# A unit vector that maximizes variance in a discrete probability distribution

this might be a dumb question, but I was working on a problem and ran into the following (sub)problem.

Suppose we have a nonnegative vector $$\pi \in \mathbf{R}^n$$ that satisfies $$\sum_{i=1}^n \pi_i = 1$$, i.e., it is a discrete probability density. We want to choose a unit vector $$v \in \mathbf{R}^n$$, $$\|v\|=1$$, where $$\| \cdot \|$$ is the Euclidean norm, such that the "variance"

$$f(v) = \sum_{i=1}^n v_i^2 \pi_i - (\sum_{i=1}^n v_i \pi_i)^2$$

is maximized. Of course there will be multiple maxima, because $$f(v) = f(-v)$$.

Is there any closed form solutions for the maximum $$v$$ or any ideas how to find it, or what is the maximum function value $$f(v)$$? Specifically, I want to show that if $$\pi_1 > 0.5$$, then any maximum $$v$$ satisfies $${\rm sign}(v_1) = -{\rm sign}(v_i)$$ for all $$i \neq 1$$.

Any tips or ideas? Thank you very much!

Define the $$n\times n$$ symmetric matrix $$M$$ with elements $$M_{ij}=\pi_i \delta_{ij}-\pi_i\pi_j.$$ We seek to maximise the quadratic form $$f(v)=\sum_{ij} v_i M_{ij} v_j$$ where $$v$$ is a vector on the unit $$n$$-sphere. An extremum is reached for an eigenvector $$v_0$$ of $$M$$ and $$f(v_0)=\mu_0$$ is the corresponding eigenvalue $$\mu_0$$ of $$M$$. For the global maximum we should take the eigenvector $$v_{\rm max}$$ with the largest eigenvalue $$\mu_{\rm max}$$.
The eigenvalues $$\mu$$ of $$M$$ are determined by the equation $$1+\sum_{i}\frac{\pi_i^2}{\mu-\pi_i}=0.$$ This follows from the Matrix determinant lemma, see this MO posting for a derivation.
The smallest eigenvalue $$\mu=0$$ has eigenvector with all elements $$1\sqrt n$$. Since the eigenvectors are orthogonal, the desired $$v_{\rm max}$$ should satisfy $$\sum_{i}(v_{\rm max})_i=0.$$
I don't think a closed form expression for $$v_{\rm max}$$ exists for arbitrary $$n$$.
For $$n=2$$ one has simply $$v_{\rm max}=(1/\sqrt 2,-1/\sqrt 2)$$ independent of $$\pi$$, but already for $$n=3$$ the expression is complicated: $$v_{\rm max}=\{\pi_2(\pi_1-\pi_3)+\sqrt X,\pi_1(\pi_3-\pi_2)-\sqrt X, \pi_3(\pi_2-\pi_1)\}$$ $$X=\pi_1^4+2 \pi_1^3 (\pi_2-1)+\pi_1^2 (3\pi_2^2 -\pi_2+1)+\pi_1 (\pi_2-1) \pi_2 (2 \pi_2+1)+(\pi_2-1)^2 \pi_2^2,$$ where $$v_{\rm max}$$ is still to be normalized to unit length. The corresponding $$\mu_{\rm max}$$ is $$\mu_{\rm max}=\sqrt{\left(\pi_1^2- (1-\pi_2)(1-\pi_3)\right)^2-3 \pi_1 \pi_2 \pi_3}-\pi_1^2-\pi_2^2-\pi_1 \pi_2+\pi_1+\pi_2.$$