The general Gaussian problem

The general Gaussian problem

• Set up the general Gaussian problem for vector binary hypothesis testing, and derive the likelihood ratio test as a difference of two quadratic forms.
• With equal covariance, derive the linear sufficient statistic T(y)=\Delta m^TC^{-1}y and the detection parameter d^2=\Delta m^TC^{-1}\Delta m.
• Diagonalize a nondiagonal covariance matrix by an orthogonal rotation (the eigenvalue problem), then reapply the test in the new coordinate system.
• With equal means, derive the energy detector T(y)=\sum y_k^2, recognise the chi-square distribution, and compute P_D,P_F.
• Apply this to symmetric hypotheses, build a dual energy detector, and compute the error probability by numerically integrating two independent chi-square densities.

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Definition: when a problem is "general Gaussian"

A hypothesis-testing problem is called a general Gaussian problem if the conditional density f_{Y|H_j}(y|H_j) is a Gaussian density function for every j; similarly, an estimation problem is general Gaussian if f_{Y|\Theta}(y|\theta) is Gaussian for every \theta (Barkat, pp. 503 to 504). This is exactly the framework used implicitly in the Gaussian examples throughout chapters 5 and 6 (modules 21, 22) — now fully generalised to a multidimensional vector observation instead of one or a few scalar samples.

The vector observation: Y=X+N

Consider H_1{:}\,Y=X+N, H_0{:}\,Y=N, with Y,X,N all K\times1 vectors. Each hypothesis is characterised by a mean vector m_1=E[Y|H_1], m_0=E[Y|H_0], and a covariance matrix C_1=E[(Y-m_1)(Y-m_1)^T|H_1], C_0=E[(Y-m_0)(Y-m_0)^T|H_0] (Barkat, pp. 503 to 504). The multivariate Gaussian density is f_{Y|H_j}(y|H_j)=\dfrac{1}{(2\pi)^{K/2}|C_j|^{1/2}}\exp\left[-\tfrac12(y-m_j)^TC_j^{-1}(y-m_j)\right].

The likelihood ratio test reduces to a difference of quadratic forms

Substituting the Gaussian density into the likelihood ratio \Lambda(y) and taking the logarithm, the likelihood ratio test becomes \tfrac12(y-m_0)^TC_0^{-1}(y-m_0)-\tfrac12(y-m_1)^TC_1^{-1}(y-m_1)\gtrless\gamma, with \gamma=\ln\eta+\tfrac12(\ln|C_1|-\ln|C_0|) (Barkat, pp. 504 to 505). This is the common starting point for every general Gaussian problem: the test is always a difference of quadratic forms, and how it simplifies further depends on the constraint placed on m_j,C_j.

The three standard cases to be studied

Evaluating the difference of quadratic forms further depends on the constraint: (1) equal covariance C_1=C_0=C (parts 2, 3) — the test reduces to a linear sufficient statistic; (2) equal mean m_1=m_0 (part 4) — the test reduces to a chi-square energy detector; (3) symmetric hypotheses, equal zero mean but the signal-noise roles swapped between two halves of the vector (part 5) — a dual energy detector (Barkat, pp. 505 to 506, 517 to 528).

Connection to chapter 5: the same framework, now vectorial

Examples 5.1, 5.2 (module 21) are special cases of part 2 below with K=1 or independent equal-variance samples: the \hat m_{ml} and threshold \gamma there are exactly the one-dimensional projection of the linear sufficient statistic T(y)=\Delta m^TC^{-1}y to be derived in part 2. Generalising to a vector allows handling real radar/sonar problems where several samples are correlated through a covariance matrix that is not a multiple of the identity.

Self-check, part 1

• When is a problem called \"general Gaussian\"?
• What form does the log-likelihood-ratio test of the general Gaussian problem reduce to?
• What are the three standard cases studied in this chapter?

Hints: when the conditional density is Gaussian for every hypothesis/parameter; a difference of two quadratic forms in y; equal covariance, equal mean, and symmetric.

When C_1=C_0=C: the quadratic terms cancel

Substituting C_1=C_0=C into the quadratic test and expanding, the y^TC^{-1}y terms in both quadratic forms are identical and cancel each other, leaving only a term linear in y: (m_1^T-m_0^T)C^{-1}y\gtrless\gamma_1 (Barkat, pp. 505 to 507). Setting \Delta m=m_1-m_0, the test becomes T(y)=\Delta m^TC^{-1}y\gtrless\gamma_1 — the sufficient statistic T(y) is a dot product, linear in y.

