Difference between revisions of "Seg18. The Correlation Matrix"

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(problem 1)
(problem 1)
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                           \end{bmatrix}</math>
 
                           \end{bmatrix}</math>
  
<math>x = cos\theta, y = sin\theta</math>
+
And since x,y is on the circle, thus <math>x = cos\theta, y = sin\theta</math>
  
 
Thus, the matrix would be <math> \begin{bmatrix}
 
Thus, the matrix would be <math> \begin{bmatrix}

Revision as of 16:04, 18 March 2013

Skilled problem

problem 1

Random points i are chosen uniformly on a circle of radius 1, and their <math>(x_i,y_i)</math> coordinates in the plane are recorded. What is the 2x2 covariance matrix of the random variables <math>X</math> and <math>Y</math>? (Hint: Transform probabilities from <math>\theta</math> to <math>x</math>. Second hint: Is there a symmetry argument that some components must be zero, or must be equal?)

The matrix would be <math> \begin{bmatrix}

                          Cov(X,X) & Cov(X,Y) \\
                          Cov(Y,X) & Cov(Y,Y) \\
                          \end{bmatrix}</math>

Since the sample space is symmetric and the sampling is uniform, it's easy to see the mean for x,y are both 0. And the covarience of X and X is it's variance. So the matrix is actually:

<math> \begin{bmatrix}

                          <x^2> & <xy> \\
                          <yx> & <y^2> \\
                          \end{bmatrix}</math>

And since x,y is on the circle, thus <math>x = cos\theta, y = sin\theta</math>

Thus, the matrix would be <math> \begin{bmatrix}

                                \pi & 0 \\
                                0 & \pi \\
                          \end{bmatrix}</math>