ldots, M\}$, $i\neq k$. $\mathbf{A}$ is an $N\times(M+k)$ matrix.

For such singular $\mathbf{A}$ matrices, one can solve $\min\limits_{\mathbf{b}}\lVert\mathbf{a}-\mathbf{A}\mathbf{b}\rVert^2$ using a Householder transformation [30]. That is, through the use of a unitary matrix $\mathbf{U}$ that brings $\mathbf{A}$ into a triangular form. Then $\lVert\mathbf{z}-\mathbf{A}\mathbf{b}\rVert^2=\lVert\mathbf{U}\mathbf{z}-(\mathbf{U}\mathbf{A})\mathbf{b}\rVert^2$.

Let $\mathbf{A}$ be partitioned in the form $\mathbf{A}=(\mathbf{A}_1\vline\mathbf{A}_2)$, where $\mathbf{A}_1$ is a regular $N\times M$ matrix, and $\mathbf{A}_2$ is an $N\times k$ matrix with rank $\mathbf{A}_2=1$. The corresponding partitioning is applied to the other terms:

\begin{align*}
\mathbf{b}=&\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\mathbf{b}_1\\
\hline\\
\mathbf{b}_2
\IEEEstrut
\end{IEEEeqnarraybox}\right)&\mathbf{y}:=&\mathbf{U}\mathbf{z}&\mathbf{y}=&\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\mathbf{y}_1\\
\hline\\
\mathbf{y}_2
\IEEEstrut
\end{IEEEeqnarraybox}\right)
\end{align*}

where $\mathbf{b}_1\in\mathbb{C}^M$, $\mathbf{b}_2\in\mathbb{C}^k$, $\mathbf{y}_1\in\mathbb{C}^M$, and $\mathbf{y}_2\in\mathbb{C}^{N-M}$. Thus, $\mathbf{U}\mathbf{A}$ has the form

\begin{equation*}
\mathbf{U}\mathbf{A}=\left(\begin{array}{cccc|cccc}
*		&* 		&\ldots	&*		&*		&* 		&\ldots	&*\\
0		&*		&\ldots	&*		&*		&* 		&\ldots	&*\\
\vdots	&\vdots	&\ddots	&\vdots	&\vdots	&\vdots	&\ddots	&\vdots\\
0		&0		&\ldots  	&*		&*		&*		&\ldots  	&*\\
0		&0		&\ldots  	&0		&0		&0		&\ldots  	&0\\
\vdots	&\vdots	&\ddots	&\vdots	&\vdots	&\vdots	&\ddots	&\vdots\\
0		&0		&\ldots	&0		&0		&0		&\ldots	&0
\end{array}\right)=\left(\begin{array}{c|c}
\mathbf{B}_1&\mathbf{B}_2\\
\mathbf{0}&\mathbf{0}
\end{array}\right)
\end{equation*}

where the asterisks denotes nonzero values. Consequently,

\begin{equation*}
\lVert\mathbf{U}\mathbf{z}-\mathbf{u}\mathbf{A}\mathbf{b}\rVert^2=\left\lVert\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\mathbf{y}_1-\mathbf{B}_1\mathbf{b}_1-\mathbf{B}_2\mathbf{b}_2\\
\mathbf{y}_2
\IEEEstrut
\end{IEEEeqnarraybox}\right)\right\rVert^2.
\end{equation*}

This expression is minimized for $\mathbf{y}_1=\mathbf{B}_1\mathbf{b}_1+\mathbf{B}_2\mathbf{b}_2$ (For example, $\mathbf{b}_2=\mathbf{0}$, $\mathbf{b}_1=\mathbf{B}_1^{-1}\mathbf{y}_1$) and has the minimal value $\lVert\mathbf{y}_2\rVert^2$, which is determined by the Householder transformation $\mathbf{U}$. However, $\mathbf{U}$ is only determined by $\mathbf{A}_1$.

Proof of (4.1-7)

Since $\boldsymbol{\Gamma}=\mathbf{I}-\mathbf{A}\left(\mathbf{A}^*\mathbf{A}\right)^{-1}\mathbf{A}^*$ and $\mathbf{A}$ is infinitely differentiable, it remains to show that $(\mathbf{A}^*\mathbf{A})^{-1}$ is infinitely differentiable.

\begin{equation*}
\mathbf{S}\left(\boldsymbol{\omega}\right):=\mathbf{A}^*(\boldsymbol{\omega})\mathbf{A}(\boldsymbol{\omega})
\end{equation*}

and consider

\begin{equation*}
\mathbf{F}:{\left(V^M \backslash \gamma\right) \times \mathbb{C}^{M\times M}\rightarrow \mathbb{C}^{M\times M}\atop (\boldsymbol{\omega},\mathbf{T})\mapsto \mathbf{S}(\boldsymbol{\omega})\mathbf{T}-\mathbf{I}}.
\end{equation*}

Then $\mathbf{F}$ implicitly defines the function $\mathbf{S}^{-1}(\boldsymbol{\omega})$ because

\begin{equation*}
\mathbf{F}(\boldsymbol{\omega},\mathbf{T})=\mathbf{0}\Longleftrightarrow \mathbf{T}=\mathbf{S}^{-1}(\boldsymbol{\omega}).
\end{equation*}

For real differentiability, let $\mathbf{F}$ be written as a real function:

\begin{equation*}
\mathbf{F}(\boldsymbol{\omega},\mathbf{T})=\left(\begin{IEEEeqnarraybox}[\relax][c]{c'c}
\IEEEstrut
\text{Re } \mathbf{S}&-\text{Im } \mathbf{S}\\
\text{Im } \mathbf{S}&\text{Re }  \mathbf{S}
\IEEEstrut
\end{IEEEeqnarraybox}\right)\left(\begin{IEEEeqnarraybox}[\relax][c]{c'c}
\IEEEstrut
\text{Re }  \mathbf{T}&-\text{Im } \mathbf{T}\\
\text{Im } \mathbf{T}&\text{Re }  \mathbf{T}
\IEEEstrut
\end{IEEEeqnarraybox}\right)-\left(\begin{IEEEeqnarraybox}[\relax][c]{c'c}
\IEEEstrut
\mathbf{I}&\mathbf{0}\\
\mathbf{0}&\mathbf{I}
\IEEEstrut
\end{IEEEeqnarraybox}\right).
\end{equation*}

Thus, $\mathbf{F}$ is infinitely differentiable with regard to $(\boldsymbol{\omega},\text{Re }\mathbf{T},\text{Im }\mathbf{T})$ and the Jacobian determinants are

\begin{equation*}
\det\left(\frac{\partial F_{ij}}{\partial t_{kl}}\right)_{i,j=1\ldots 2M\atop k,l=1\ldots 2M}=\det\left(
\begin{IEEEeqnarraybox}[\relax][c]{c'c'c'c}
\IEEEstrut
\mathbf{S}&\mathbf{0}&\ldots&\mathbf{0}\\
\mathbf{0}&\mathbf{S}&\ldots&\mathbf{0}\\
\vdots	&\vdots	&\ddots&\vdots\\
\mathbf{0}&\mathbf{0}&\ldots&\mathbf{S}
\IEEEstrut
\end{IEEEeqnarraybox}\right)
\end{equation*}

because $\mathbf{S}$ is regular on $V^M\backslash \gamma$.

