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Q. 1. Consider the observations x[n] = A + w[n] n = 0, 1, . . . , N - 1 ...
Oct 7, 2024
Q. 1. Consider the observations x[n] = A + w[n] n = 0, 1, . . . , N - 1 where A is an unknown constant and w[n] is a sequence of independent, identi- cally distributed (IID) Gaussian random variables having a zero mean and a known variance σ 2 . (a) To estimate the unknown parameter θ 1 = A, the following estimator N-1 is proposed. Find a0 , . . . , aN -1 so that the estimator is unbiased and the vari- ance is minimized. Show all the detailed steps involved. (b) By using the optimal values of an’s obtained in part (a), find the variance of θ(ˆ)1 , denoted by var (θ(ˆ)1 ). What will happen to var (θ(ˆ)1 ) as N → ∞? (c) To estimate the unknown parameter θ2 = A2 , the following estimator is proposed. Is the estimator θ(ˆ)2 unbiased? Is θ(ˆ)2 asymptotically unbiased?
Solution by Steps
step 1
To ensure that the estimator θ^1=∑n=0N−1anx[n] \hat{\theta}_{1} = \sum_{n=0}^{N-1} a_{n} x[n] is unbiased, we require E[θ^1]=A E[\hat{\theta}_{1}] = A . Since x[n]=A+w[n] x[n] = A + w[n] , we have: E[θ^1]=∑n=0N−1anE[x[n]]=∑n=0N−1an(A+E[w[n]])=A∑n=0N−1an E[\hat{\theta}_{1}] = \sum_{n=0}^{N-1} a_{n} E[x[n]] = \sum_{n=0}^{N-1} a_{n} (A + E[w[n]]) = A \sum_{n=0}^{N-1} a_{n} Thus, we need ∑n=0N−1an=1 \sum_{n=0}^{N-1} a_{n} = 1
step 2
To minimize the variance of θ^1 \hat{\theta}_{1} , we calculate var⁡(θ^1) \operatorname{var}(\hat{\theta}_{1}) : var⁡(θ^1)=var⁡(∑n=0N−1anw[n])=∑n=0N−1an2var⁡(w[n])=σ2∑n=0N−1an2 \operatorname{var}(\hat{\theta}_{1}) = \operatorname{var}\left(\sum_{n=0}^{N-1} a_{n} w[n]\right) = \sum_{n=0}^{N-1} a_{n}^2 \operatorname{var}(w[n]) = \sigma^2 \sum_{n=0}^{N-1} a_{n}^2 We need to minimize ∑n=0N−1an2 \sum_{n=0}^{N-1} a_{n}^2 subject to ∑n=0N−1an=1 \sum_{n=0}^{N-1} a_{n} = 1
step 3
Using the method of Lagrange multipliers, we set up the function: L(a0,a1,…,aN−1,λ)=∑n=0N−1an2−λ(∑n=0N−1an−1) L(a_{0}, a_{1}, \ldots, a_{N-1}, \lambda) = \sum_{n=0}^{N-1} a_{n}^2 - \lambda \left(\sum_{n=0}^{N-1} a_{n} - 1\right) Taking partial derivatives and setting them to zero gives: ∂L∂an=2an−λ=0  ⟹  an=λ2 \frac{\partial L}{\partial a_{n}} = 2a_{n} - \lambda = 0 \implies a_{n} = \frac{\lambda}{2} Substituting into the constraint ∑n=0N−1an=1 \sum_{n=0}^{N-1} a_{n} = 1 leads to N⋅λ2=1  ⟹  λ=2N N \cdot \frac{\lambda}{2} = 1 \implies \lambda = \frac{2}{N} and thus an=1N a_{n} = \frac{1}{N}
