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Data sampled at two time instances giving bivariate Gaussian vector $X=(X_1,X_2)^T$ with

$f(x_1,x_2)=\exp(-(x_1^2+1.8x_1x_2+x_2^2)/0.38)/2\pi \sqrt{0.19}$

Data measured in noisy environment with vector: $(Y_1,Y_2)^T=(X_1,X_2)^T+(W_1,W_2)^T$

where $W_1,W_2$ are both $i.i.d.$ with $\sim N (0,0.2)$.

I have found correlation coefficient of $X_1,X_2$, $\rho=-0.9$ and $X_1,X_2 \sim N(0,1)$

Question: How to design filter to obtain MMSE estimator of $X_1$ from $Y$ vector and calculate MSE of this estimator?

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You didn't mention that $W$ is independent of $X$. People often leave that out. I'd prefer to include it. –  Michael Hardy Oct 25 '11 at 3:39

2 Answers 2

up vote 2 down vote accepted

What you need is $\mathbb{E}(X_1 \mid Y_1, Y_2)$. We have $$ \operatorname{var}\begin{bmatrix} X_1 \\ Y_1 \\ Y_2 \end{bmatrix} = \left[\begin{array}{r|rr} 1 & 1 & -0.9 \\ \hline1 & 1.02 & -0.9 \\ -0.9 & -0.9 & 1.02 \end{array}\right]= \begin{bmatrix} \Sigma_{11} & \Sigma_{12} \\ \Sigma_{12}^\top & \Sigma_{22} \end{bmatrix}. $$ So the conditional expected value is $$ \mathbb{E}(X_1) + \Sigma_{12} \Sigma_{22}^{-1} \left( \begin{bmatrix} Y_1 \\ Y_2 \end{bmatrix} - \mathbb{E}\begin{bmatrix} Y_1 \\ Y_2 \end{bmatrix}. \right) $$ See: http://en.wikipedia.org/wiki/Multivariate_normal_distribution#Conditional_distributions

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This looks like homework but here goes. Since everything is Gaussian, the MMSE estimator for $X_1$ is the mean of the conditional pdf of $X_1$ given $(Y_1, Y_2)$ and the mean square error is the conditional variance of this. Do you know how to find the conditional pdf (hint: it is also Gaussian)

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$f(x_1|Y_1,Y_2)=f(x_1,(y_1,y_2))/f(y_1,y_2)$. So $f(y_1,y_2)$ is $N(0,1.2)$, but how to find the upper part pdf? –  Ab. Oct 25 '11 at 3:46

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