Thus, the implicit function theorem holds that $\mathbf{S}^{-1}(\boldsymbol{\omega})$ is likewise infinitely differentiable, though the derivatives from $\mathbf{S}^{-1}$ and $\mathbf{S}$ can indeed be singular matrices (For example, $\mathbf{S}_{\omega_i\omega_k}$ ($i\neq k$) is singular).

Computation of the First and Second Derivatives of $Q$

* Lemma 1

\begin{equation*}
\frac{\partial}{\partial u_i}(\mathbf{A}^*)=\left(\frac{\partial \mathbf{A}}{\partial u_i}\right)^*\quad \text{for $i=1,\ldots, M$}
\end{equation*}

An analogous statement can be made for $v_i$. The proof is through substitution.

* Lemma 2

For all regular differentiable matrices $\mathbf{M}(\mathbf{u})$ and $ \mathbf{M}^{-1}(\mathbf{u})$, the following holds

\begin{equation*}
\frac{\partial}{\partial u_i}(\mathbf{M}^{-1})=-\mathbf{M}^{-1}\frac{\partial \mathbf{M}}{\partial u_i}\mathbf{M}^{-1}.
\end{equation*}

\begin{equation*}
\mathbf{M}\mathbf{M}^{-1}=\mathbf{I}\Longrightarrow \frac{\partial\mathbf{M}}{\partial u_i}\mathbf{M}^{-1}+\mathbf{M}\frac{\partial\mathbf{M}^{-1}}{\partial u_i}=\mathbf{0}.
\end{equation*}

In the following the abbreviation $f_{u_i}:=\frac{\partial f}{\partial u_i}$. Then one can write

\begin{equation*}
\end{equation*}

\begin{align}
\boldsymbol{\Gamma}_{u_i}=&-\frac{\partial}{\partial u_i}\mathbf{A}\left(\mathbf{A}^*\mathbf{A}\right)^{-1}\mathbf{A}^*\notag\\
\end{align}

In particular, because $\boldsymbol{\Gamma}\mathbf{A}=\mathbf{0}$,

\begin{align*}
\mathbf{A}^*\boldsymbol{\Gamma}_{u_i}\mathbf{A}=\mathbf{0}\\
\trace\boldsymbol{\Gamma}_{u_i}=0
\end{align*}

\begin{align}
\E\left\{Q_{u_i}\right\}=&\trace\boldsymbol{\Gamma}_{u_i}\E\left\{\mathbf{z}\mathbf{z}^*\right\}\notag\\
\end{align}

so that

\begin{equation*}
\left.\E\left\{Q_{ui}\right\}\right|_{\bar{\boldsymbol{\omega}}}=\mathbf{0}
\end{equation*}

\begin{align*}
Q_{u_i}=&-2\text{Re }\mathbf{z}^*\mathbf{A}\left(\mathbf{A}^*\mathbf{A}\right)^{-1}\mathbf{A}_{u_i}^*\boldsymbol{\Gamma}\mathbf{z}\\
=&-2\text{Re }\hat{\mathbf{b}}^*\mathbf{A}^*_{u_i}\left(\mathbf{z}-\mathbf{A}\hat{\mathbf{b}}\right)
\end{align*}

\begin{align*}
\hat{\mathbf{b}}:=&\left(\mathbf{A}^*\mathbf{A}\right)^{-1}\mathbf{A}^*\mathbf{z}\\
\mathbf{A}_{u_i}=&\left(\begin{IEEEeqnarraybox}[\relax][c]{c'c'c'c'c}
\IEEEstrut
0		&\ldots	&-jx_ia_{11}	&\ldots	&0\\
\vdots	&\ddots	&\vdots		&\ddots	&0\\
0		&\ldots	&-jx_Na_{Ni}	&\ldots	&0
\IEEEstrut
\end{IEEEeqnarraybox}\right)
\end{align*}

where only the $i$th column of $\mathbf{A}_{u_i}$ is nonzero. Thus,

\begin{align*}
\hat{\mathbf{b}}^*\mathbf{A}^*_{u_i}=j\hat{b}_i^*\mathbf{a}_i^*\mathbf{D}_x\\
\mathbf{D}_x=\diag\left(x_k \right)
\end{align*}

and then

\begin{align}
Q_{u_i}=&-2\text{Re }j\hat{b}^*_i\left(\mathbf{a}_i^*\mathbf{D}_x\mathbf{z}-\mathbf{a}_i^*\mathbf{D}_x\mathbf{A}\hat{\mathbf{b}}\right)\notag\\
=&2\text{Im }\hat{b}_i^*\left(\mathbf{a}_i^*\mathbf{D}_x\mathbf{z}-\mathbf{a}_i^*\mathbf{D}_x\mathbf{A}\hat{\mathbf{b}}\right)\notag\\
Q_{v_i}=&2\text{Im }\hat{b}_i^*\left(\mathbf{a}_i^*\mathbf{D}_y\mathbf{z}-\mathbf{a}_i^*\mathbf{D}_y\mathbf{A}\hat{\mathbf{b}}\right)\label{eqA-1-4}\\
Q_{\mathbf{u}}=&2\text{Im }\diag(\hat{b}_i^*)\left(\mathbf{A}^*\mathbf{D}_x\mathbf{z}-\mathbf{A}^*\mathbf{D}_x\mathbf{A}\hat{\mathbf{b}}\right)\notag
\end{align}

and $Q_{\mathbf{v}}$ is the same as $Q_{\mathbf{u}}$ with the appropriate substitutions.

