Exercise Solution 3.16

Exercise Solution 3.16

This is an application for our multi-dimensional identity [3.31]. This may not be immediately obvious, but let’s present our problem in a slightly different manner.

Define a 2-dimensional random vector Z as having uncorrelated components Z1 and Z2. Each has mean 0. Like the first two components of X, they have standard deviations of 5 and 4 respectively. Now set:

[s1]

where the μi are the unspecified means of the components of X.

We have not altered our characterization of X in any way. We have simply placed the components of X on an equal footing. Instead of defining X3 in terms of X1 and X2, we define all three components as a linear polynomial of Z. Now we apply [3.31]

[s2]

[s3]

Based upon this covariance matrix, we obtain

[s4]

 
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