Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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:

share|cite|improve this answer

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)

share|cite|improve this answer
$f(x_1|Y_1,Y_2)=f(x_1,(y_1,y_2))/f(y_1,y_2)$. 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.