For the second derivative, the following holds:

\begin{equation*}
Q_{\omega_i\omega_k}=\mathbf{z}^*\boldsymbol{\Gamma}_{\omega_i\omega_k}\mathbf{z}\quad(\omega_i=u_i\text{ or }v_i)
\end{equation*}

\begin{align}
\boldsymbol{\Gamma}_{\omega_i\omega_k}=&-\boldsymbol{\Gamma}\left(\mathbf{A}_{\omega_i}\mathbf{A}^{-1}\mathbf{A}_{\omega_k}^*+\mathbf{A}_{\omega_k}\mathbf{S}^{-1}\mathbf{A}_{\omega_i}^*\right)\boldsymbol{\Gamma}\notag\\
\end{align}

\begin{equation*}
\mathbf{S}:=\mathbf{A}^*\mathbf{A}.
\end{equation*}

In particular, because $\boldsymbol{\Gamma}\mathbf{A}=\mathbf{0}$,

\begin{align}
\left.\mathbf{s}^*\boldsymbol{\Gamma}_{\omega_i\omega_k}\mathbf{s}\right|_{\bar{\boldsymbol{\omega}}}=&\mathbf{b}^*\mathbf{A}^*\boldsymbol{\Gamma}_{\omega_i\omega_k}\mathbf{A}\mathbf{b}\notag\\
=&2\text{Re }\left(b_k^*\mathbf{a}^*_{k,\omega_k}\boldsymbol{\Gamma}\mathbf{a}_{i,\omega_i}b_i\right)\notag\\
=&2\text{Re }\left(b_k^*\mathbf{a}^*_k\mathbf{D}\boldsymbol{\Gamma}\mathbf{D}\mathbf{a}_ib_i\right)\quad (\text{$\mathbf{D}=\mathbf{D}_x$ or $\mathbf{D}_y$}).\label{eqA-1-6}
\end{align}

\begin{align}
\mathbf{a}_i^*\boldsymbol{\Gamma}_{\omega_i\omega_k}\mathbf{a}_k=&\delta_{ik}2\text{Re }\mathbf{a}^*_{i,\omega_i}\boldsymbol{\Gamma}\mathbf{a}_{i,\omega_i}\notag\\
=&\delta_{ik}2\text{Re }\mathbf{a}_i^*\mathbf{D}\boldsymbol{\Gamma}\mathbf{D}\mathbf{a}_i\label{eqA-1-7}
\end{align}

\begin{equation}
\trace\boldsymbol{\Gamma}_{\omega_i,\omega_k}=0\label{eqA-1-8}
\end{equation}

because then the first two terms in (A.1-5) vanish.

§.§ Proof of Theorem (4.3-2):

Given for $\mathbf{z}=\mathbf{s}+\mathbf{n}$, $\mathbf{z}\sim\mathcal{N}(\mathbf{s},\mathbf{I})$:

\begin{align*}
\E\left\{Q\right\}=&\E\left\{\mathbf{s}^*\boldsymbol{\Gamma}\mathbf{s}+\mathbf{n}^*\boldsymbol{\Gamma}\mathbf{s}+\mathbf{s}^*\boldsymbol{\Gamma}\mathbf{n}+\mathbf{n}^*\boldsymbol{\Gamma}\mathbf{n}\right\}\\
\end{align*}

\begin{equation*}
\trace\boldsymbol{\Gamma}=\trace\mathbf{I}-\trace\left(\mathbf{A}^*\mathbf{A}\right)^{-1}\mathbf{A}^*\mathbf{A}.
\end{equation*}

At the point $\boldsymbol{\omega}_0$, where $Q_{\boldsymbol{\omega}}(\boldsymbol{\omega}_0)=\mathbf{0}$, the matrix of the second F-tensor is

\begin{equation*}
\mathbf{S}(\boldsymbol{\omega}_0)=\mathbf{s}^*\boldsymbol{\Gamma}_{\boldsymbol{\omega}\boldsymbol{\omega}}(\boldsymbol{\omega}_0)\mathbf{s}.
\end{equation*}

Furthermore using (3.1-10) for $\boldsymbol{\vartheta}=(\mathbf{u},\mathbf{v},\text{Re }b_1,\text{Im }b_2,\ldots. \text{Re }g_m,\text{Im }b_M)$,

\begin{align}
F_{ik}=&\int\frac{\partial \ln p}{\partial \vartheta_i}(\mathbf{z};\mathbf{s}(\boldsymbol{\vartheta}))\frac{\partial \ln p}{\partial \vartheta_k}(\mathbf{z};\mathbf{s}(\boldsymbol{\vartheta}))p(\mathbf{z};\mathbf{s}(\boldsymbol{\vartheta}))d\mathbf{z}\notag\\
=&-\int\frac{\partial^2 \ln p}{\partial \vartheta_i\partial \vartheta_k}(\mathbf{z};\mathbf{s}(\boldsymbol{\vartheta}))p(\mathbf{z};\mathbf{s}(\boldsymbol{\vartheta}))d\mathbf{z}\label{eqA-2-1}
\end{align}

\begin{equation*}
\frac{\partial^2 \ln p}{\partial \vartheta_i \partial \vartheta_k}=-\frac{1}{p^2}\frac{\partial p}{\partial\vartheta_i}\frac{\partial p}{\partial\vartheta_k}+\frac{1}{p}\frac{\partial^2 p}{\partial \vartheta_i\partial \vartheta_k}
\end{equation*}

\begin{equation*}
\int\frac{\partial^2 p}{\partial \vartheta_i \partial \vartheta_k}d\mathbf{z}=0
\end{equation*}

assuming that of the integrals exist.

\begin{equation*}
\end{equation*}

one must find the derivatives of

\begin{equation*}
\tilde{Q}(\boldsymbol{\vartheta})=\left(\mathbf{z}-\mathbf{s}\left(\boldsymbol{\vartheta}\right)\right)^*\left(\mathbf{z}-\mathbf{s}\left(\boldsymbol{\vartheta}\right)\right)
\end{equation*}

\begin{equation*}
\mathbf{s}\left(\boldsymbol{\vartheta}\right)=\mathbf{A}\mathbf{b}
\end{equation*}

\begin{align*}
\tilde{Q}_{\vartheta_i}=&-2\text{Re }\mathbf{s}^*_{\vartheta_i}(\mathbf{z}-\mathbf{s})\\
\tilde{Q}_{\vartheta_i\vartheta_k}=&-2\text{Re}\left[\mathbf{s}_{\vartheta_i\vartheta_k}(\mathbf{z}-\mathbf{s})-\mathbf{s}^*_{\vartheta_i}\mathbf{s}_{\vartheta_k}\right]
\end{align*}

and also for $\mathbf{z}\sim\mathcal{N}(\mathbf{s},\mathbf{I})$

\begin{equation*}
\E\left\{\tilde{Q}_{\vartheta_i\vartheta_k}\right\}=F_{ik}=2\text{Re }\mathbf{s}^*_{\vartheta_i}\mathbf{s}_{\vartheta_k}
\end{equation*}

\begin{equation*}
\mathbf{F}=2\text{Re }\mathbf{s}^*_{\boldsymbol{\vartheta}}\mathbf{s}_{\boldsymbol{\vartheta}}
\end{equation*}

\begin{equation*}
\left(\mathbf{s}_{\boldsymbol{\vartheta}}\right)_{i,k}=\frac{\partial s_i}{\partial \vartheta_k}.
\end{equation*}

