P1

Show the NP detector for a Gaussian rank one signal as a Gaussian random process whose covariance matrix is $C_s=\sigma^2_Ahh^T$ embedded in WGN with variance $\sigma^2$ can be written as

Also, determine $P_{FA}$ and $P_D$.
Hint: The test statistic is a scaled $\chi^2_1$ random variable under $\mathcal{H}_0$ and $\mathcal{H}_1$

Problem Setup

We are tasked with deriving the Neyman-Pearson (NP) detector for a Gaussian rank-one signal embedded in white Gaussian noise (WGN). The covariance matrix of the signal is given by:

Observations:

• Signal Model:

  • Under ( \mathcal{H}_0 ): Noise only, ( \mathbf{x} \sim \mathcal{N}(0, \sigma^2 \mathbf{I}) )
  • Under ( \mathcal{H}_1 ): Signal + Noise, ( \mathbf{x} \sim \mathcal{N}(0, \sigma^2 \mathbf{I} + \sigma_A^2 \mathbf{h} \mathbf{h}^T) )

• Test Statistic:

  $T'(\mathbf{x}) = \left( \sum_{n=0}^{N-1} x[n] h[n] \right)^2 = \left( \mathbf{h}^T \mathbf{x} \right)^2$

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Step 1: Distribution of the Test Statistic

Under ( \mathcal{H}_0 ):

• Since ( \mathbf{x} \sim \mathcal{N}(0, \sigma^2 \mathbf{I}) ), the projection ( \mathbf{h}^T \mathbf{x} ) is Gaussian:

  $\mathbf{h}^T \mathbf{x} \sim \mathcal{N}(0, \sigma^2 \|\mathbf{h}\|^2)$

• Thus, the test statistic ( T'(\mathbf{x}) ) follows a scaled chi-squared distribution with 1 degree of freedom:

  $T'(\mathbf{x}) \sim \sigma^2 \|\mathbf{h}\|^2 \chi_1^2$

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Under ( \mathcal{H}_1 ):

• The mean of ( \mathbf{h}^T \mathbf{x} ) is:

  $\mu = \sqrt{\|\mathbf{h}\|^2} \sigma_A^2$

• The variance is:

  $\text{Var}(\mathbf{h}^T \mathbf{x}) = \sigma^2 \|\mathbf{h}\|^2$

• Hence, ( \mathbf{h}^T \mathbf{x} \sim \mathcal{N}(\mu, \sigma^2 |\mathbf{h}|^2) ), and ( T'(\mathbf{x}) ) follows a noncentral chi-squared distribution:

  $T'(\mathbf{x}) \sim \sigma^2 \|\mathbf{h}\|^2 \chi_1^2(\lambda)$

  where the noncentrality parameter is:

  $\lambda = \frac{\sigma_A^2 \|\mathbf{h}\|^2}{\sigma^2}$

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Step 2: Determine ( P_{FA} ) and ( P_D )

Probability of False Alarm (( P_{FA} )):

• Under ( \mathcal{H}_0 ), ( T'(\mathbf{x}) \sim \sigma^2 |\mathbf{h}|^2 \chi_1^2 )
• For a threshold ( \eta ), the probability of false alarm is:
  $P_{FA} = P(T'(\mathbf{x}) > \eta | \mathcal{H}_0) = Q_{\chi_1^2}\left(\frac{\eta}{\sigma^2 \|\mathbf{h}\|^2}\right)$

Probability of Detection (( P_D )):

• Under ( \mathcal{H}_1 ), ( T'(\mathbf{x}) \sim \sigma^2 |\mathbf{h}|^2 \chi_1^2(\lambda) )
• For the same threshold ( \eta ), the probability of detection is:
  $P_D = P(T'(\mathbf{x}) > \eta | \mathcal{H}_1) = Q_{\chi_1^2(\lambda)}\left(\frac{\eta}{\sigma^2 \|\mathbf{h}\|^2}\right)$

Here, ( Q{\chi_1^2} ) and ( Q{\chi_1^2(\lambda)} ) are the complementary cumulative distribution functions (CCDFs) of the central and noncentral chi-squared distributions, respectively.

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Summary of Results

1. The test statistic ( T'(\mathbf{x}) ) is:

   $T'(\mathbf{x}) = \left( \mathbf{h}^T \mathbf{x} \right)^2$

2. Under ( \mathcal{H}_0 ):

   $T'(\mathbf{x}) \sim \sigma^2 \|\mathbf{h}\|^2 \chi_1^2$

3. Under ( \mathcal{H}_1 ):

   $T'(\mathbf{x}) \sim \sigma^2 \|\mathbf{h}\|^2 \chi_1^2(\lambda), \quad \lambda = \frac{\sigma_A^2 \|\mathbf{h}\|^2}{\sigma^2}$

4. Probabilities:

   • False Alarm:
     $P_{FA} = Q_{\chi_1^2}\left(\frac{\eta}{\sigma^2 \|\mathbf{h}\|^2}\right)$
   • Detection:
     $P_D = Q_{\chi_1^2(\lambda)}\left(\frac{\eta}{\sigma^2 \|\mathbf{h}\|^2}\right)$

