| Line 73: | Line 73: | ||
==Solution 2== | ==Solution 2== | ||
| − | + | Assume | |
| − | <math> | + | <math>Y=\left(\begin{array}{c}Y_i \\ Y_j\end{array} \right)=A\left(\begin{array}{c}X_i \\ X_j\end{array} \right)=\left(\begin{array}{c}a_{11}X_i+a_{12}X_j \\ a_{21}X_i+a_{22}X_j\end{array} \right)</math>. |
| + | |||
| + | Then | ||
| + | |||
| + | <math>\begin{array}{l}E(Y_iY_j)=E[(a_{11}X_i+a_{12}X_j)(a_{21}X_i+a_{22}X_j)]\\ | ||
| + | =a_{11}a_{21}\sigma^2+a_{12}a_{22}\sigma^2+(a_{11}a_{21}+a_{22}a_{11})E(X_iX_j) | ||
| + | \end{array}</math> | ||
| + | |||
| + | For <math>|i-j|\geq1</math>, E(X_i,X_j)=0. Therefore, <math>a_{11}a_{21}+a_{12}a_{22}=0</math>. | ||
| + | |||
| + | One solution can be | ||
| + | |||
| + | <math>A=\left(\begin{array}{cc} | ||
| + | 1 & -1\\ | ||
| + | 1 & 1 | ||
| + | \end{array} \right)</math>. | ||
| − | |||
<font color="red"><u>'''Critique on Solution 2:'''</u> | <font color="red"><u>'''Critique on Solution 2:'''</u> | ||
| − | + | 1. <math>E(Y_iY_j)=0</math> is not the condition for the two random variables to be independent. | |
| + | 2. "For <math>|i-j|\geq1</math>, E(X_i,X_j)=0" is not supported by the given conditions. | ||
</font> | </font> | ||
Revision as of 19:00, 3 November 2014
Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
August 2013
Part 2
Let $ X_1,X_2,... $ be a sequence of jointly Gaussian random variables with covariance
$ Cov(X_i,X_j) = \left\{ \begin{array}{ll} {\sigma}^2, & i=j\\ \rho{\sigma}^2, & |i-j|=1\\ 0, & otherwise \end{array} \right. $
Suppose we take 2 consecutive samples from this sequence to form a vector $ X $, which is then linearly transformed to form a 2-dimensional random vector $ Y=AX $. Find a matrix $ A $ so that the components of $ Y $ are independent random variables You must justify your answer.
Solution 1
Suppose
$ A=\left(\begin{array}{cc} a & b\\ c & d \end{array} \right) $.
Then the new 2-D random vector can be expressed as
$ Y=\left(\begin{array}{c}Y_1 \\ Y_2\end{array} \right)=A\left(\begin{array}{c}X_i \\ X_j\end{array} \right)=\left(\begin{array}{c}aX_i+bX_j \\ cX_i+dX_j\end{array} \right) $
Therefore,
$ \begin{array}{l}Cov(Y_1,Y_2)=E[(aX_i+bX_j-E(aX_i+bX_j))(cX_i+dX_j-E(cX_i+dX_j))] \\ =E[(aX_i+bX_j-aE(X_i)-bE(X_j))(cX_i+dX_j-cE(X_i)-dE(X_j))] \\ =E[acX_i^2+adX_iX_j-acX_iE(X_i)-adX_iE(X_j)+bcX_iX_j+bdX_j^2-bcX_jE(X_i)\\ -bdX_jE(X_j)-acX_iE(X_i)-adX_jE(X_i)+acE(X_i)^2+adE(X_i)E(X_j)\\ -bcX_iE(X_j)-bdX_jE(X_j)+bcE(X_i)E(X_j)+bdE(X_i)^2]\\ =E(ac(X_i-E(X_i))^2+(ad+bc)(X_i-E(X_i)(X_j-E(X_j))+bd(X_j-E(X_j))^2]\\ =(ac)Cov(X_i,X_i)+(ad+bc)Cov(X-i,X_j)+(bd)Cov(X_j,X_j)\\ =ac\sigma^2+(ad+bc)\rho\sigma^2+bd\sigma^2 \end{array} $
Let the above formula equal to 0 and $ a=b=d=1 $, we get $ c=-1 $.
Therefore, a solution is
$ A=\left(\begin{array}{cc} 1 & 1\\ -1 & 1 \end{array} \right) $.
Solution 2
Assume
$ Y=\left(\begin{array}{c}Y_i \\ Y_j\end{array} \right)=A\left(\begin{array}{c}X_i \\ X_j\end{array} \right)=\left(\begin{array}{c}a_{11}X_i+a_{12}X_j \\ a_{21}X_i+a_{22}X_j\end{array} \right) $.
Then
$ \begin{array}{l}E(Y_iY_j)=E[(a_{11}X_i+a_{12}X_j)(a_{21}X_i+a_{22}X_j)]\\ =a_{11}a_{21}\sigma^2+a_{12}a_{22}\sigma^2+(a_{11}a_{21}+a_{22}a_{11})E(X_iX_j) \end{array} $
For $ |i-j|\geq1 $, E(X_i,X_j)=0. Therefore, $ a_{11}a_{21}+a_{12}a_{22}=0 $.
One solution can be
$ A=\left(\begin{array}{cc} 1 & -1\\ 1 & 1 \end{array} \right) $.
Critique on Solution 2:
1. $ E(Y_iY_j)=0 $ is not the condition for the two random variables to be independent. 2. "For $ |i-j|\geq1 $, E(X_i,X_j)=0" is not supported by the given conditions.