\begin{equation*}
\mathbf{s}_{\omega_i}=\mathbf{a}_{i,\omega_i}b_i
\end{equation*}

\begin{equation*}
\mathbf{s}_{\alpha_i}=\mathbf{a}_i
\end{equation*}

for $\alpha_i=\text{Re }b_i$ and

\begin{equation*}
\mathbf{s}_{\beta_i}=i\mathbf{a}_i
\end{equation*}

for $\beta_i=\text{Im }b_i$. With that, $\mathbf{F}$ has the form

\begin{equation*}
\mathbf{F}=2\text{Re }\left(\begin{array}{c|cc}
\mathbf{B}^*\mathbf{A}^*_{\boldsymbol{\omega}}\mathbf{A}_{\boldsymbol{\omega}}\mathbf{B}&\mathbf{B}^*\mathbf{A}^*_{\boldsymbol{\omega}}\mathbf{A}&j\mathbf{B}^*\mathbf{A}^*_{\boldsymbol{\omega}}\mathbf{A}\\
\hline
\mathbf{A}^*\mathbf{A}_{\boldsymbol{\omega}}\mathbf{B}&\mathbf{A}^*\mathbf{A}&j\mathbf{A}^*\mathbf{A}\\
\end{array}\right)
\end{equation*}

\begin{align*}
\mathbf{B}:=&\diag\limits_{k=1}^M(b_k)&\left(\mathbf{A}_{\boldsymbol{\omega}}\right)_{ik}=&\frac{\partial a_{ik}}{\partial \omega_k}.
\end{align*}

The matrix $\mathbf{F}$ can be simplified to

\begin{equation*}
\mathbf{F}=\left(\begin{array}{c|cc}
\text{Re }\mathbf{M}&\text{Re }\mathbf{R}&-\text{Im }\mathbf{R}\\
\hline
\text{Re }\mathbf{R}^T&\text{Re }\mathbf{G}&-\text{Im} \mathbf{G}\\
\text{Im }\mathbf{R}^T&\text{Im}\mathbf{G}&\text{Re }\mathbf{G}
\end{array}\right)
\end{equation*}

\begin{align*}
\mathbf{M}:=&2\mathbf{B}^*\mathbf{A}^*_{\boldsymbol{\omega}}\mathbf{A}_{\boldsymbol{\omega}}\mathbf{B}&\mathbf{R}:=&2\mathbf{B}^*\mathbf{A}^*_{\boldsymbol{\omega}}\mathbf{A}_{\boldsymbol{\omega}}&\mathbf{G}:=&2\mathbf{A}^*\mathbf{A}.
\end{align*}

It is to be shown that the block corresponding to $\text{Re }\mathbf{M}$ in $\mathbf{F}^{-1}$ equals $\mathbf{S}^{-1}\left(\boldsymbol{\omega}_0\right)$. For every regular matrix, one can write

\begin{equation}
\left(\begin{IEEEeqnarraybox}[\relax][c]{c'c}
\IEEEstrut
\mathbf{M}&\mathbf{R}\\
\mathbf{R}'&\mathbf{G}
\IEEEstrut
\end{IEEEeqnarraybox}\right)^{-1}=\left(\begin{IEEEeqnarraybox}[\relax][c]{c'c}
\IEEEstrut
\mathbf{E}&\mathbf{F}\\
\mathbf{F}'&\mathbf{H}
\IEEEstrut
\end{IEEEeqnarraybox}\right)\rightarrow\mathbf{E}^{-1}=\mathbf{M}-\mathbf{R}\mathbf{G}^{-1}\mathbf{R}^T.\label{eqA2-2}
\end{equation}

Thus, it can be shown that

\begin{equation*}
\mathbf{s}^*\boldsymbol{\Gamma}_{\boldsymbol{\omega}\boldsymbol{\omega}}\mathbf{s}=\text{Re }\mathbf{M}-\left(\text{Re }\mathbf{R},-\text{Im }\mathbf{R}\right)\left(\begin{IEEEeqnarraybox}[\relax][c]{c'c}
\IEEEstrut
\text{Re }\mathbf{G}&-\text{Im } \mathbf{G}\\
\text{Im }\mathbf{G}&\text{Re }\mathbf{G}
\IEEEstrut
\end{IEEEeqnarraybox}\right)^{-1}\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\text{Re }\mathbf{R}^T\\
\text{Im }\mathbf{R}^T
\IEEEstrut
\end{IEEEeqnarraybox}\right).
\end{equation*}

Now there is

\begin{equation*}
\left(\begin{IEEEeqnarraybox}[\relax][c]{c'c}
\IEEEstrut
\text{Re }\mathbf{G}&-\text{Im } \mathbf{G}\\
\text{Im }\mathbf{G}&\text{Re }\mathbf{G}
\IEEEstrut
\end{IEEEeqnarraybox}\right)^{-1}=\left(\begin{IEEEeqnarraybox}[\relax][c]{c'c}
\IEEEstrut
\text{Re }(\mathbf{G}^{-1})&-\text{Im }(\mathbf{G}^{-1})\\
\text{Im }(\mathbf{G}^{-1})&\text{Re }(\mathbf{G}^{-1})
\IEEEstrut
\end{IEEEeqnarraybox}\right)
\end{equation*}

due to the real representation of complex numbers $\alpha+j\beta$ in $2\times 2$ matrices $\left(\begin{IEEEeqnarraybox}[\relax][c]{c'c}
\IEEEstrut
\alpha&-\beta\\
\beta&\alpha
\IEEEstrut
\end{IEEEeqnarraybox}\right)$.

Thus, it can be shown that

\begin{equation*}
\left(\begin{IEEEeqnarraybox}[\relax][c]{c'c}
\IEEEstrut
\text{Re }\mathbf{R}&-\text{Im }\mathbf{R}
\IEEEstrut
\end{IEEEeqnarraybox}\right)\left(\begin{IEEEeqnarraybox}[\relax][c]{c'c}
\IEEEstrut
\text{Re }\mathbf{G}&-\text{Im } \mathbf{G}\\
\text{Im }\mathbf{G}&\text{Re }\mathbf{G}
\IEEEstrut
\end{IEEEeqnarraybox}\right)^{-1}\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\text{Re }\mathbf{R}^T\\
\text{Im }\mathbf{R}^T
\IEEEstrut
\end{IEEEeqnarraybox}\right)=\text{Re }\mathbf{R}\mathbf{G}^{-1}\mathbf{R}^*
\end{equation*}

\begin{align}
\text{Re }\left(\mathbf{M}-\mathbf{R}\mathbf{G}^{-1}\mathbf{R}^*\right)=&2\text{Re }\left(\mathbf{B}^*\mathbf{A}^*_{\boldsymbol{\omega}}\mathbf{A}_{\boldsymbol{\omega}}\mathbf{B}-\mathbf{B}^*\mathbf{A}^*_{\boldsymbol{\omega}}\mathbf{A}\left(\mathbf{A}^*\mathbf{A}\right)^{-1}\mathbf{A}^*\mathbf{A}_{\boldsymbol{\omega}}\mathbf{B}\right)\notag\\
=&2\text{Re }\mathbf{B}^*\mathbf{A}^*_{\boldsymbol{\omega}}\boldsymbol{\Gamma}\mathbf{A}_{\boldsymbol{\omega}}\mathbf{B}\notag\\
\end{align}

where the last step uses (A.1-6).

