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Suppose I have this covariance matrix of regression coefficient for $\hat{\beta_1}$ and $\hat{\beta_2}$

\begin{pmatrix} &5 &-3\\ &-3& 0.5\\ \end{pmatrix}

And I'm going to test $\begin{cases} H_0:\beta_1+\beta_2=1\\ H_1:\beta_1+\beta_2\neq 1 \end{cases}$

If I am going to test with t-ratio I need to compute $se(\hat{\beta_1}-\hat{\beta_2}-1)$ so the square root of the variance:

$VAR(\hat{\beta_1}-\hat{\beta_2}-1)=VAR(\hat{\beta_1})+VAR(\hat{\beta_2})+2cov(VAR(\hat{\beta_1}),VAR(\hat{\beta_1}))=5+0.5-2*3<0$

This is an error depending on the covariance matrix which is not positive definite (in fact real part of the eigenvalues are not all positive). AM I right?

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  • $\begingroup$ How did you get this covariance matrix? $\endgroup$ – EdM Oct 15 '15 at 16:21
  • $\begingroup$ On an exercise, but I think that there is obviously an error on the covariance matrix. $\endgroup$ – Marco Oct 16 '15 at 6:39
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covariance matrix is always non-negative definite if your design matrix is full rank. Just compute your t-ratio as: t-value=(β1+β2-1)/sqrt([1 1] * covb * [1 1]')

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  • $\begingroup$ 1) Every covariance matrix is positive semi-definite. 2) sqrt([1 1] * covb * [1 1]' is Negative! $\endgroup$ – Marco Oct 16 '15 at 6:38

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