Problem Let $Y_i, X, V_i$ be scalar random variables $\forall i=1,2,\dots, p$, defined as follow:

  • $X\sim \mathcal{N}(\bar{x}, \sigma_x^2)$, i.e. $X$ is a Gaussian variables with expected value $\bar{x}$ and variance $\sigma_x^2$;
  • $V_i \sim \mathcal{U}([-\delta, \delta])\quad \forall i=1,2,\dots, p$, i.e. all $V_i$ variables are uniformly distribuited over the interval $[-\delta, \delta]$;
  • $V_i,\dots,V_p$ are independent each other;
  • $V_i \perp X \quad \forall i=1,2,\dots, p$, i.e. all $V_i$ variables are uncorrelated with $X$;
  • $Y_i\triangleq X + V_i \quad \forall i=1,2,\dots, p$.

For the following linear model $$Y\triangleq CX+V $$ where

$$Y\triangleq\begin{bmatrix} Y_1 \\ \vdots \\ Y_p \end{bmatrix} \quad V\triangleq\begin{bmatrix} V_1 \\ \vdots \\ V_p \end{bmatrix} \qquad C\triangleq\begin{bmatrix} 1 \\ \vdots \\ 1 \end{bmatrix}\in\mathbb{R}^p$$

prove, under the hypotesis of $\sigma_x^2\to\infty$ (no a priori information), that

  1. the MMSE estimation $\hat{x}_\text{MMSE}$ of $X$ based on $Y$ is $$\hat{x}_\text{MMSE}=\frac{y_++y_-}{2}$$ where $y_+\triangleq\max_{1\leq i \leq p}Y_i$ and $y_-\triangleq\min_{1\leq i \leq p}Y_i$;
  2. the MSE $\hat{\sigma}_x^2$ of the previous estimation is $$\hat{\sigma}_x^2=\frac{6\sigma_V^2}{(p+1)(p+2)}$$ where $\sigma_V^2\triangleq\delta^2/3$

Proposed solution of point 2 Since equations (1) holds, we can observe that the maximum and the minimum observations in $Y=[Y_1 \dots Y_p]^\text{T}$ are \begin{equation}Y_+=X+V_+ \qquad Y_-=X+V_-\end{equation} where the random variables $V_+$ and $V_-$ are respectively the maximum and the minimum components in the noise vector $V$, i.e. $V_+\triangleq\max_{1\leq i \leq p}V_i$, $V_-\triangleq\min_{1\leq i \leq p}V_i$. It turns out that the MMSE estimator can be rewritten as \begin{equation}\hat{x}_\text{MMSE}(Y)=X+\mathcal{V}\end{equation} where we have introduced the random variable \begin{equation}\mathcal{V}\triangleq \frac{V_++V_-}{2}\end{equation} the estimation error is $\tilde{X} \triangleq X-\hat{x}_\text{MMSE}(Y)=\mathcal{V}$, thus, since the MSE (Mean Square Error) of an estimator is the expected value of $\tilde{X}$, follows \begin{equation}{\hat{\sigma}_X}^2 \triangleq\mathbb{E}_{X,Y}[\tilde{X}^2]=\mathbb{E}_\mathcal{V}[\mathcal{V} ^2]\end{equation} by studing $\mathcal{V}$ we can prove the explicit expression for ${\hat{\sigma}_X}^2$.

Question I'm not sure about this last equation because the two expected value are defined over two different PDF, i.e. $$\mathbb{E}_{X,Y}[\tilde{X}^2]\triangleq\iint_{\mathbb{R}^2} (x-\hat{x}_\text{MMSE}(y))^2 f_{X,Y} (x,y)\,\text{d}x\,\text{d}y$$ $$\mathbb{E}_\mathcal{V}[\mathcal{V}^2]\triangleq\int_{\mathbb{R}} \nu^2 f_\mathcal{V} (\nu)\,\text{d}\nu$$ Is it correct my derivation?


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