§.§ Proof of Theorem (5.1-3):

Relating to (i)

It is sufficient, as in Chapter <ref> to show the linear independence of three components of three vectors $\mathbf{a}(u_1)$, $\mathbf{a}(u_2)$, $\mathbf{a}(u_3)$. Letting the element positions be $x_1=0$, $x_2=d$, $x_3=rd$ with $r>1$, then it is to be shown that

\begin{equation*}
\left|\begin{IEEEeqnarraybox}[\relax][c]{c'c'c}
\IEEEstrut
\IEEEstrut
\end{IEEEeqnarraybox}\right|\neq 0.
\end{equation*}

Without loss of generality, one can choose $u_1=0$ (eliminating $e^{-jdu_1}$ and $e^{-jdru_1}$), so that with $\gamma_i:=e^{-jdu_i}$, one gets

\begin{equation*}
\left|\begin{IEEEeqnarraybox}[\relax][c]{c'c'c}
\IEEEstrut
\IEEEstrut
\end{IEEEeqnarraybox}\right|=0\Longleftrightarrow \frac{\gamma_2^r-1}{\gamma_2-1}=\frac{\gamma_3^r-1}{\gamma_3-1}
\end{equation*}

Additionally, considering the function

\begin{equation*}
f(u)=e^{-j\frac{r-1}{2}du}\frac{\sin r\frac{d}{2}u}{\sin \frac{d}{2}u}
\end{equation*}

one can then write

\begin{equation*}
f(u_i)=\frac{\gamma^r_i-1}{\gamma_i-1},\quad i=2,3.
\end{equation*}

\begin{equation*}
f(u_2)=f(u_3)\Longleftrightarrow\left(\frac{r-1}{2}du_2=\frac{r-1}{2}du_2+2\pi k,k\in\mathbb{Z}\right)
\end{equation*}

and $(u_2=-u_3)$ if $u$ falls within the beamwidth of the $3$-element pattern, $|u|\leq \frac{\lambda}{rd}$.

For $u=u_2=-u_3$ and $d$ in centimeters $\frac{2\pi}{\lambda}d$ instead of $d$, compare to (2.1-2) then $\frac{2\pi}{\lambda}(r-1)du=2\pi k$, meaning that $u=\frac{\lambda}{(r-1)d}k$. Because $|u|\leq\frac{\lambda}{rd}$, this condition is not fulfilled, and so the determinant is never $0$.

Relating to (ii)

For the Forward Equivalence $\Longrightarrow$:

Given three directions $\omega_0,\omega_1,\omega_2\in \Omega$ with the same elevation $v$, meaning that $\omega_i=(u_i,v)$, then

\begin{align*}
\mathbf{a}(\omega_i)=\mathbf{D}\left(\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
\vdots\\
\IEEEstrut
\end{IEEEeqnarraybox}\right)\\
\end{align*}

where $\mathbf{D}=\diag\left(e^{-jy_kv}\right)$ so that

\begin{equation*}
\sum_{i=0}^2b_i\mathbf{a}(\omega_i)=\mathbf{0}\Longleftrightarrow\sum_{i=0}^2b_i\mathbf{D}\mathbf{a}_x(u_i)=\mathbf{0}.
\end{equation*}

For the Reverse Equivalence $\Longleftarrow$:

Assuming that $\mathbf{x},\mathbf{y}$ are not weakly $2$-regular on $\Omega$, there exists $\omega_0,\omega_1,\omega_2\in\Omega$ so that $\mathbf{a}(\omega_0)=b_1\mathbf{a}(\omega_1)+b_2\mathbf{a}(\omega_2)$. Without loss of generality, it can be assumed that $\omega_0=(0,v)$, $\omega_1=(u_1,0)$, $\omega_2=(u_2,0)$, and that the coordinate system for $\mathbf{x}, \mathbf{y}$ are appropriately chosen as in Fig. <ref>. Then, for all $k=1\ldots N$:

\begin{equation*}
\end{equation*}

if $b_i=\beta_ie^{j\psi_i}$, and respectively

\begin{equation*}
\end{equation*}

with $\xi_1=x_ku_i-y_kv+\psi_i$. For a given $\beta_1$, $\beta_2$, these equations have only two solutions $(\mu_1,\mu_2)$ and $(-\mu_1,-\mu_2)$ as in Fig. <ref>.

\begin{equation*}
\left\{\begin{IEEEeqnarraybox}[\relax][c]{c'l}
\IEEEstrut
x_ku_1-y_kv+\psi_1=\mu_1+2\pi l&\\
x_ku_2-y_kv+\psi_2=\mu_2+2\pi l&(l=\ldots-2,-1,0,1\ldots)
\IEEEstrut
\end{IEEEeqnarraybox}\right.
\end{equation*}

\begin{equation*}
\left\{\begin{IEEEeqnarraybox}[\relax][c]{c}
\IEEEstrut
x_ku_1-y_kv+\psi_1=-\mu_1+2\pi l\\
x_ku_2-y_kv+\psi_2=-\mu_2+2\pi l
\IEEEstrut
\end{IEEEeqnarraybox}\right.
\end{equation*}

meaning that

\begin{equation*}
x_k=\frac{\pm(\mu_1-\mu_2)-(\psi_1-\psi_2)}{u_1-u_2}+\frac{2\pi l}{u_1-u_2}
\end{equation*}

independently of $v$. Thus, within a multiple of $2\pi$, such a non-regular positioning only has two different $x$ (and $y$) coordinates, (because $\left(\begin{IEEEeqnarraybox}[\relax][c]{c'c}
\IEEEstrut
\IEEEstrut
\end{IEEEeqnarraybox}\right)$ is regular if $u_1\neq u_2$).
These $x$ (or respectively $y$) coordinates are, however, not $2$-regular, because for the direction $u=0$, $u_1,u_2$, the associated direction vectors $\mathbf{a}(u)$ are also linearly independent due to (*) (given the same $\beta_1,\beta_2,\psi_1\psi_2$).

