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Let $X=(X_1, X_2,..., X_n)$ be random variables $$ v_{ij} = cov(X_i, X_j) = E(X_i, X_j) - E(X_i)(X_j) $$

Show that the det of v is zero iff there are $a_1, a_2,..., a_n $ and b such that

$$ P(a_1X_1 + a_2X_2 +... + a_nX_n = b) = 1 $$

I'm not sure how to even begin this problem, so any leads would be great.

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2 Answers 2

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The determinant is zero if and only if the matrix is singular if and only if there exists $a$ such that $$ Ca=0 $$ where $C$ is the covariance matrix. The variance of $$ y = a_1X_1 + a_2X_2 +... + a_nX_n $$ is $$ a^t C a $$ which is zero.

So $y$ is almost surely constant and we are done.

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Hint: Try computing $\text{Var}(a_1X_1+a_2X_2+\dots+a_nX_n)$. What does it mean if a random variable has zero variance?

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