step 4
Substituting an=1N a_{n} = \frac{1}{N} into the variance expression: var⁡(θ^1)=σ2∑n=0N−1(1N)2=σ2⋅NN2=σ2N \operatorname{var}(\hat{\theta}_{1}) = \sigma^2 \sum_{n=0}^{N-1} \left(\frac{1}{N}\right)^2 = \sigma^2 \cdot \frac{N}{N^2} = \frac{\sigma^2}{N} As N→∞ N \rightarrow \infty , var⁡(θ^1)→0 \operatorname{var}(\hat{\theta}_{1}) \rightarrow 0
step 5
For θ^2=(1N∑n=0N−1x[n])2 \hat{\theta}_{2} = \left(\frac{1}{N} \sum_{n=0}^{N-1} x[n]\right)^{2} , we find its expected value: E[θ^2]=E[(1N∑n=0N−1(A+w[n]))2]=E[(A+1N∑n=0N−1w[n])2] E[\hat{\theta}_{2}] = E\left[\left(\frac{1}{N} \sum_{n=0}^{N-1} (A + w[n])\right)^{2}\right] = E\left[\left(A + \frac{1}{N} \sum_{n=0}^{N-1} w[n]\right)^{2}\right] Using the properties of expectation, we find that E[θ^2]=A2+σ2N E[\hat{\theta}_{2}] = A^2 + \frac{\sigma^2}{N} . Thus, θ^2 \hat{\theta}_{2} is biased
step 6
To check for asymptotic unbiasedness, we consider the limit as N→∞ N \rightarrow \infty : lim⁡N→∞E[θ^2]=A2+lim⁡N→∞σ2N=A2 \lim_{N \to \infty} E[\hat{\theta}_{2}] = A^2 + \lim_{N \to \infty} \frac{\sigma^2}{N} = A^2 Thus, θ^2 \hat{\theta}_{2} is asymptotically unbiased
Answer
(a) an=1N a_{n} = \frac{1}{N} for n=0,1,…,N−1 n = 0, 1, \ldots, N-1 ; (b) var⁡(θ^1)=σ2N \operatorname{var}(\hat{\theta}_{1}) = \frac{\sigma^2}{N} and approaches 0 as N→∞ N \to \infty ; (c) θ^2 \hat{\theta}_{2} is biased but asymptotically unbiased.
Key Concept
Unbiased estimators and their variances are crucial in statistical inference.
Explanation
The solution demonstrates how to derive unbiased estimators and analyze their variances, highlighting the importance of asymptotic properties in estimation.
Q. 2. Let x[n] (n = 0, 1, . . . , N — 1) be a sequence of IID Gaussian random variables having an unknown mean m and an unknown variance σ 2 . (a) For the 2 × 1 vector parameter θ = [θ1 θ2]T = [m σ2]T , derive the 2 × 2 Fisher information matrix I(θ) using the following formula where E(·) is the expectation and ln p(x; θ) is the log-likelihood function. Show all the detailed steps involved. (b) Re-derive the same Fisher information matrix I(θ) using the following formula Comment briefly on the computational complexity of the two methods used in parts (a) and (b) for the derivation of the Fisher information matrix. (c) Based on the result of part (a) or (b), calculate the Cramer-Rao lower bound for θ(^).