For $|u_1-u_2|\leq BW\leq\frac{\lambda}{D}$ ($D$ being the aperture diameter in centimeters) there is no $2\pi$ repetition, because then with

\begin{equation*}
\frac{2\pi}{\lambda}|x_k-x_l|=\frac{2\pi r}{|u_1-u_2|}
\end{equation*}

with $k,l,r\in\mathbb{Z}$, and $x_k$ in centimeters, it must be that

\begin{equation*}
|x_k-x_l|\geq rD.
\end{equation*}

Relating to (iii)

It is clear that $\mathbf{a}(\omega_0+\omega)=\mathbf{D}_0\mathbf{a}(\omega)$ with $\mathbf{D}_0=\diag e^{-j(x_ku_0+y_kv_0)}$.

§.§ The First and Second Moments of $Q$

\begin{align*}
\mathbf{Q}=&\mathbf{z}^*\boldsymbol{\Gamma}\mathbf{z}\\
=&\trace \boldsymbol{\Gamma}\mathbf{z}\mathbf{z}^*.
\end{align*}

\begin{align}
\E\left\{Q\right\}=&\E\{\trace \boldsymbol{\Gamma}\mathbf{z}\mathbf{z}^*\}\label{EqA4-1}\\
\end{align}

because the integral can be pulled into the sum.

Lemma (A.4-2):

Let $\mathbf{A},\mathbf{B}\in\mathbb{R}^{N\times N}, \mathbf{C}, \mathbf{D}\in\mathbb{C}^{N\times N},\mathbf{u}\in\mathbb{R}^N,\mathbf{w}\in\mathbb{C}^N$.

* $\mathbf{n}\sim\mathcal{N}_{\mathbb{R}^N}(\mathbf{0},\mathbf{R})\Longrightarrow\E\left\{\mathbf{n}^T\mathbf{A}\mathbf{n}\mathbf{n}^T\mathbf{u}\right\}=\mathbf{0}$ and $\E\left\{\mathbf{n}^T\mathbf{A}\mathbf{n}\mathbf{n}^T\mathbf{B}\mathbf{n}\right\}=\trace(\mathbf{A}\mathbf{R})\trace(\mathbf{B}\mathbf{R})+\trace(\mathbf{A}\mathbf{R}\mathbf{B}^T\mathbf{R})+\text{sp}(\mathbf{A}\mathbf{R}\mathbf{B}\mathbf{R})$.

* $\mathbf{n}\sim\mathcal{N}_{\mathbb{C}^N}(\mathbf{0},\mathbf{R})\Longrightarrow\E\left\{\mathbf{n}^*\mathbf{C}\mathbf{n}\mathbf{n}^*\mathbf{w}\right\}=0$ and $\E\left\{\mathbf{n}^*\mathbf{C}\mathbf{n}\mathbf{n}^*\mathbf{D}\mathbf{n}\right\}=\trace(\mathbf{C}\mathbf{R})\trace(\mathbf{D}\mathbf{R})+\trace(\mathbf{C}\mathbf{R}\mathbf{D}\mathbf{R})$.

Proof: It suffices to prove the solution for $\mathbf{n}\sim\mathcal{N}_{\mathbb{C}^N}(\mathbf{0},\mathbf{I})$, because then for $\tilde{\mathbf{n}}=\mathbf{L}\mathbf{n}$ with $\mathbf{L}\mathbf{L}^*=\mathbf{R}$ holds $\tilde{\mathbf{n}}\sim\mathcal{N}(\mathbf{0},\mathbf{R})$, so that the statements for $\mathbf{L}^*\mathbf{A}\mathbf{L}$, $\mathbf{L}\mathbf{u}$ can be applied.

Initially, let $\mathbf{n}\sim\mathcal{N}_{\mathbb{R}^N}\left(\mathbf{0},\mathbf{I}\right)$ then

\begin{equation*}
\mathbf{n}^T\mathbf{D}\mathbf{n}\mathbf{n}^T\mathbf{u}=\sum_{\nu,\mu,\lambda}n_\nu D_{\nu\mu}n_{\mu}n_{\lambda}u_{\lambda}
\end{equation*}

where if two indices are the same, the expected value of the remaining factors is zero and if three indices are the same, then the third moment is zero. Thus, for real as well as for complex $\mathbf{n}$, the third moment is always zero.

\begin{align*}
\mathbf{n}^T\mathbf{A}\mathbf{n}\mathbf{n}^T\mathbf{B}\mathbf{n}=&\sum_{\nu,\mu}n_\nu A_{\nu\mu}n_{\mu}\sum_{\lambda,\tau}n_{\lambda}B_{\lambda\tau}n_{\tau}\\
=&\sum_{\nu}n^4\nu A_{\nu\nu}B_{\nu\nu}+\sum_{\nu\neq\lambda}n_{\nu}^2A_{\nu\nu}n_{\lambda}^2B_{\lambda\lambda}+\underbrace{\sum_{\nu,\lambda\neq\tau}n_{\nu}^2A_{\nu\nu}n_{\lambda}B_{\lambda\tau}n_{\tau}}_{\text{expected value $=0$}}+\underbrace{\sum_{\lambda,\nu\neq\mu}n_\nu A_{\nu\mu}n_{\mu}n_{\lambda}^2B_{\lambda\lambda}}_{\text{expected value $=0$}}\notag\\
&+\underbrace{\sum_{\nu\neq\mu,\lambda\neq\tau}n_\nu A_{v\mu}n_{\mu}n_{\lambda}B_{\lambda\tau}n_{\tau}}_{\text{Expected value $\neq0$ only if $\nu=\lambda\wedge\mu=\tau$ or $\nu=\tau\wedge\mu=\lambda$}}.
\end{align*}

\begin{align*}
\E\left\{\mathbf{n}^T\mathbf{A}\mathbf{n}\mathbf{n}^T\mathbf{B}\mathbf{n}\right\}=&\sum_{\nu}3A_{\nu\nu}B_{\nu\nu}+\sum_{\nu\neq\lambda}A_{\nu\nu}B_{\lambda\lambda}+\sum_{\nu\neq\mu}A_{\nu\mu}B_{\nu\mu}+\sum_{\nu\neq\mu}A_{\nu\mu}B_{\mu\nu}\\
\end{align*}

so Condition 1 of the lemma holds.

Now let $\mathbf{n}=\mathbf{a}+j\mathbf{b}$ with $\mathbf{a},\mathbf{b}\sim\mathcal{N}_{\mathbb{R}^N}(\mathbf{0},\mathbf{I})$ Then

\begin{align*}
\mathbf{a}^T\mathbf{C}\mathbf{a}\mathbf{a}^T\mathbf{D}\mathbf{a}+\mathbf{a}^T\mathbf{C}\mathbf{a}\mathbf{b}^T\mathbf{D}\mathbf{b}-j\underbrace{\mathbf{a}^T\mathbf{C}\mathbf{a}(\mathbf{b}^T\mathbf{D}\mathbf{a}-\mathbf{a}^T\mathbf{D}\mathbf{b})}\notag\\
\end{align*}

where the braced terms (all of the complex terms) have an expected value of zero.

