# Determinant of the variance-covariance matrix

1. Oct 10, 2009

### kingwinner

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

a.s.
d=(c1)(X1)+(c2)(X2) (i.e. (c1)(X1)+(c2)(X2) is equal to some constant d almost surely)
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I don't understand the last part. Why is it true? How can we prove it?

Any help is appreciated!:)

2. Oct 10, 2009

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

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 $$D$$ in terms of $$c_1, c_2$$ and the elements of $$\sigma$$.

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