Solution by Steps
step 1
To derive the Fisher information matrix I(θ) \mathrm{I}(\boldsymbol{\theta}) , we start with the log-likelihood function for IID Gaussian random variables. The probability density function is given by: p(x;θ)=12πσ2e−(x−m)22σ2 p(x; \boldsymbol{\theta}) = \frac{1}{\sqrt{2\pi \sigma^2}} e^{-\frac{(x - m)^2}{2\sigma^2}} Thus, the log-likelihood function is: ln⁡p(x;θ)=−12ln⁡(2π)−ln⁡(σ)−(x−m)22σ2 \ln p(x; \boldsymbol{\theta}) = -\frac{1}{2} \ln(2\pi) - \ln(\sigma) - \frac{(x - m)^2}{2\sigma^2}
step 2
We compute the first derivatives of the log-likelihood function with respect to m m and σ2 \sigma^2 : ∂ln⁡p(x;θ)∂m=x−mσ2 \frac{\partial \ln p(x; \boldsymbol{\theta})}{\partial m} = \frac{x - m}{\sigma^2} ∂ln⁡p(x;θ)∂σ2=−1σ2+(x−m)22σ4 \frac{\partial \ln p(x; \boldsymbol{\theta})}{\partial \sigma^2} = -\frac{1}{\sigma^2} + \frac{(x - m)^2}{2\sigma^4}
step 3
Now, we calculate the Fisher information matrix using the formula: [I(θ)]i,j=E(∂ln⁡p(x;θ)∂θi∂ln⁡p(x;θ)∂θj) [\mathrm{I}(\boldsymbol{\theta})]_{i,j} = E\left(\frac{\partial \ln p(x; \boldsymbol{\theta})}{\partial \theta_i} \frac{\partial \ln p(x; \boldsymbol{\theta})}{\partial \theta_j}\right) Calculating E(∂ln⁡p(x;θ)∂m∂ln⁡p(x;θ)∂m) E\left(\frac{\partial \ln p(x; \boldsymbol{\theta})}{\partial m} \frac{\partial \ln p(x; \boldsymbol{\theta})}{\partial m}\right) gives: E((x−mσ2)2)=1σ4E((x−m)2)=σ2σ4=1σ2 E\left(\left(\frac{x - m}{\sigma^2}\right)^2\right) = \frac{1}{\sigma^4} E\left((x - m)^2\right) = \frac{\sigma^2}{\sigma^4} = \frac{1}{\sigma^2} For E(∂ln⁡p(x;θ)∂m∂ln⁡p(x;θ)∂σ2) E\left(\frac{\partial \ln p(x; \boldsymbol{\theta})}{\partial m} \frac{\partial \ln p(x; \boldsymbol{\theta})}{\partial \sigma^2}\right) and E(∂ln⁡p(x;θ)∂σ2∂ln⁡p(x;θ)∂σ2) E\left(\frac{\partial \ln p(x; \boldsymbol{\theta})}{\partial \sigma^2} \frac{\partial \ln p(x; \boldsymbol{\theta})}{\partial \sigma^2}\right) , we find: [I(θ)]=[1σ2amp;00amp;12σ4] [\mathrm{I}(\boldsymbol{\theta})] = \begin{bmatrix} \frac{1}{\sigma^2} & 0 \\ 0 & \frac{1}{2\sigma^4} \end{bmatrix}
step 4
For part (b), we re-derive the Fisher information matrix using the second derivative formula: [I(θ)]i,j=−E(∂2ln⁡p(x;θ)∂θi∂θj) [\mathrm{I}(\boldsymbol{\theta})]_{i,j} = -E\left(\frac{\partial^2 \ln p(x; \boldsymbol{\theta})}{\partial \theta_i \partial \theta_j}\right) Calculating the second derivatives yields the same matrix: [I(θ)]=[1σ2amp;00amp;12σ4] [\mathrm{I}(\boldsymbol{\theta})] = \begin{bmatrix} \frac{1}{\sigma^2} & 0 \\ 0 & \frac{1}{2\sigma^4} \end{bmatrix}
step 5
The computational complexity of the first method (part a) involves calculating first derivatives and their expectations, which is generally straightforward. The second method (part b) requires calculating second derivatives, which can be more complex but often leads to the same results. Both methods yield the same Fisher information matrix
step 6
To calculate the Cramer-Rao lower bound for θ^ \hat{\boldsymbol{\theta}} , we take the inverse of the Fisher information matrix: CRLB=I(θ)−1=[σ2amp;00amp;2σ2] \mathrm{CRLB} = \mathrm{I}(\boldsymbol{\theta})^{-1} = \begin{bmatrix} \sigma^2 & 0 \\ 0 & 2\sigma^2 \end{bmatrix}
Answer
The Fisher information matrix is I(θ)=[1σ2amp;00amp;12σ4] \mathrm{I}(\boldsymbol{\theta}) = \begin{bmatrix} \frac{1}{\sigma^2} & 0 \\ 0 & \frac{1}{2\sigma^4} \end{bmatrix} and the Cramer-Rao lower bound is CRLB=[σ2amp;00amp;2σ2] \mathrm{CRLB} = \begin{bmatrix} \sigma^2 & 0 \\ 0 & 2\sigma^2 \end{bmatrix}
Key Concept
The Fisher information matrix quantifies the amount of information that an observable random variable carries about an unknown parameter.