Using Part 1 of the lemma, one gets

\begin{align*}
\end{align*}

for $\mathbf{a}\mathbf{b}\sim\mathcal{N}_{\mathbb{R}^N}(\mathbf{0},\frac{1}{2}\mathbf{I})$. Consequently, $\mathbf{a}+j\mathbf{b}\sim\mathcal{N}_{\mathbb{C}^N}(\mathbf{0},\mathbf{I})$. One thus has

\begin{equation*}
\E\left\{\mathbf{n}^*\mathbf{C}\mathbf{n}\mathbf{n}^*\mathbf{D}\mathbf{n}\right\}=\trace(\mathbf{C})\trace(\mathbf{D})+\trace\mathbf{C}\mathbf{D}
\end{equation*}

which proves Part 2 of the lemma. Thus, one can write

\begin{align*}
\E\left\{Q^2\right\}=&\E\left\{\mathbf{z}^*\boldsymbol{\Gamma}\mathbf{z}\mathbf{z}^*\boldsymbol{\Gamma}\mathbf{z}\right\}\quad\text{for $\mathbf{z}\sim\mathcal{N}_{\mathbb{C}^N}(\mathbf{0},\mathbf{R})$}\\
\end{align*}

\begin{align}
\var\left\{\mathbf{Q}\right\}=&\E\left\{Q^2\right\}-\E\left\{Q\right\}^2\notag\\
\end{align}

§.§ The Probability Distribution of Averaged Hermite Forms of Complex Normally Distributed Random Variables

\begin{equation*}
\end{equation*}

where $\mathbf{b}_i\sim\mathcal{N}_{\mathbb{C}^N}(\mathbf{0},\mathbf{B})$, $\mathbf{G},\mathbf{B}$ are positive-definite $N\times N$ Hermitian matrices.

The characteristic function of $q$ is according to Goodman [5]

\begin{align*}
\Psi_q(\theta)=&\frac{1}{|\mathbf{I}-j\theta\mathbf{B}\mathbf{G}|^K}\\
\end{align*}

when $\lambda_1,\ldots,\lambda_N$ are the eigenvalues from $\mathbf{B}\mathbf{G}$ (or $\mathbf{G}\mathbf{B}$ or $\mathbf{L}^*\mathbf{G}\mathbf{L}$ for $\mathbf{B}=\mathbf{L}\mathbf{L}^*$). Continuing the simplification,

\begin{align*}
\Psi_q(\theta)=\frac{\prod_{i=1}^N\mu_i^K}{\prod_{i=1}^N(\mu_i-j\theta)^K}
\end{align*}

with $\mu_i=\frac{1}{\lambda_i}$. Thus,

\begin{align*}
p_q(x)=&\frac{1}{2\pi}\int_{-\infty}^\infty e^{-j\theta x}\Psi_q(\theta)d\theta\\
=&\frac{\prod\mu_i^K}{2\pi j}\int_{-\infty}^\infty\frac{e^{-j\theta x}}{\prod_{i=1}^K(\mu_i-j\theta)^K}jd\theta\\
=&\frac{\prod\mu_i^K}{2\pi j}\lim\limits_{\epsilon\rightarrow\infty}\int\limits_{c_\epsilon}\frac{e^{-xz}}{\prod (\mu_i-z)^K}dz
\end{align*}

where $c_\epsilon$ is the way shown in Fig. <ref>

where $c_\epsilon=c_1+c_2$ with $c_1=jt,t\in(-\epsilon,\epsilon)$ and $c_2=\epsilon e^{-jt},t\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, and $\int\limits_{c_2}$ goes to zero for $\epsilon\rightarrow\infty$. Thus,

\begin{equation*}
p_q(x)=\prod_{i=1}^N\mu_i^K\sum_{k=1}^N\text{Residual }(f,\mu_k)\cdot\text{number of rotations }(c,\mu_k)
\end{equation*}

\begin{equation*}
\end{equation*}

with rotation number $(c,\mu_k)=-1$ for all $k=1,\ldots, N$ (negative rotation direction) and residual $\left(\frac{g(z)}{(z-a)^K},a\right)=\frac{g^{(k-1)}(a)}{(k-1)!}$, the function $p_q$ can finally be computed as:

\begin{equation*}
p_q(x)=\left.\prod_{i=1}\mu_i^K\frac{(-1)^{NK+1}}{(K-1)!}\sum_{l=1}^N\frac{\partial^{K-1}}{\partial z^{K-1}}\frac{e^{-xz}}{\prod_{i\neq l}(z-\mu_l)^K}\right|_{z=\mu_l}.
\end{equation*}

Thus, for $N=1$:

\begin{equation*}
\left.\frac{\partial^{K-1}}{\partial z^{K-1}}e^{-xz}\right|_{\mu}=(-x)^{K-1}e^{-x\mu}
\end{equation*}

\begin{equation*}
\end{equation*}

$K=1$; $N$ Arbitrary

\begin{equation*}
\text{Res }\left(\frac{e^{-xz}}{\prod_l(z-\mu_l)},\mu_k\right)=\frac{e^{-x\mu_k}}{\prod_{l\neq k}(\mu_k-\mu_l)}.
\end{equation*}

\begin{equation*}
p_{N,1}(x)=-\prod_{i=1}^N\mu_i\sum_{k=1}^N\frac{e^{-x\mu_k}}{\prod_{l\neq k}(\mu_k-\mu_l)}
\end{equation*}

and for $N=2$

\begin{equation*}
\end{equation*}

\begin{align*}
\text{Res }\left(\frac{e^{-xz}}{(z-\mu_2)^2},\mu_1\right)=-\frac{e^{-x\mu_1}}{(\mu_1-\mu_2)^3}(2+(\mu_1-\mu_2)x)\\
\text{Res }\left(\frac{e^{-xz}}{(z-\mu_1)^2},\mu_2\right)=-\frac{e^{-x\mu_2}}{(\mu_2-\mu_1)^3}(2-(\mu_1-\mu_2)x)
\end{align*}

\begin{equation*}
\end{equation*}

\begin{align*}
\text{Res }\left(\frac{e^{-xz}}{(z-\mu_2)^3},\mu_1\right)=&\left.\frac{1}{2}\left(\frac{e^{-xz}}{(z-\mu_2)^3}\right)''\right|_{z=\mu_1}\\
\end{align*}

\begin{equation*}
\text{Res }\left(\frac{e^{-xz}}{(z-\mu_1)^3},\mu_2\right)=-\frac{1}{2}\frac{e^{-x\mu_2}}{(\mu_1-\mu_2)^3}\left(\frac{12}{(\mu_1-\mu_2)^2}-\frac{6x}{\mu_1-\mu_2}+x^2\right).
\end{equation*}

\begin{align*}
\end{align*}

§.§ The Computation of the Integral in (7.2-3)

One only has to compute the integrals in (7.2-4, i) because part (ii) follows from (7.2-2, iv).

