Convexity of a function 3

Question

I have following function: \begin{equation} \lambda(\xi;\mu) = \mu^{T}\left\{M(\xi) - M(\xi)JM(\xi)\right\}\mu, \end{equation} where $M(\xi) ={\rm{diag}}(\xi_{1},\ldots,\xi_{K})$, and $ J = \mathbf{1}\mathbf{1}^{T}$, where $\mathbf{1}$ is a vector of all one's of length $K$. Additionally $\sum_{i=1}^{K}\xi_{i} = 1$ and $\xi_{i}\geq0$ for $i=1,\ldots,K$.

Is $\lambda(\xi;\mu)$ convex in $\mu$?

I tried by showing $A = M(\xi) - M(\xi)JM(\xi)$ is positive sem-definite matrix but stuck.

Isn't $M(\xi)$ the covariance matrix for a multinomial distribution? — kimchi lover, Dec 27 '17 at 12:57

score 0 · Answer 1 · answered Dec 27 '17 at 13:55

As @kimchi lover suggested, $A$ is $K$ times the covariance matrix of a multinomial random vector having probability vector $\zeta$ (column vector of $\zeta_i's$). As such, it must be positive semi-definite. Therefore, the quadratic form $\lambda$ is a convex function of $\mu$.

Details: See the formula for covariance (called Var) of multinomial random vector at https://en.wikipedia.org/wiki/Multinomial_distribution#Matrix_notation . Take $K = n$ and $\zeta = p$, and note that $MJM = \zeta\zeta'$

@Satya Prakash Does this answer your question? I think it does. — Mark L. Stone, Dec 28 '17 at 12:24

Convexity of a function 3

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