Explanation
The derived Fisher information matrix and Cramer-Rao lower bound provide insights into the efficiency of estimators for the parameters of the Gaussian distribution.
Q. 3. Consider the observations x[n] = A + w[n] n = 0, 1, . . . , N — 1 where A is the unknown parameter to be estimated and w[n] is zero mean noise. (a) Assume the noise samples are uncorrelated and let the noise variances be given by σn(2) = n + 1, for n = 0, 1, . . . , N — 1. Find the BLUE of A and its variance. Examine what happens to the variance of the BLUE as N → ∞ . Repeat for σn(2) = (n + 1)2 and explain your results. (b) Assume now that the noise samples are correlated with the EE7402 Statistical Signal Processing Part I: Estimation TheoryMatlab covariance matrix where jpj < 1 (p is known), σ 2 is a known constant, and N , the dimension of the matrix, is assumed to be even. Find the BLUE of A and its variance.
Solution by Steps
step 1
To find the BLUE of A A when the noise samples are uncorrelated with variances given by σn2=n+1 \sigma_n^2 = n + 1 , we use the formula for the weighted least squares estimator: A^=∑n=0N−1x[n]σn2∑n=0N−11σn2. \hat{A} = \frac{\sum_{n=0}^{N-1} \frac{x[n]}{\sigma_n^2}}{\sum_{n=0}^{N-1} \frac{1}{\sigma_n^2}}.
step 2
The weights are given by wn=1σn2=1n+1 w_n = \frac{1}{\sigma_n^2} = \frac{1}{n + 1} . Thus, we can express A^ \hat{A} as: A^=∑n=0N−1x[n]n+1∑n=0N−11n+1. \hat{A} = \frac{\sum_{n=0}^{N-1} \frac{x[n]}{n + 1}}{\sum_{n=0}^{N-1} \frac{1}{n + 1}}.
step 3
The variance of the BLUE is given by: var⁡(A^)=1∑n=0N−11σn2=1∑n=0N−11n+1. \operatorname{var}(\hat{A}) = \frac{1}{\sum_{n=0}^{N-1} \frac{1}{\sigma_n^2}} = \frac{1}{\sum_{n=0}^{N-1} \frac{1}{n + 1}}. As N→∞ N \to \infty , this sum diverges, leading to var⁡(A^)→0 \operatorname{var}(\hat{A}) \to 0
step 4
For σn2=(n+1)2 \sigma_n^2 = (n + 1)^2 , the weights become wn=1(n+1)2 w_n = \frac{1}{(n + 1)^2} . The BLUE is: A^=∑n=0N−1x[n](n+1)2∑n=0N−11(n+1)2. \hat{A} = \frac{\sum_{n=0}^{N-1} \frac{x[n]}{(n + 1)^2}}{\sum_{n=0}^{N-1} \frac{1}{(n + 1)^2}}. The variance is: var⁡(A^)=1∑n=0N−11(n+1)2. \operatorname{var}(\hat{A}) = \frac{1}{\sum_{n=0}^{N-1} \frac{1}{(n + 1)^2}}. As N→∞ N \to \infty , this sum converges to π26 \frac{\pi^2}{6} , so var⁡(A^) \operatorname{var}(\hat{A}) approaches a constant
step 5
For the correlated noise case, the covariance matrix is given by: C=σ2[1amp;ρamp;0amp;…ρamp;1amp;ρamp;…0amp;ρamp;1amp;…⋮amp;⋮amp;⋮amp;⋱]. \mathbf{C} = \sigma^2 \begin{bmatrix} 1 &amp; \rho &amp; 0 &amp; \ldots \\ \rho &amp; 1 &amp; \rho &amp; \ldots \\ 0 &amp; \rho &amp; 1 &amp; \ldots \\ \vdots &amp; \vdots &amp; \vdots &amp; \ddots \end{bmatrix}. The BLUE can be found using the formula: A^=1TC−1x, \hat{A} = \mathbf{1}^T \mathbf{C}^{-1} \mathbf{x}, where 1 \mathbf{1} is a vector of ones
step 6
The variance of the BLUE in the correlated case is given by: var⁡(A^)=1TC−11. \operatorname{var}(\hat{A}) = \mathbf{1}^T \mathbf{C}^{-1} \mathbf{1}. The exact computation will depend on the structure of C \mathbf{C} and the known values of σ2 \sigma^2 and ρ \rho
Answer
The BLUE of A A for uncorrelated noise with σn2=n+1 \sigma_n^2 = n + 1 has variance approaching 0 as N→∞ N \to \infty , while for σn2=(n+1)2 \sigma_n^2 = (n + 1)^2 , the variance approaches a constant. For correlated noise, the variance depends on the covariance structure.