T(y) is Gaussian: only the mean and variance are needed

T(Y) is a linear combination of the Gaussian variables Y_k, so it is itself a Gaussian random variable. The conditional mean is E[T(Y)|H_j]=\Delta m^TC^{-1}m_j; the variance is \text{var}[T(Y)|H_j]=\Delta m^TC^{-1}CC^{-1}\Delta m=\Delta m^TC^{-1}\Delta m — independent of which hypothesis (Barkat, p. 507). Because T(Y) is Gaussian with the same variance under both hypotheses, the entire test performance (exactly the ROC framework of module 21) depends on a single number: the detection parameter.

The detection parameter d^2=\Delta m^TC^{-1}\Delta m

Generalising the definition of d met in module 21, d^2\triangleq\{E[T|H_1]-E[T|H_0]\}^2/\text{var}[T|H_0]. Substituting the expressions above, the numerator is (\Delta m^TC^{-1}\Delta m)^2 and the denominator is \Delta m^TC^{-1}\Delta m, giving d^2=\Delta m^TC^{-1}\Delta m (Barkat, pp. 507 to 508). d^2 measures how distinguishable the two hypotheses are: the test performance (the ROC curve, the error probability) depends only on d, on no other detail of m_j,C.

A special case: equal variance, diagonal

If C=\sigma^2I (independent components, equal variance), C^{-1}=(1/\sigma^2)I, so T(y)=\Delta m^Ty/\sigma^2 — just the dot product of \Delta m and y, divided by \sigma^2. The detection parameter reduces to d^2=|\Delta m|^2/\sigma^2, i.e. d=|\Delta m|/\sigma — the Euclidean distance between the two mean vectors, divided by the common standard deviation (Barkat, pp. 508 to 509, Fig. 9.1).

Measured: d^2 and the mean/variance of T(y) match simulation

Measured: with K=4, C=2I (diagonal, equal variance \sigma^2=2), \Delta m=[1,2,-1,0.5]^T: d^2=|\Delta m|^2/\sigma^2 = 3.125, matching the general formula \Delta m^TC^{-1}\Delta m = 3.125. A Monte Carlo simulation of 200,000 samples under each hypothesis: the mean of T(Y) under H_0 is 0.0037 (theory 0) and under H_1 is 3.1225 (matching \Delta m^TC^{-1}m_1); the measured variance of T(Y) under both hypotheses is approximately 3.1206, matching d^2 since C^{-1}=I/\sigma^2 makes the variance equal \Delta m^TC^{-1}\Delta m=d^2 itself.

Measured: the minimum error probability by Q(d/2), exactly module 21's framework

With equal priors and the minimum-error-probability criterion, module 21's result applies directly: P(\varepsilon)=Q(d/2). Measured: with a nondiagonal C (a covariance matrix with cross-correlation), d = 1.2536, and the Monte Carlo simulated error probability is 0.2654, matching Q(d/2) = 0.2654 — confirming the formula d^2=\Delta m^TC^{-1}\Delta m holds even when C has cross-correlation, not just the simple diagonal case.

Self-check, part 2

• Why do the quadratic-in-y terms cancel when C_1=C_0?
• What is the general formula for the detection parameter d^2?
• With C=\\sigma^2I, what does d reduce to?

Hints: because both quadratic forms share the same C^{-1} so the y^TC^{-1}y terms are identical; d^2=\Delta m^TC^{-1}\Delta m; d=|\Delta m|/\sigma.

When the components are correlated: a change of coordinates is needed

If C is nondiagonal, the components Y_k are not statistically independent. To separate them, a new coordinate system with orthonormal axes \{\Phi_k\} must be found such that the projected components Y_k'=\Phi_k^TY are independent — i.e. the covariance in the new system is diagonal: E[(Y_j'-m_j)(Y_k'-m_k)]=\lambda_k\delta_{jk} (Barkat, pp. 511 to 512). This condition only holds when \Phi_j^TC\Phi_k=\lambda_k\delta_{jk}, i.e. C\Phi_k=\lambda_k\Phi_k — the eigenvalue problem.

The modal matrix and the norm-preserving rotation

Assembling the orthonormal eigenvectors into the modal matrix M=[\Phi_1\ \Phi_2\ \cdots\ \Phi_K]; since C is real symmetric, M^{-1}=M^T. The similarity transformation gives M^TCM=\text{diag}(\lambda_1,\ldots,\lambda_K) (Barkat, pp. 512 to 513). The vector in the new system is y'=M^Ty (or y=My'); since M is orthogonal, this is a pure rotation, so the norm of y' equals the norm of y in the original system — no information lost, only the viewpoint changed.

