On Minimax Detection of Gaussian Stochastic Sequences with Imprecisely Known Means and Covariance Matrices

Abstract

We consider the problem of detecting (testing) Gaussian stochastic sequences (signals) with imprecisely known means and covariance matrices. An alternative is independent identically distributed zero-mean Gaussian random variables with unit variances. For a given false alarm (1st-kind error) probability, the quality of minimax detection is given by the best miss probability (2nd-kind error probability) exponent over a growing observation horizon. We study the maximal set of means and covariance matrices (composite hypothesis) such that its minimax testing can be replaced with testing a single particular pair consisting of a mean and a covariance matrix (simple hypothesis) without degrading the detection exponent. We completely describe this maximal set.

Appendix: Proof of Lemma 1

Let $\boldsymbol{\xi}_n$ be a Gaussian random vector with the distribution $\boldsymbol{\xi}_n \sim{\mathcal{N}}({\bf0},\boldsymbol{I}_n)$, and let $\boldsymbol{A}_n$ be a symmetric $n\times n$ matrix with eigenvalues $\{a_i\}$. Consider the quadratic form $(\boldsymbol{\xi}_n,\boldsymbol{A}_n\boldsymbol{\xi}_n)$. There exists an orthogonal matrix $\boldsymbol{T}_n$ such that $\boldsymbol{T}_n'\boldsymbol{A}_n\boldsymbol{T}_n=\boldsymbol{B}_n$, where $\boldsymbol{B}_n$ is the diagonal matrix with diagonal entries $\{a_i\}$ [6, Section 4.7]. Since $\boldsymbol{T}_n\boldsymbol{\xi}_n \sim{\mathcal{N}}({\bf0},\boldsymbol{I}_n)$, the quadratic forms $(\boldsymbol{\xi}_n,\boldsymbol{A}_n\boldsymbol{\xi}_n)$ and $(\boldsymbol{\xi}_n,\boldsymbol{B}_n\boldsymbol{\xi}_n)$ have the same distributions. Therefore, by equation (12) we have

$$\ln\frac{p^{}_{\boldsymbol{I}_n}}{p^{}_{\boldsymbol{a}_n,\boldsymbol{M}_n}} (\boldsymbol{y}_n)\stackrel{d}{=} \frac{1}{2}\bigl[\ln|\boldsymbol{M}_n|+(\boldsymbol{a}_n,\boldsymbol{M}_n^{-1}\boldsymbol{a}_n)+\eta_n\bigr],$$

(71)

where

$$\eta_n= (\boldsymbol{y}_n, (\boldsymbol{M}_n^{-1}- \boldsymbol{I})\boldsymbol{y}_n) -2(\boldsymbol{y}_n,\boldsymbol{M}_n^{-1}\boldsymbol{a}_n).$$

(72)

Introduce the quantity (see (31))

$$\alpha_{\mu}=\operatorname{\mathbf P}\nolimits_{\boldsymbol{I}_n}\biggl\{\ln\frac{p^{}_{\boldsymbol{I}_n}}{p^{}_{\boldsymbol{a}_n,\boldsymbol{M}_n}}(\boldsymbol{x}_n)\le D(\boldsymbol{I}_n\,\|\,\boldsymbol{a}_n,\boldsymbol{M}_n)-\mu\biggr\}.$$

(73)

Then by (71), (72), and (17) or $\alpha_{\mu}$ from (73) we have

$$\begin{aligned}[b]\alpha_{\mu}&\le \operatorname{\mathbf P}\nolimits_{\boldsymbol{\xi}_n}\Biggl\{\Biggl| (\boldsymbol{\xi}_n, (\boldsymbol{M}_n^{-1}-\boldsymbol{I})\boldsymbol{\xi}_n) - 2(\boldsymbol{\xi}_n,\boldsymbol{M}_n^{-1}\boldsymbol{a}_n)-\sum_{i=1}^n\Bigl(\frac{1}{\lambda_i}-1\Bigr)\Biggr|>2\mu\Biggr\}\\ &=\operatorname{\mathbf P}\nolimits_{\boldsymbol{\xi}_n}\Biggl\{\Biggl|\sum_{i=1}^n\Bigl(\frac{1}{\lambda_i}-1\Bigr)(\xi_i^2-1)-2(\boldsymbol{\xi}_n,\boldsymbol{M}_n^{-1}\boldsymbol{a}_n)\Bigr|>2\mu\Biggr\}\le P_1+P_2,\end{aligned}$$