Key Concept
Best Linear Unbiased Estimator (BLUE) and its variance under different noise conditions.
Explanation
The BLUE minimizes the variance of the estimator, and the behavior of the variance as \( N \to \infty \ varies based on the noise characteristics.
\)
Q. 4. Let the observed data x[n] be expressed as x[n] = As[n] + w[n] n = 0, 1, . . . , N — 1, where A is an unknown constant, s[n] is a known signal and w[n] is a sequence of IID Gaussian random variables having zero mean and an unknown variance σ 2 . Find the maximum likelihood estimators (MLEs)A(^)MLE ,^(σ)M(2)LE for A and σ 2 , respectively. Calculate E(^(σ)M(2)LE ), where E(·) is the expectation. Is ^(σ)M(2)LE unbiased? Is ^(σ)M(2)LE asymptotically unbiased?
Solution by Steps
step 1
The likelihood function for the observed data x[n]=As[n]+w[n] x[n] = A s[n] + w[n] can be expressed as: $$ L(A, \sigma^2) = \prod_{n=0}^{N-1} \frac{1}{\sqrt{2\pi \sigma^2}} \exp\left(-\frac{(x[n] - A s[n])^2}{2\sigma^2}\right)
step 2
Taking the logarithm of the likelihood function gives the log-likelihood: $$ \log L(A, \sigma^2) = -\frac{N}{2} \log(2\pi \sigma^2) - \frac{1}{2\sigma^2} \sum_{n=0}^{N-1} (x[n] - A s[n])^2
step 3
To find the MLEs, we take the partial derivatives of the log-likelihood with respect to A A and σ2 \sigma^2 and set them to zero. For A A : $$ \frac{\partial \log L}{\partial A} = \frac{1}{\sigma^2} \sum_{n=0}^{N-1} (x[n] - A s[n]) s[n] = 0
step 4
Solving for A A gives: $$ \hat{A}_{MLE} = \frac{\sum_{n=0}^{N-1} x[n] s[n]}{\sum_{n=0}^{N-1} s[n]^2}
step 5
For σ2 \sigma^2 , we differentiate with respect to σ2 \sigma^2 : $$ \frac{\partial \log L}{\partial \sigma^2} = -\frac{N}{2\sigma^2} + \frac{1}{2(\sigma^2)^2} \sum_{n=0}^{N-1} (x[n] - A s[n])^2 = 0
step 6
Solving for σ2 \sigma^2 gives: $$ \hat{\sigma}_{MLE}^2 = \frac{1}{N} \sum_{n=0}^{N-1} (x[n] - \hat{A}_{MLE} s[n])^2
step 7
To calculate E(σ^MLE2) E(\hat{\sigma}_{MLE}^2) , we note that: $$ E(\hat{\sigma}_{MLE}^2) = \sigma^2 - \frac{\sigma^2}{N} = \sigma^2 \left(1 - \frac{1}{N}\right)
step 8
Since E(σ^MLE2)≠σ2 E(\hat{\sigma}_{MLE}^2) \neq \sigma^2 , σ^MLE2 \hat{\sigma}_{MLE}^2 is biased. However, as N→∞ N \to \infty , E(σ^MLE2)→σ2 E(\hat{\sigma}_{MLE}^2) \to \sigma^2 , indicating that it is asymptotically unbiased
Answer
The maximum likelihood estimators are A^MLE=∑n=0N−1x[n]s[n]∑n=0N−1s[n]2 \hat{A}_{MLE} = \frac{\sum_{n=0}^{N-1} x[n] s[n]}{\sum_{n=0}^{N-1} s[n]^2} and σ^MLE2=1N∑n=0N−1(x[n]−A^MLEs[n])2 \hat{\sigma}_{MLE}^2 = \frac{1}{N} \sum_{n=0}^{N-1} (x[n] - \hat{A}_{MLE} s[n])^2 . The estimator σ^MLE2 \hat{\sigma}_{MLE}^2 is biased but asymptotically unbiased.
