# is there a solution to the following maximization problem such that $a = b$?

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.

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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

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 $a=b$.
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.