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P2

A rank one signal is a random signal whose mean is zero and whose covariance matrix has rank one. As such the covariance matrix can be written as $C_s=uu^T$, where $u$ is an $N\times 1$ vector.
Show that the signal $s[n]=Ah[n]$ for $n=0,1,\dots,N-1$ where $h[n]$ is a deterministic sequence and $A$ is a random variable with $E(A)=0$ and $\text{var}(A)=\sigma^2_A$ is a rank one signal

Problem Statement

A rank-one signal satisfies the following:
1. Mean $E[s[n]] = 0$,
2. Covariance matrix $\mathbf{C}_s$ of rank one:

where $\mathbf{u}$ is an $N \times 1$ vector.

We need to verify that the signal $s[n] = A h[n]$ is a rank-one signal, where:

• $h[n]$ is a deterministic sequence,
• $A$ is a random variable with:
  $E[A] = 0 \quad \text{and} \quad \text{Var}(A) = \sigma_A^2.$

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Step 1: Expressing the Signal

The signal $s[n]$ is defined as:

In vector form, this can be written as:

where:

• $\mathbf{s} = [s[0], s[1], \ldots, s[N-1]]^T$,
• $\mathbf{h} = [h[0], h[1], \ldots, h[N-1]]^T$.

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Step 2: Mean of the Signal

The mean of $\mathbf{s}$ is given by:

Using the property of expectation and the fact that $A$ is a random variable with $E[A] = 0$:

Thus, the mean of the signal is zero:

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Step 3: Covariance Matrix of the Signal

The covariance matrix of $\mathbf{s}$ is defined as:

Substitute $\mathbf{s} = A \mathbf{h}$:

Simplify:

Since $E[A^2] = \text{Var}(A) = \sigma_A^2$ (because $E[A] = 0$):

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Step 4: Rank of the Covariance Matrix

The matrix $\mathbf{C}_s = \sigma_A^2 \mathbf{h} \mathbf{h}^T$ is a rank-one matrix because it is formed as the outer product of the vector $\mathbf{h}$ with itself. Specifically:
1. The rank of $\mathbf{h} \mathbf{h}^T$ is 1 if $\mathbf{h} \neq \mathbf{0}$.
2. Scaling by $\sigma_A^2$ does not change the rank.

Thus, $\mathbf{C}_s$ is a rank-one matrix.

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Conclusion

The signal $s[n] = A h[n]$ satisfies:
1. $E[s[n]] = 0$,
2. The covariance matrix $\mathbf{C}_s = \sigma_A^2 \mathbf{h} \mathbf{h}^T$ is of rank one.

Therefore, $s[n] = A h[n]$ is a rank-one signal.

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P3

We observe two independent samples $x[n]$ for $n = 0, 1$ from the exponential PDF:

where $\lambda$ is unknown and $\lambda > 0$.

The hypothesis testing problem is:

• $H_0: \lambda = \lambda_0$
• $H_1: \lambda > \lambda_0$

We aim to determine:
1. If a uniformly most powerful (UMP) test exists,
2. The test statistic $T(x)$,
3. $P_{FA}$ as a function of the threshold.

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Step 1: Likelihood Ratio Test (LRT)

Joint PDF

The two samples $x[0]$ and $x[1]$ are independent and identically distributed (IID). The joint PDF is:

Likelihood Ratio

The likelihood ratio (LR) is:

Simplify:

Taking the logarithm (monotonic transformation), we can write:

where $S = x[0] + x[1]$ is the sufficient statistic.

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Step 2: Uniformly Most Powerful (UMP) Test

Neyman-Pearson Lemma

For a one-sided hypothesis test ($H_1: \lambda > \lambda_0$), the Neyman-Pearson Lemma ensures that a UMP test exists if the likelihood ratio $\Lambda(x)$ is a monotonic function of the sufficient statistic $S = x[0] + x[1]$.

From the likelihood ratio:

• $\Lambda(x)$ increases as $S$ decreases (since $(\lambda_1 - \lambda_0) > 0$).

Thus, the UMP test exists, and the test statistic is:

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Step 3: Test Threshold and Decision Rule

The decision rule for the UMP test is:

• Reject $H_0$ if $T(x) < \eta$, where $\eta$ is the threshold.

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Step 4: False Alarm Probability ($P_{FA}$)

Under $H_0$, $\lambda = \lambda_0$, and the sufficient statistic $S = x[0] + x[1]$ is the sum of two IID exponential random variables. The sum of two exponential random variables follows a Gamma distribution:

The cumulative distribution function (CDF) is:

The false alarm probability is:

Substitute the CDF:

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Summary

1. UMP Test: The UMP test exists.
2. Test Statistic: $T(x) = x[0] + x[1]$.
3. False Alarm Probability:
   $P_{FA} = 1 - \exp(-\lambda_0 \eta)(1 + \lambda_0 \eta).$

The threshold $\eta$ can be set to achieve a desired $P_{FA}$.