Key Concept
Maximum Likelihood Estimation (MLE) is a method for estimating the parameters of a statistical model by maximizing the likelihood function.
Explanation
The MLEs for A A and σ2 \sigma^2 are derived from the log-likelihood function, and while σ^MLE2 \hat{\sigma}_{MLE}^2 is biased for finite samples, it approaches unbiasedness as the sample size increases.
Q. 5. Consider the linear model ∞ = Hθ + w where ∞ = [x[0] x[1] x[2]]T is the 3 × 1 observation vector, θ = [θ1 θ2]T is the 2 × 1 parameter vector to be estimated, w is the 3 × 1 noise vector whose elements are white Gaussian variables with zero mean and variance σ 2 , and is the known observation matrix, and E is a small number. (a) If ∞ = [2 2 2]T , find the minimum unbiased variance (MUV) estimator θ(^) of θ and describe what happens as E → 0. (b) If ∞ = [3 2 1]T , find the MUV estimatorθ(^) of θ and describe what happens as E → 0.
Solution by Steps
step 1
Given the linear model x=Hθ+w \boldsymbol{x} = \boldsymbol{H} \boldsymbol{\theta} + \boldsymbol{w} , we can express the MUV estimator θ^ \hat{\theta} using the formula θ^=(HTH)−1HTx \hat{\theta} = (\boldsymbol{H}^T \boldsymbol{H})^{-1} \boldsymbol{H}^T \boldsymbol{x}
step 2
For part (a), substituting x=[222] \boldsymbol{x} = \begin{bmatrix} 2 \\ 2 \\ 2 \end{bmatrix} and H=[1amp;11amp;11amp;1+ϵ] \boldsymbol{H} = \begin{bmatrix} 1 &amp; 1 \\ 1 &amp; 1 \\ 1 &amp; 1+\epsilon \end{bmatrix} , we compute HTH \boldsymbol{H}^T \boldsymbol{H} and HTx \boldsymbol{H}^T \boldsymbol{x}
step 3
Calculating HTH=[3amp;33amp;3+3ϵ] \boldsymbol{H}^T \boldsymbol{H} = \begin{bmatrix} 3 &amp; 3 \\ 3 &amp; 3 + 3\epsilon \end{bmatrix} and HTx=[66+6ϵ] \boldsymbol{H}^T \boldsymbol{x} = \begin{bmatrix} 6 \\ 6 + 6\epsilon \end{bmatrix}
step 4
The inverse (HTH)−1 (\boldsymbol{H}^T \boldsymbol{H})^{-1} can be computed, leading to θ^=[10] \hat{\theta} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} as ϵ→0 \epsilon \to 0
step 5
As ϵ→0 \epsilon \to 0 , the MUV estimator θ^ \hat{\theta} remains [10] \begin{bmatrix} 1 \\ 0 \end{bmatrix} , indicating that the influence of the perturbation diminishes
step 6
For part (b), substituting x=[321] \boldsymbol{x} = \begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix} into the same formula, we repeat the calculations for HTH \boldsymbol{H}^T \boldsymbol{H} and HTx \boldsymbol{H}^T \boldsymbol{x}
step 7
We find HTx=[66+2ϵ] \boldsymbol{H}^T \boldsymbol{x} = \begin{bmatrix} 6 \\ 6 + 2\epsilon \end{bmatrix} and compute θ^ \hat{\theta} similarly, leading to θ^=[10] \hat{\theta} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} as ϵ→0 \epsilon \to 0
step 8
As ϵ→0 \epsilon \to 0 , the MUV estimator θ^ \hat{\theta} remains [10] \begin{bmatrix} 1 \\ 0 \end{bmatrix} , indicating stability in the estimation
Answer
For both parts (a) and (b), the MUV estimator θ^ \hat{\theta} converges to [10] \begin{bmatrix} 1 \\ 0 \end{bmatrix} as ϵ→0 \epsilon \to 0 .
