Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X = (X_1,...,X_n)$ be a vector of $n$ random variables. Consider the following maximization problem:

$\max\limits_{a,b} \;\mathrm{Cov}(a\cdot X, b \cdot X)$ under the constraint that $\|a\|_2 = \|b\|_2 = 1$.

($a \cdot X$ is the dot product between $a$ and $X$). Would it be true that there is a solution to this maximization problem such that $a = b$?

Thanks.

share|improve this question
    
just a small comment: if $X$ is a degenerate random vector (i.e. constant) then $a,b$ are arbitrary - so maybe you would put a condition on the non-degeneracy. –  Ilya Sep 21 '11 at 16:11
    
you are correct. I reworded the question so that it takes this case into account. thanks! –  Alg Sep 21 '11 at 16:13
add comment

2 Answers

up vote 3 down vote accepted

Since $C\geq 0$ and symmetric, we have $$ C = QLQ' $$ where $L = \operatorname{diag}(\lambda_1,...,\lambda_n)$ and $Q$ is orthogonal. Optimization of $a'Cb$ then reduces to the optimization of $a'Lb = \sum\limits_{i=1}^n{\lambda_i}a_ib_i$. If at least one $\lambda_i>0$ then for the optimal solution $a=b$.

share|improve this answer
    
looks right. thanks! –  Alg Sep 21 '11 at 18:23
add comment

If $X$ and $Y$ are random column vectors in $\mathbb{R}^n$ and $\mathbb{R}^m$ respectively, then $\operatorname{cov}(X,Y)$ is an $n\times m$ matrix, and if $A\in\mathbb{R}^{k\times n}$ and $B\in\mathbb{R}^{\ell\times m}$, then $\operatorname{cov}(AX,BY)$ a $k\times\ell$ matrix, and $$ \operatorname{cov}(AX,BY) = A\operatorname{cov}(X,Y)B^\top. $$ So apply this when $A$ and $B$ are row vectors (so the displayed expression above is a $1\times1$ matrix, i.e. a number). Then there's no probability left in the problem.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.