The question is as follows:
Have$$z_i = y(x_i) + w_i,\quad i = 1, 2, 3$$ where $x_1<x_2<x_3$ and $z_1<z_2>z_3$. $\space$ Assume the unknown function $y(x)$ can be approximated with parabolic interpolation.
Find the location of the estimate of the maximum point $y(x)$ $$\hat x_m = f(x_1,x_2,x_3,z_1,z_2,z_3)$$
Assume $$E[w] = 0; \space \space E[ww^T]=P$$ find the variance associated with the estimate $\hat x_m$ using first-order series expansion.
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I was able to solve part 1 by considering the model:
$$ z(i) = a{x_{(i)}}^2+bx_{(i)}+c+w$$ Then with three measurements, can write the measurement equation as:
$$\begin{pmatrix}z_1 \\\ z_2 \\\ z_3\end{pmatrix} = \begin{pmatrix} {x_1}^2 & x_1 & 1\\\ {x_2}^2 & x_2 & 1\\\ {x_3}^2 & x_3 & 1 \end{pmatrix} \begin{pmatrix} a \\\ b \\\ c \end{pmatrix} + \begin{pmatrix} 1\\\ 1\\\ 1\end{pmatrix}w $$
Denote this as $$Z = XA+W$$ The parabola parameter matrix $A$ can then be written as $$A = X^{-1}Z - X^{-1}W$$ where $\hat A$ can be found as $$ \hat A = E[A] = X^{-1}Z - X^{-1}E[W]$$
Its terms can be found through matrix operations assuming $E[W]=0$. The maximum point of $y(x)$ occurs at $-\frac{b}{2a}$, which can be written as
$$\hat x_m = -\frac{b}{2a}= \frac{z_1({x_3}^2-{x_2}^2) + z_2({x_1}^2-{x_3}^2) + z_3(-{x_1}^2+{x_2}^2)} {2(z_1(x_3-x_2) + z_2(x_1-x_3) + z_3(-x_1+x_2))}$$
But I can't figure out how to find the variance of the estimate. I thought about finding the covariance matrix of $\hat A$, which would give me the variance of $a, b$ and $c$. Assuming the terms are independent, I can calculate the variance of $\hat x$. But the covariance would be $$E[(\hat A-\bar A)(\hat A-\bar A)^T]$$ but since $$\hat A = \bar A = E[A]$$ the covariance matrix would just be $0$ which doesn't makes sense.
Would really appreciate it if someone could point me in the right direction. Thanks in advance!
I denote vectors by small letters. The estimated coefficients are given by \begin{align} \hat{\mathbf{a}}=\mathbf{X}^{-1}\mathbf{z}, \end{align} whereas the true coefficients are given by \begin{align} \mathbf{a}=\mathbf{X}^{-1}(\mathbf{z}-\mathbf{w}), \end{align} where \begin{align} \mathbf{X}=\left[\begin{array}{ccc}x_1^2&x_1&1\\x_2^2&x_2&1\\x_3^2&x_3&1\end{array}\right],~ \mathbf{a}=\left[\begin{array}{c}a\\b\\c\end{array}\right],~ \mathbf{z}=\left[\begin{array}{c}z_1\\z_2\\z_3\end{array}\right],~ \mathbf{w}=\left[\begin{array}{c}w_1\\w_2\\w_3\end{array}\right]. \end{align} We see that $\text{var}(\mathbf{z})=\text{var}(\mathbf{w})$. A variance of the obtained parameters $\hat{\mathbf{a}}$ is given by \begin{align} \text{var}(\hat{\mathbf{a}})=\text{var}(\mathbf{X}^{-1}\mathbf{z})=\mathbf{X}^{-1}\text{var}(\mathbf{z})\mathbf{X}^{-\text{T}}=\mathbf{X}^{-1}\underbrace{\text{var}(\mathbf{w})}_{=P\mathbf{I}}\mathbf{X}^{-\text{T}}=P\mathbf{X}^{-1}\mathbf{X}^{-\text{T}}, \end{align} where we have used formula $\text{var}(\mathbf{Ay})=\mathbf{A}\text{var}(\mathbf{y})\mathbf{A}^{\text{T}}$ and $\text{var}(\mathbf{w})=P$. Therefore we get \begin{align} \text{var}(a)&=P(\mathbf{X}^{-1}\mathbf{X}^{-\text{T}})_{11},\\ \text{var}(b)&=P(\mathbf{X}^{-1}\mathbf{X}^{-\text{T}})_{22}. \end{align} Finally, we would like to evaluate $\text{var}\left(-\frac{2a}{b}\right)$, which is not possible directly, but only approximately. By the use of the first order Taylor expansion $-\frac{2a}{b}\approx -\frac{2a_0}{b_0}-\frac{2}{b_0}a+\frac{2a_0}{b_0^2}b$ and the formula $\text{var}(a\mathbf{x}+b)=a^2\text{var}(\mathbf{x})$ we obtain the following approximation \begin{align} \text{var}(\hat{x}_{\text{m}})=\text{var}\left(-\frac{2a}{b}\right)\approx\,& \text{var}\left(-\frac{2}{b_0}a+\frac{2a_0}{b_0^2}b\right)=\frac{4}{b_0^2}\text{var}(a)+\frac{4a_0^2}{b_0^4}\text{var}(b)= \nonumber\\ &\frac{4P}{b_0^2}\left[(\mathbf{X}^{-1}\mathbf{X}^{-\text{T}})_{11}+\frac{a_0^2}{b_0^2}(\mathbf{X}^{-1}\mathbf{X}^{-\text{T}})_{22}\right]. \end{align}