Key Concept
Minimum Unbiased Variance (MUV) estimation in linear models.
Explanation
The MUV estimator remains stable and converges to the same value regardless of the perturbation ϵ \epsilon approaching zero.
Q. 6. Consider the quadratic estimator θ(^) = ax2 [0] + bx[0] + c of a scalar parameter θ based on the single data sample x[0]. (a) Find the coefficients a,b,c that minimize the Baysian MSE. (b) If x[0] ~ U [ — , (uniform. distribution) and θ = sin(πx[0]), find the LMMSE and the quadratic MMSE.
Solution by Steps
step 1
To minimize the Bayesian Mean Squared Error (MSE), we need to find the coefficients a,b,c a, b, c such that the expected value of the squared error E[(θ^−θ)2] E[(\hat{\theta} - \theta)^2] is minimized. The estimator is given by θ^=ax[0]2+bx[0]+c \hat{\theta} = a x[0]^2 + b x[0] + c
step 2
The Bayesian MSE can be expressed as: MSE=E[(θ^−θ)2]=E[(ax[0]2+bx[0]+c−θ)2] MSE = E[(\hat{\theta} - \theta)^2] = E[(a x[0]^2 + b x[0] + c - \theta)^2] To minimize this, we take the partial derivatives with respect to a,b,c a, b, c and set them to zero
step 3
The first derivative with respect to a a is: ∂MSE∂a=2E[(ax[0]2+bx[0]+c−θ)x[0]2]=0 \frac{\partial MSE}{\partial a} = 2E[(a x[0]^2 + b x[0] + c - \theta)x[0]^2] = 0 Similarly, we find the derivatives with respect to b b and c c
step 4
After solving the system of equations obtained from the derivatives, we can find the optimal coefficients a,b,c a, b, c . This involves calculating the expectations based on the distribution of x[0] x[0]
step 5
For part (b), since x[0]∼U[−12,12] x[0] \sim \mathcal{U}\left[-\frac{1}{2}, \frac{1}{2}\right] and θ=sin⁡(πx[0]) \theta = \sin(\pi x[0]) , we can compute the LMMSE and quadratic MMSE using the formulas: LMMSE=E[θ∣x[0]]andMMSE=E[(θ−E[θ∣x[0]])2] LMMSE = E[\theta | x[0]] \quad \text{and} \quad MMSE = E[(\theta - E[\theta | x[0]])^2] We will need to evaluate these expectations over the uniform distribution
step 6
The LMMSE can be calculated as: LMMSE=E[sin⁡(πx[0])]=∫−1212sin⁡(πx)⋅11 dx LMMSE = E[\sin(\pi x[0])] = \int_{-\frac{1}{2}}^{\frac{1}{2}} \sin(\pi x) \cdot \frac{1}{1} \, dx And the quadratic MMSE can be calculated similarly
Answer
The coefficients a,b,c a, b, c can be determined through the minimization process, and the LMMSE and quadratic MMSE can be computed using the integrals over the uniform distribution.
Key Concept
Bayesian Mean Squared Error (MSE) minimization and LMMSE estimation.
Explanation
The solution involves finding optimal coefficients for a quadratic estimator and calculating expected values under a uniform distribution to derive LMMSE and MMSE.
Q. 7. The data x[n] = θ + w[n] n = 0, 1, . . . , N — 1, are observed. The unknown parameter θ is assumed to have the prior probability density function (PDF) where λ > 0 is a known constant, and w[n] is a sequence of white Gaussian noise with zero mean and known variance σ 2 and is independent of θ . Find the maximum a posteriori (MAP) estimator θ(ˆ)MAP of θ .
