I'm trying to find the optimal probability of detection/false alarm for the following detection task. Given $N$ signal samples with $d$ dimensions (independent channels) each, assign the samples to one of two distributions: $$ H_0: \mathcal{N}(0, \sigma^2I^d) \\ H_1: \mathcal{N}(\boldsymbol{\mu}, \sigma^2I^d) $$ The value of each of the $d$ components of $\boldsymbol{\mu}$ is unknown but it is known that the signal power is bounded such that $\boldsymbol{\mu}^T\boldsymbol{\mu} \leq C$. Since we don't know the values of the entries in $\boldsymbol{\mu}$, a Neyman-Pearson optimal (and even uniformly most powerful) test doesn't exist, so I have to resort to a generalized likelihood ratio test. Under $H_0$, the squared magnitudes of the signal follow a central $\chi^2$ and under $H_1$ a non-central $\chi^2$ distribution. However, I'm not sure how many degrees of freedom the distributions should have.

In Kay Vol. 2, Section 7.4.1 it is noted that the distributions should have $N$ degrees of freedom, but I think that the book assumes a scalar and not a vector-valued signal. Extrapolating from this, my assumption is that the correct degrees of freedom should be $d \cdot N$. However, this would mean that the detection of the signal is more difficult with higher dimensionality. This surprises me as in the Neyman-Pearson setting, the dimensionality of the signal doesn't make a difference (it's always possible to rotate the problem to linearly classify it assuming (like here) isotropic Gaussians). Or is the correct number of DOF still $N$ because we are really only estimating a single value (the magnitude)? In this case, why is the number of DOF for complex noise $2N$? (Kay Vol. 2, Section 13.6.6)

Could someone explain the line of thought here?

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    $\begingroup$ I think your intuition of $d\ N$ degrees of freedom is correct. It is common for 2D "vectors" of data to be defined as temporal-samples-by-spatial-samples (say, $N$x$M$) from an array of sensors. If a deterministic signal is imparted on that array, you simply have $NM$ number of samples of a measurement with different temporal/spatial correlations between the samples. The detection theory aspect doesn't understand the relation between those samples (and doesn't need to). It simply sees it as a collection of $NM$ values, so you could just permute $\mu$ to be $NM$x$1$ and follow Kay's method. $\endgroup$
    – Ash
    Commented Oct 10, 2022 at 18:55
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    $\begingroup$ Further, recall that each sample is perturbed by AGWN, and thus is independent of all other samples. This means that all samples have a degree of freedom. If your noise is complex and is non-symmetric in the real/imaginary plane, that gives you an additional degree of freedom per sample. $\endgroup$
    – Ash
    Commented Oct 10, 2022 at 19:01
  • $\begingroup$ Regarding your second comment, Kay gives 2N DOF for circularly symmetric complex noise. I suppose the idea follows from your first comment though, we can just flatten the real/imag part into the same vector and get 2dN DOF in total. $\endgroup$
    – Sami
    Commented Oct 10, 2022 at 19:42
  • $\begingroup$ The increase in DOF still seems strange though, it would mean that a high dimensional signal is much harder to be detected… could it be that we only consider 1 DOF because energy detection is equivalent to stating "0 vs. Non-zero" independent of dimensionality because the latter can be assumed whenever even a single component of the vector is significantly non-zero? Still doesn’t fully explain the complex case though… $\endgroup$
    – Sami
    Commented Oct 10, 2022 at 19:46
  • $\begingroup$ Yes, the problem boils down to an energy detector. Figure 7.3 describes the probability of detection as a function of deflection coefficient, the ratio of signal energy to noise energy. Probability of detection only suffers if noise energy prevails. Also, you can find the explanation of the $2N$ for complex [linear] models in Appendix 13A. The two parameters are the real and imaginary part of the expected value of the PDF $\mathcal{C}\mathcal{N}([\alpha, \beta]^T, \textbf{C})$. The distribution itself is symmetric about the expected value, but not the origin unless $[\alpha, \beta]^T=[0,0]^T$. $\endgroup$
    – Ash
    Commented Oct 12, 2022 at 5:56


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