(74)

where

$$\begin{aligned}P_1&=\operatorname{\mathbf P}\nolimits_{\boldsymbol{\xi}_n}\Biggl\{\Biggl|\sum_{i=1}^n\Bigl(\frac{1}{\lambda_i}-1\Bigr)(\xi_i^2-1)\Biggr|>\mu\Biggr\},\\ P_2&=\operatorname{\mathbf P}\nolimits_{\boldsymbol{\xi}_n}\bigl\{\bigl|(\boldsymbol{\xi}_n,\boldsymbol{M}_n^{-1}\boldsymbol{a}_n)\bigl|>\mu/2\bigr\}.\end{aligned}$$

(75)

To estimate $P_1$ in (75), we use the following result [13, Section III.5.15]: let $\zeta_1,\ldots,\zeta_n$ be independent random variables with $\operatorname{\mathbf E}\nolimits\zeta_i=0$, $i=1,\ldots,n$. Then for any $1\le p\le 2$ we have

$$\operatorname{\mathbf E}\nolimits\Biggl|\sum_{i=1}^n\zeta_i\Biggr|^p\le 2\sum_{i=1}^n\operatorname{\mathbf E}\nolimits|\zeta_i|^p.$$

(76)

Therefore, using Chebychev's inequality and (76) for $P_1$, we obtain

$$\begin{aligned}[b]P_1&\le\mu^{-p}\operatorname{\mathbf E}\nolimits\Biggl|\sum_{i=1}^n\Bigl(\frac{1}{\lambda_i}-1\Bigr)(\xi_i^2-1)\Biggr|^p\\ &\le2\mu^{-p}\sum_{i=1}^n\Bigl|\frac{1}{\lambda_i}-1\Bigr|^p\operatorname{\mathbf E}\nolimits |\xi_i^2-1|^p\\ &\le2\mu^{-p}\sum_{i=1}^n\Bigl|\frac{1}{\lambda_i}-1\Bigr|^p\bigl(\operatorname{\mathbf E}\nolimits|\xi^2-1|^2\bigr)^{p/2}\\ &\le 2 \mu^{-p}6^{p2}\sum_{i=1}^n\Bigl|\frac{1}{\lambda_i}-1\Bigr|^p\\ &\le12\mu^{-p}\sum_{i=1}^n\Bigl|\frac{1}{\lambda_i}-1\Bigr|^p.\end{aligned}$$

(77)

To estimate $P_2$ in (74) and (75), note that

$$(\boldsymbol{\xi}_n,\boldsymbol{M}_n^{-1}\boldsymbol{a}_n)\sim\mathcal{N}(0,\|\boldsymbol{M}_n^{-1}\boldsymbol{a}_n\|),$$

and so

$$(\boldsymbol{\xi}_n,\boldsymbol{M}_n^{-1}\boldsymbol{a}_n) \stackrel{d}{=}\|\boldsymbol{M}_n^{-1}\boldsymbol{a}_n\|\xi.$$

Therefore, using the standard bound

$$\operatorname{\mathbf P}\nolimits(|\xi|\ge z)\le e^{-z^2/2},\quad z\ge 0,$$

we obtain ($\xi_i \sim \mathcal{N}(0,1)$)

$$P_2=\operatorname{\mathbf P}\nolimits_{\boldsymbol{\xi}_n}\bigl\{|(\boldmath{\xi}_n,\boldsymbol{M}_n^{-1}\boldsymbol{a}_n)|>\mu/2\bigr\}\le e^{-\mu^2/(8\|\boldsymbol{M}_n^{-1}\boldsymbol{a}_n\|^2)}.$$

(78)

For the condition $\alpha_{\mu}\le\alpha$ to be satisfied, we choose $\mu$ so that $\max\{P_1,P_2\}\le\alpha/2$. For that, by (77) and (78), it suffices to take $\mu$ satisfying (32).