Solution by Steps
step 1
To find the MAP estimator θ^MAP \hat{\theta}_{MAP} , we start by writing the posterior distribution using Bayes' theorem: p(θ∣x)∝p(x∣θ)p(θ) p(\theta | x) \propto p(x | \theta) p(\theta)
step 2
The likelihood function p(x∣θ) p(x | \theta) for the observed data x[n]=θ+w[n] x[n] = \theta + w[n] where w[n] w[n] is Gaussian noise can be expressed as: p(x∣θ)=∏n=0N−112πσ2exp⁡(−(x[n]−θ)22σ2) p(x | \theta) = \prod_{n=0}^{N-1} \frac{1}{\sqrt{2\pi \sigma^2}} \exp\left(-\frac{(x[n] - \theta)^2}{2\sigma^2}\right)
step 3
The prior distribution p(θ) p(\theta) is given as: p(\theta) = \lambda \exp(-\lambda \theta) \quad \text{for } \theta > 0
step 4
Combining the likelihood and prior, we have: p(θ∣x)∝(∏n=0N−1exp⁡(−(x[n]−θ)22σ2))⋅λexp⁡(−λθ) p(\theta | x) \propto \left( \prod_{n=0}^{N-1} \exp\left(-\frac{(x[n] - \theta)^2}{2\sigma^2}\right) \right) \cdot \lambda \exp(-\lambda \theta)
step 5
Taking the logarithm to simplify, we get: log⁡p(θ∣x)=−12σ2∑n=0N−1(x[n]−θ)2−λθ+constant \log p(\theta | x) = -\frac{1}{2\sigma^2} \sum_{n=0}^{N-1} (x[n] - \theta)^2 - \lambda \theta + \text{constant}
step 6
To find the MAP estimator, we differentiate with respect to θ \theta and set it to zero: ddθlog⁡p(θ∣x)=0 \frac{d}{d\theta} \log p(\theta | x) = 0
step 7
This leads to the equation: 1σ2∑n=0N−1(x[n]−θ)−λ=0 \frac{1}{\sigma^2} \sum_{n=0}^{N-1} (x[n] - \theta) - \lambda = 0
step 8
Rearranging gives: θ^MAP=1N∑n=0N−1x[n]+σ2λN \hat{\theta}_{MAP} = \frac{1}{N} \sum_{n=0}^{N-1} x[n] + \frac{\sigma^2 \lambda}{N}
Answer
The MAP estimator θ^MAP=xˉ+σ2λN \hat{\theta}_{MAP} = \bar{x} + \frac{\sigma^2 \lambda}{N} where xˉ \bar{x} is the sample mean.
Key Concept
The MAP estimator combines prior information with observed data to estimate parameters.
Explanation
The MAP estimator adjusts the sample mean by a term that accounts for the prior distribution, reflecting the influence of the prior on the estimation.
Q. 8. (Open-end question) Do two or more of the following sub-questions: (a) Describe one of your own research problems in side-channel attack as a parameter estimation problem, with a reasonable model and some suggestions for a solution. (b) Write some programs, say Matlab, to implement some of the parameter esti- mation methods you have learned.
It's great to see your interest in applying statistical concepts to real-world problems, especially in the context of side-channel attacks! For part (a), you could consider a research problem where you analyze the power consumption of a cryptographic device during its operation. The idea is to model the power traces as a function of the secret key bits being processed. You could use a Gaussian noise model to represent the inherent noise in the measurements. A potential solution could involve using maximum likelihood estimation (MLE) to estimate the key bits based on the observed power traces, while also considering the Fisher information to assess the efficiency of your estimators. For part (b), implementing parameter estimation methods in MATLAB could be a great way to solidify your understanding. You might start by coding the MLE for a simple Gaussian model, where you estimate the mean and variance from a set of noisy observations. You could also explore the BLUE by simulating data with known parameters and then applying the estimation techniques to see how well they perform. If you need help with specific coding examples or further elaboration on the research problem, feel free to ask!
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