In the new coordinates: exactly reapplying the part-2 result

After the rotation, the covariance \Lambda is diagonal with entries \lambda_k (generally different); applying the unequal-variance result of part 2 (replacing \sigma_k^2 with \lambda_k): the sufficient statistic is T(y')=\sum_{k=1}^K\Delta m_k'y_k'/\lambda_k, and the detection parameter is d^2=\sum_{k=1}^K(\Delta m_k')^2/\lambda_k — the same formula d^2=\Delta m^TC^{-1}\Delta m, just written in the diagonalizing coordinate system (Barkat, pp. 509 to 511, reapplied at pp. 513 to 514).

Measured: diagonalizing a specific 2\times2 covariance

Measured: with C=\begin{bmatrix}2&0.5\\0.5&1\end{bmatrix}, solving the eigenvalue problem gives \lambda_1 = 0.7929, \lambda_2 = 2.2071; checking orthonormality \Phi_1^T\Phi_2 = 0.00e+00 (near zero) and \Phi_1^T\Phi_1 = 1 (equal to one). The largest off-diagonal element of M^TCM after diagonalization is 2.04e-17 (near zero), confirming the diagonalization succeeded.

Measured: d^2 computed in the original and diagonalized systems agree

Measured: with \Delta m=[1.5,\,1.0]^T and the same C above, the detection parameter computed directly as \Delta m^TC^{-1}\Delta m in the original system is 1.5714; computed as \sum_k(\Delta m_k')^2/\lambda_k after rotating to the diagonalized system is 1.5714 — agreeing exactly, confirming d^2 is invariant under the rotation, exactly as the theory predicts (Barkat, pp. 513 to 514).

Self-check, part 3

• What mathematical problem does diagonalizing C correspond to?
• What special property does the transformation y'=M^Ty have when M is orthogonal?
• Does the detection parameter d^2 change when computed in the diagonalized system instead of the original one?

Hints: the eigenvalue problem C\Phi=\lambda\Phi; it is a rotation, preserving the vector norm; no, d^2 is invariant under the rotation.

When m_1=m_0: only the quadratic term remains

Setting m_1=m_0=m (without loss of generality, choose m=0 since m does not affect the decision), the quadratic test reduces to T(y)=y^T(C_0^{-1}-C_1^{-1})y\gtrless\gamma_2 (Barkat, pp. 517 to 519). This corresponds to the physical problem H_1{:}\,Y_k=S_k+N_k, H_0{:}\,Y_k=N_k — only noise under H_0, both signal and noise under H_1, with signal and noise independent.

Independent signal components, equal variance: reducing to energy

If the noise is white (C_n=\sigma_n^2I=C_0) and the signal components are independent, equal variance (C_s=\sigma_s^2I), then C_1=C_s+C_n=(\sigma_s^2+\sigma_n^2)I. Substituting into the test, the scalar coefficients fold into the threshold, leaving a surprisingly simple sufficient statistic: T(y)=\sum_{k=1}^Ky_k^2 — the sum of squared observations, i.e. the energy of the observation vector (Barkat, pp. 519 to 520).

T(Y) follows a chi-square distribution

Since Y_k are independent, identically Gaussian distributed, each Y_k^2 has a Gamma distribution with \alpha=1/2, \beta=2\sigma^2; the sum of K independent variables of the same form has a multiplying characteristic function, giving T(Y)=\sum Y_k^2 a Gamma distribution with \alpha=K/2 — exactly the chi-square distribution with K degrees of freedom, scaled by \sigma^2 (Barkat, pp. 520 to 522). The variance \sigma^2 is \sigma_n^2 under H_0 and \sigma_s^2+\sigma_n^2 under H_1.

P_D, P_F from two chi-square densities

With f_{T|H_j}(t|H_j) the chi-square density with K degrees of freedom scaled by \sigma_j^2 (\sigma_0^2=\sigma_n^2, \sigma_1^2=\sigma_s^2+\sigma_n^2), P_F=\int_{\gamma_3}^\infty f_{T|H_0}(t|H_0)\,dt and P_D=\int_{\gamma_3}^\infty f_{T|H_1}(t|H_1)\,dt (Barkat, pp. 521 to 522). This is exactly the survival function of the scaled chi-square distribution — computable directly with standard statistics packages, no hand integration needed.

Measured: theoretical P_D,P_F match simulation

Measured: with K=6, \sigma_n^2=1, \sigma_s^2=3, threshold \gamma_3=8: the theoretical P_F (chi-square survival function) = 0.2381, a Monte Carlo simulation of 300,000 trials gives 0.2366; the theoretical P_D = 0.9197, simulated 0.9196 — agreeing within statistical error, confirming the chi-square formula.

Self-check, part 4

• When m_1=m_0, which term in the quadratic test disappears?
• With independent equal-variance signal, white noise, what does the sufficient statistic reduce to?
• What distribution does T(Y)=\\sum_ky_k^2 follow?

