1
$\begingroup$

I need to construct a Neyman Pearson statistic for detecting a complex signal with known amplitude and unknown phase in additive complex Gaussian noise. My hypotheses are:

\begin{align} \mathcal{H}_0 &: R_i=W_i,& R_i \sim \mathcal{CN} (0,1) \\ \mathcal{H}_1 &: R_i=W_i + A,&R_i \sim \mathcal{CN}(A,1) \end{align}

alternatively,

\begin{align} \mathcal{H}_0 &: R_i=W_i,& W_i \sim \mathcal{CN} (0,1) \\ \mathcal{H}_1 &: R_i -A=W_i,&W_i \sim \mathcal{CN}(0,1) \end{align}

where $A = A_0e^{j\theta_0}$ is the signal, and $i=1,2,...N$. The signal A occurs over a known region of the data, say $m_1$ to $m_2$, where $ 0 \leq m_1 \lt N, m_1 \lt m_2 \leq N$ and as such the statistical model is to be applied only over that region.

My first steps are marginalizing the nuisance parameters $\theta_0$, and calculating the statistic, assuming $\theta_0 \sim \mathcal{U}(0,2\pi)$:

$$ \frac{f(\mathbf r|H_1)}{f(\mathbf r|H_0)} = \frac{\displaystyle \prod_{i=1}^{N}\int\limits_0^{2\pi} \frac{1}{2\pi} \frac{1}{\pi} \exp{\left(-\lVert r_i - A_0e^{j\theta_0}\rVert^2\right)}\,\mathrm d\theta_0 }{\displaystyle\prod_{i=1}^N \frac{1}{\pi} \exp{\left(-\lVert r_i\rVert^2\right)}}\text,$$

with $\mathbf r=(r_1,r_2,\ldots,r_N)$ being the observation.

Is my construction correct? I can't see how to proceed from here that will allow me to construct a statistic to compare against a threshold. Any suggestions?

$\endgroup$
  • $\begingroup$ In your formulation of $H_1$, what is circularly-symmetric normal? $A_i$ or $(Y_i-A_i)$? Is it possible that your $A_i=A_0e^{i\theta_0}$ is missing an imaginary unit ($j$) in the exponent? $Y_i$ and $A_i$ are independent, right? $\endgroup$ – Marcus Müller Jul 1 at 7:37
  • $\begingroup$ Isn't there an integral sign missing in $f(Y|H_0)$? The $\mathrm d\theta_0$ in your integral should be after the exponential, seeing that this is the integrand? $\endgroup$ – Marcus Müller Jul 1 at 7:42
  • $\begingroup$ Fixed the question. Yes the whole thing is under the integral sign. And I don't see why $f(Y|H_0)$ should have an integral, since it does not have the $\theta_0$ parameter. $Y_i -A_i$ is the circular symmetric variable here. $\endgroup$ – Razor Jul 1 at 9:16
  • $\begingroup$ While you're right about the integral, this raises a lot of questions. So, let's start small: what is the relationship of the elements in $Y$? I think there's still too much confusion in your notation. For example, your $f(Y|H_1)$ can only be the product of densities of $Y_i-A_i$ if these are independent. But in that case, there's nothing to be done here: in both you $H_0$ and $H_1$ you get the exact same statistic, iid $~\mathcal{CN}(0,1)$, and Neyman-Pearson can't do anything about that. $\endgroup$ – Marcus Müller Jul 1 at 9:41
  • 1
    $\begingroup$ It looks fine, thanks. $\endgroup$ – Razor Jul 2 at 8:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.