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When constructing a detection test between two Gaussian vectors, I have $H_0: \mathcal{N}(0, \sigma_n^2)$ and $H_1: \mathcal{N}(0,\sigma_n^2 + \sigma_s^2)$, and $\sigma_n^2$ is unknown under finite samples collected, while $\sigma_s^2$ in known. Since both hypotheses are composite, I construct the GLRT while looking for ML estimate of $\sigma_n^2$ under both hypothesis. Unfortunately, I come up with both $f_0(\widehat{\sigma_w^2})$ and $f_1(\widehat{\sigma_w^2})$ to be equal and hence the GLRT is inconclusive. Am I making a mistake, or is there another way of constructing a detection test here?

King008
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1 Answers1

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While I'm not quite sure of the exact model you have in mind, I found strange that you say that the test is inconclusive, because—although asymptotic—the statistic has a well defined distribution: we have $$-2\log \lambda\xrightarrow{\;\mathcal{D}\;}\chi^2_r,$$ where $r$ is the difference between the number of free parameters in $H_1$ and $H_0$.

So you will always have a conclusion, be it rejecting $H_0$ or not.

But it's looking at that distribution that I saw the problem. If $\sigma^2_n$ is unknown and $\sigma^2_s$ is known, then you have one free parameter in both $H_0$ and $H_1$ with leaves you with $r=0$, that is, the GLRT does not apply. Might it be the case that it is the opposite case? Or maybe are both unknown parameters?

Alejandro Nasif Salum
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  • I am considering $L$ samples. For the construction of GLRT, I find the ML estimate of $\sigma_n^2$ under both hypothesis $H_0$ and $H_1$, giving $\widehat{\sigma_{n,0}^2} = \frac{\sum_{l=1}^{L}y_l^2}{L}$ and $\widehat{\sigma_{n,1}^2} = \frac{\sum_{l=1}^{L}y_l^2}{L} - \sigma_s^2$, respectively. Thus I get $\ln f_0(\widehat{\sigma_{n,0}^2}) = \ln f_1(\widehat{\sigma_{n,1}^2}) = -\frac{L}{2} - \frac{L}{2} \ln \frac{2 \pi \sum_{l=1}^{L}y_l^2}{L}$, which leads to an inconclusive GLRT. – King008 Aug 20 '18 at 06:01
  • Ok, two comments:
    1. as I said, the statistic is $-2\log \left(\frac{L_1}{L_0}\right)$, so in your case this would be $0$. Since the distribution is asymptotically a $\chi^2$, and $H_0$ is rejected for large values of the statistic, if everything else were fine the conclusion would not be inconclusive: it would be a decisive 'do not reject $H_0$', no matter how big you choose your $\alpha$.
    2. But the thing here is that these hypotheses do not allow a GLRT, since the statistic would be a $\chi^2_0$; and you can't have $0$ deg. of freedom. That is actually why you end up with a $0$ statistic.
     – Alejandro Nasif Salum Aug 22 '18 at 04:36
  • It is not that the GLRT is inconclusive: it is not suited to treat this pair of hypotheses. – Alejandro Nasif Salum Aug 22 '18 at 04:38
  • The thing is that these two hypotheses overlap. For instance: let's say that $\sigma^2_s=3$. This means that $H_0$ states that the distribution has any possible variance ($\sigma^2=\sigma^2_n>0$), while $H_1$ says that $\sigma^2=\sigma^2_n+3$, that is, the variance of the distribution is any value greater than three ($\sigma^2>3$). That means that a value like $\sigma^2=4$ makes both $H_0$ and $H_1$ true (in on case you would say that $\sigma^2_n=4$ and in the other one that $\sigma^2_n=1$. It's like you over-parameterized the model. – Alejandro Nasif Salum Aug 22 '18 at 04:45
  • If the only thing you wanna test is whether $H_0: \sigma^2\le 3$ (I mean $3$ to say some fixed number) or $H_1: \sigma^2>3$, then the usual $\chi^2$ test based on $$\frac{L \hat\sigma^2}{\sigma^2_0}=\frac{\sum_i y_i^2}{3}\sim\chi^2_{L-1},$$ is the best choice. Here you would reject $H_0$ for large values of the statistic—that is, those greater than $\chi^2_{L-1,\alpha}$ (the point that lefts $\alpha$ probability in the right tail). . – Alejandro Nasif Salum Aug 22 '18 at 04:49
  • Just a correction: being the case that the mean is known (here $\mu=0$), the aforementioned distribution of the statistic is not $\chi^2_{L-1}$, but actually $\chi^2_L$. – Alejandro Nasif Salum Aug 22 '18 at 05:28