Hints: the linear-in-y term; the energy \sum_ky_k^2; chi-square with K degrees of freedom, scaled by the variance.

The structure of symmetric hypotheses

Consider a 2K-dimensional problem: H_1{:}\,Y_k=S_k+N_k (k\le K), Y_k=N_k (k>K); H_0{:}\,Y_k=N_k (k\le K), Y_k=S_k+N_k (k>K) — the signal appears in the first half under H_1 but in the second half under H_0 (Barkat, pp. 524 to 525). With m_1=m_0=0 and white noise C_n=\sigma_n^2I, the block covariance matrices are C_1=\text{diag}(C_s+\sigma_n^2I,\ \sigma_n^2I), C_0=\text{diag}(\sigma_n^2I,\ C_s+\sigma_n^2I) — just the two blocks swapped.

The test reduces: a difference of two half-vector energies

With independent signal, equal variance \sigma_s^2 per component, substituting the blocks into the shared quadratic test and reducing (using exactly the part-4 technique for each block) gives T(y)=\sum_{k=1}^Ky_k^2-\sum_{k=K+1}^{2K}y_k^2\gtrless\gamma_3 (Barkat, pp. 526 to 527) — directly comparing the first half's energy with the second half's. With equal priors and the minimum-error-probability criterion, the threshold \gamma_3=0 (since |C_1|=|C_0| by symmetry), so the test simplifies to: whichever half has more energy carries the signal.

The error probability: two independent chi-square variables

Set T_1=\sum_{k=1}^Ky_k^2, T_0=\sum_{k=K+1}^{2K}y_k^2 — two independent variables, each chi-square with K degrees of freedom scaled by that half's own variance. Since the problem is symmetric, P(\varepsilon|H_0)=P(\varepsilon|H_1), so P(\varepsilon)=P(T_1>T_0\,|\,H_0) (Barkat, pp. 527 to 528, Fig. 9.3). This is computed directly by numerically integrating two independent chi-square densities, with no closed combinatorial formula needed.

Measured: the error probability matches both numerical integration and simulation

Measured: with K=4, \sigma_n^2=1, \sigma_s^2=2 (so \sigma_1^2=\sigma_s^2+\sigma_n^2=3 for the signal-bearing half, \sigma_n^2=1 for the noise-only half): numerically integrating P(T_1>T_0|H_0) (using the two independent scaled chi-square densities) gives P(\varepsilon) = 0.1562; a Monte Carlo simulation of 400,000 trials directly on the original definition gives 0.1554 — closely agreeing, confirming both the numerical integration and the simulated definition give the same result.

Looking back: one family of formulas, three constraints

All five problems in the chapter — from Examples 5.1/5.2 (module 21) to the dual energy detector — originate from exactly one quadratic test \tfrac12(y-m_0)^TC_0^{-1}(y-m_0)-\tfrac12(y-m_1)^TC_1^{-1}(y-m_1)\gtrless\gamma. The constraint C_1=C_0 gives a linear statistic; the constraint m_1=m_0 gives a quadratic (energy) statistic; the symmetric constraint gives a difference of two energies. This is the core lesson: the same general Gaussian framework, differing only in the constraint placed on (m_j,C_j).

Self-check, part 5

• In the symmetric hypothesis, which half of the vector carries the signal under each hypothesis?
• What does the dual energy test directly compare?
• How is the error probability P(\\varepsilon) computed in this part, instead of using a closed combinatorial formula?

Hints: the first half under H_1, the second half under H_0 (swapped); the first half's energy against the second half's; numerically integrating two independent chi-square densities.

Takeaways

• The general Gaussian problem: the conditional density is Gaussian for every hypothesis; the log-likelihood-ratio test always reduces to a difference of quadratic forms \tfrac12(y-m_0)^TC_0^{-1}(y-m_0)-\tfrac12(y-m_1)^TC_1^{-1}(y-m_1).
• Equal covariance (C_1=C_0=C): the test linearises, T(y)=\Delta m^TC^{-1}y, and the entire performance depends on a single number d^2=\Delta m^TC^{-1}\Delta m.
• A nondiagonal covariance: diagonalize by the eigenvalue problem C\Phi=\lambda\Phi, the rotation y'=M^Ty leaves d^2 unchanged; then reapply exactly the part-2 formula.
• Equal mean (m_1=m_0): the test reduces to the energy detector T(y)=\sum y_k^2, following a chi-square distribution with K degrees of freedom, giving a closed form for P_D,P_F.
• Symmetric hypotheses: a dual energy detector directly compares the two halves' energy; the error probability is computed by numerically integrating two independent chi-square densities.