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Determinant of the variance-covariance matrix

  1. Oct 10, 2009 #1
    Let ∑ be the variance-covariance matrix of a random vector X. The first component of X is X1, and the second component of X is X2.

    Then det(∑)=0
    <=> the inverse of ∑ does not exist

    <=> there exists c≠0 such that

    d=(c1)(X1)+(c2)(X2) (i.e. (c1)(X1)+(c2)(X2) is equal to some constant d almost surely)

    I don't understand the last part. Why is it true? How can we prove it?

    Any help is appreciated!:)
  2. jcsd
  3. Oct 10, 2009 #2


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    Homework Helper

    \sigma = \begin{bmatrix} \sigma_1^2 & \sigma_{12}\\ \sigma_{21} & \sigma_2^2\end{bmatrix}

    and then write down the expression for its determinant, noting that it equals zero.

    Now, take

    D = c_1X_1 + c_2X_2

    and use the usual rules to write out the variance of [tex] D [/tex] in terms of [tex] c_1, c_2 [/tex] and the elements of [tex] \sigma [/tex].

    Compare the determinant to the expression just obtained - you should see that why the statement is true.
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