Let $X_1,...,X_n$ be iid standard gaussian random variable. Define the set $S$ as $$ S = \{(z_1,...,z_n):z_i\geq 0,\sum_{i=1}^nz_i=1\} $$ and the function $f:S\to[0,\infty)$ as $$ f(z_1,...,z_n) = E\left[\sqrt{\sum_{i=1}^nz_iX_i^2}\right]. $$ Is it true that $f$ is concave? My approach was by calculating the Hessian: $$ \frac{\partial^2 f}{\partial z_k\partial z_j} = -\frac{1}{4}E\left[\frac{(X_jX_k)^2}{\left(\sum_{i=1}^nz_iX_i^2\right)^{3/2}}\right]. $$ I couldn't prove that this Hessian is non-positive definite.
Is there another way to prove that $f$ is concave? (Of course, if it is true.)
Let $\phi(z,\omega) = \sqrt{\sum_k z_k X_k^2(\omega)}$ and note that for a fixed $\omega$ the function $\omega \mapsto \phi(z,\omega)$ is concave (composition of $\sqrt{\cdot}$ and a linear function). Hence $f(z) = E_\omega [ \phi(z,\omega) ] = \int_\Omega \phi(z,\omega) d \mu(\omega) $ is concave.