[DetnEst] Assignment 3

KBC·2024년 10월 25일

Detection and Estimation

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Given 3 data points $(t_0,x_0),\;(t_1,x_1)$ and $(t_2,x_2)$
We can have a second order polynomial, where 3 coefficients are unknown
What is the best estimate for unknown parameters $\theta_0,\theta_1,\text{and }\theta_2$ of a linear model $x_n=\theta_0+\theta_1t_n+\theta_2t^2_n+w_n$
if the data points are $(1,0),(2,0.69),(4,1.39)$ , where $\ln 2 \sim0.69$ and $\ln 4\sim1.39$

\vec{x} = H \vec{\theta} + \vec{w}\\[0.3cm] \Rightarrow \begin{bmatrix} x_0 \\ x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 1 & t_0 & t_0^2 \\ 1 & t_1 & t_1^2 \\ 1 & t_2 & t_2^2 \end{bmatrix} \begin{bmatrix} \theta_0 \\ \theta_1 \\ \theta_2 \end{bmatrix} + \begin{bmatrix} w_0 \\ w_1 \\ w_2 \end{bmatrix}\\[0.3cm] \Rightarrow H = \begin{bmatrix} 1 & t_0 & t_0^2 \\ 1 & t_1 & t_1^2 \\ 1 & t_2 & t_2^2 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 16 \end{bmatrix}\\[0.3cm] \vec{x} = \begin{bmatrix} x_0 \\ x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ \ln 2 \\ \ln 4 \end{bmatrix}

Since the $H$ is invertible : $\left( H^T H \right)^{-1} H^T = H^{-1}$ $\hat{\vec{\theta}} = H^{-1} \vec{x}\\[0.2cm] = \begin{bmatrix} \frac{8}{3} & -2 & \frac{1}{3} \\ -2 & \frac{5}{2} & -\frac{1}{2} \\ \frac{1}{3} & -\frac{1}{2} & \frac{1}{6} \end{bmatrix} \begin{bmatrix} 0 \\ \ln 2 \\ \ln 4 \end{bmatrix} = \begin{bmatrix} -\frac{5}{3} \ln 2 \\ \frac{3}{2} \ln 2 \\ -\frac{1}{6} \ln 2 \end{bmatrix} = \begin{bmatrix} -0.92 \\ 1.037 \\ -0.115 \end{bmatrix}$

Assume that we have a single sinusoidal component at $f_k=k/N$
The model as given by (4.12) is $x[n]=a_k\cos(2\pi f_k n)+b_k\sin(2\pi f_k n)+w[n],\;n=0,1,\dots,N-1$
Using the identity $A\cos w+B\sin w = \sqrt{A^2+B^2}\cos(w-\phi)$ , where $\phi = \arctan(B/A)$
We can rewrite the model as $x[n]=\sqrt{a^2_k+b^2_k}\cos(2\pi f_kn-\phi)+w[n]$
An MVUE is used for $a_k,b_k$ , so that the estimated power of the sinusoid is $\hat P=\frac{\hat a_k^2+\hat b^2_k}{2}$
A measure of detectability is $E^2(\hat P)/\text{var}(\hat P)$
Compare the measure when a sinsoid is present to the case when only noise is present or $a_k=b_k=0$
Could you use the estimated power to decide if a signal is present?

\hat{\Theta} = \begin{bmatrix} \hat{a}_k \\ \hat{b}_k \end{bmatrix} \Rightarrow C_{\hat{\Theta}} = \frac{2\sigma^2}{N} I\\[0.3cm] \hat{a}_k \sim \mathcal{N}(a_k, \frac{2\sigma^2}{N}), \quad \hat{b}_k \sim \mathcal{N}(b_k, \frac{2\sigma^2}{N})\\[0.3cm] E\left[ \hat{p}_k^2 \right] = E\left[ \frac{\hat{a}_k^2 + \hat{b}_k^2}{2} \right]\\[0.3cm] = \frac{1}{2} \left( \text{Var}(\hat{a}_k) + E^2[\hat{a}_k] + \text{Var}(\hat{b}_k) + E^2[\hat{b}_k] \right)\\[0.3cm] = \frac{1}{2} \left( 2 \cdot \frac{2\sigma^2}{N} + a_k^2 + b_k^2 \right)\\[0.3cm] = \frac{2\sigma^2}{N} + \frac{a_k^2 + b_k^2}{2}\\[0.3cm] E^2[\hat{p}_k^2] = \left( \frac{2\sigma^2}{N} + \frac{a_k^2 + b_k^2}{2} \right)^2\\[0.3cm] \text{Var}[\hat{p}_k] = \text{Var} \left( \frac{\hat{a}_k^2 + \hat{b}_k^2}{2} \right) = \frac{1}{4} \left( \text{Var}(\hat{a}_k^2) + \text{Var}(\hat{b}_k^2) \right)

For the Gaussian $\text{If } X \sim \text{Gaussian Distribution } (\mu, \sigma^2), \quad \text{Var}[X^2] = E[X^4] - E^2[X^2]\\[0.3cm] E[X^4] = \mu^4 + 6\mu^2 \sigma^2 + 3\sigma^4, \quad E^2[X^2] = (\mu^2 + \sigma^2)^2\\[0.3cm] \text{Var}[X^2] = 4\mu^2 \sigma^2 + 2\sigma^4$
Then

\text{Var}(\hat{a}_k^2) = 4a_k^2 \frac{2\sigma^2}{N} + 2 \frac{4\sigma^4}{N^2} \quad (b_k \text{도 동일})\\[0.3cm] \text{Var}(\hat{p}_k) = \frac{1}{4} \left( 4a_k^2 \frac{2\sigma^2}{N} + 8 \frac{\sigma^4}{N^2} + 4b_k^2 \frac{2\sigma^2}{N} + 8 \frac{\sigma^4}{N^2} \right)\\[0.3cm] = \frac{(a_k^2 + b_k^2)}{4} \frac{\sigma^2}{N} + \frac{4\sigma^4}{N^2} + \frac{2\sigma^2 (a_k^2 + b_k^2)}{N} + \frac{4\sigma^4}{N^2}\\[0.3cm] \Rightarrow \frac{E^2[\hat{p}_k^2]}{\text{Var}(\hat{p}_k)} = \frac{(a_k^2 + b_k^2)^2/4 + 2\sigma^2 (a_k^2 + b_k^2)/N + 4\sigma^4/N^2}{(a_k^2 + b_k^2) \frac{\sigma^2}{2N} + 4 \frac{\sigma^4}{N^2}}

Consequently $\text{If } a_k, b_k \to 0, \quad \frac{E^2[\hat{p}_k^2]}{\text{Var}(\hat{p}_k)} \to \frac{4\sigma^4/N^2}{4\sigma^4/N^2} = 1\\[0.3cm] \text{If not}, \quad \frac{E^2[\hat{p}_k^2]}{\text{Var}(\hat{p}_k)} > 1$

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