For $M=1$, according to (7.2.2, ii)

\begin{align*}
\beta_{1,K}=&K\text{\ae}\int_{-\infty}^{U_\alpha}\phi(x)\frac{\mu^K}{(K-1)!}(K\text{\ae}(U_\alpha-x))^{K-1}e^{-K\text{\ae}(U_\alpha-x)\mu}dx\\
=&\frac{K\text{\ae}\mu^K}{(K-1)!}(-1)^{K-1}\frac{\partial^{K-1}}{\partial \mu^{K-1}}\underbrace{\int_{-\infty}^{U_{\alpha}}\phi(x)e^{-\mu K\text{\ae}(U_{\alpha}-x)}dx}_{=:I}
\end{align*}

\begin{equation*}
\text{\ae}=\sigma^2\sqrt{\frac{N-\hat{M}}{K}}.
\end{equation*}

Regarding the integral $I$,

\begin{align*}
I=&e^{-\tau U_{\alpha}}\int_{-\infty}^{U_{\alpha}}\phi(x)e^{\tau x} dx,\quad(\tau=\mu K \text{\ae})\\
=&e^{-\tau U_{\alpha}}\left[\left.\frac{1}{\tau}e^{\tau x}\phi(x)\right|_{-\infty}^{U_{\alpha}}-\frac{1}{\tau}\int_{-\infty}^{u_\alpha}e^{\tau x}\phi'(x)dx\right]\quad\text{integration by parts}\\
=&\frac{1-\alpha}{\tau}-\frac{e^{-\tau U_{\alpha}}}{\tau\sqrt{2\pi}}\underbrace{\int_{-\infty}^{U_{\alpha}}e^{-\frac{x^2}{2}+\tau x-\frac{\tau^2}{2}}dx}_{=\int_{-\infty}^{U_{\alpha}-\tau}e^{-\frac{x^2}{2}}dx}e^{\frac{\tau^2}{2}}\\
=&\frac{1}{\tau}\left(1-\alpha-e^{\frac{\tau^2}{2}-\tau U_{\alpha}}\phi(U_{\alpha}-\tau)\right)\\
\end{align*}

\begin{equation*}
\beta_{11}=1-\alpha-e^{-\frac{U_{\alpha}^2}{2}}e^{\frac{(U_{\alpha}-\mu\text{\ae})^2}{2}}\phi(U_{\alpha}-\mu\text{\ae}).
\end{equation*}

For $K=2$:

\begin{align*}
\frac{\partial I}{\partial \mu}=&K\text{\ae}\frac{\partial I}{\partial \tau}\\
\end{align*}

\begin{equation*}
\beta_{12}=2\text{\ae}\mu I-2\text{\ae}\mu e^{-\frac{U_{\alpha}^2}{2}}\left[(U_{\alpha}-\tau)e^{\frac{(U_{\alpha}-\tau)^2}{2}}\phi(U_{\alpha}-\tau)+\frac{1}{\sqrt{2\pi}}\right].
\end{equation*}

For $K=3$:

\begin{align*}
\frac{\partial^2I}{\partial \mu^2}=&(K\text{\ae})^2\frac{\partial^2 I}{\partial \tau^2}\quad\text{define $\Delta=U_{\alpha}-\tau$}\\
=&(K\text{\ae})^2\left[\frac{\frac{\partial}{\partial \tau}e^{-\frac{U_{\alpha}^2}{2}}\left(\Delta e^{\frac{\Delta^2}{2}}\phi(\Delta)+\frac{1}{\sqrt{2\pi}}\right)-\frac{\partial I}{\partial \tau}}{\tau}-\frac{e^{-\frac{U_{\alpha}^2}{2}}\left(\Delta e^{\frac{\Delta^2}{2}}\phi(\Delta)+\frac{1}{\sqrt{2\pi}}\right)-I-I}{\tau^2}\right]\\
=&(K\text{\ae})^2\left[\frac{2 I}{\tau^2}-\frac{2e^{-\frac{U_{\alpha}^2}{2}}\left(\Delta e^{\frac{\Delta^2}{2}}\phi(\Delta)+\frac{1}{\sqrt{2\pi}}\right)}{\tau^2}+\frac{e^{-\frac{U_{\alpha}^2}{2}}}{\tau}\left(\Delta_{\tau}e^{\frac{\Delta^2}{2}}\phi(\Delta)+\Delta^2\Delta_\tau e^{\frac{\Delta^2}{2}}\phi(\Delta)+\Delta e^{\frac{\Delta^2}{2}}e^{-\frac{\Delta^2}{2}}\frac{\Delta_\tau}{\sqrt{2\pi}}\right)\right]\\
\end{align*}

\begin{align*}
\beta_{13}=&K\text{\ae}\mu\left[I-\frac{K\text{\ae}\mu}{2}e^{-\frac{U_{\alpha}}{2}}\left(e^{\frac{\Delta^2}{2}}\phi(\Delta)(1+\Delta^2)+\frac{\Delta}{\sqrt{2\pi}}\right)\right]\\
=&1-\alpha-\psi f(\Delta)-\frac{\tau^2}{2}\psi\left(f(\Delta)(1+\Delta^2)+\frac{\Delta}{\sqrt{2\pi}}\right)
\end{align*}

\begin{align*}
\psi=&e^{-\frac{U_{\alpha}^2}{2}}.
\end{align*}

§ RÉSUMÉ

14 March 1948				   Born in Burg on the island of Fehmarn

1954–1958				   Attended primary school (Volkschule) in Hanau am Main

1958–1966				   Attended linguistic secondary school (neusprachliches Gymnasium) in Neuß am Rhein

5 Nov. 1966				   Passed the Abitur on the linguistic secondary school in Neuß am Rhein

1 Jan. 1967 – 1 Jan. 1969		   Service with the German military

April 1969					   Beginning of a degree program in Mathematics with a minor in Physics at the University

of Cologne

16 July 1975				   Passed the primary test for the degree of Diplom in Mathematics at the University of

Since 1 Sep. 1975			   Researcher at the Research Institute for Radio and Mathematics (Forschungsinstitut für

Funk and Mathematik) in Wachtberg-Werthhoven

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