Please help me show that $f(w)$ is strictly concave in $w\in[0,\infty)$:
$f(w)=\sum_{j=1}^N P_j (w)\cdot u_j $
where
$P_j (w)=\sqrt{w}\int _{-\infty}^{\infty}\Pi_{k\neq j}\{\Phi[\sqrt{w}(v-u_k)]\}\phi[\sqrt w(v-u_j)]dv $
and $\Phi$ and $\phi $ are the standard normal CDF and PDF respectively, and $u_j \in \mathbb{R}$ $\forall j\in \{1,...,N\}$ where there exists some $j^*$ such $u_{j^*}>u_j$ for all $j\neq j^*$.
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Physical Interpretation: There are $N$ possible states with associated outcomes $u_j$ for each state $j$. The state that realizes will be the state that has the highest noisy outcome $\tilde{u}_j=u_j +\epsilon_j$ where the $\epsilon_j$ are distributed $i.i.d.$ $N(0,\frac{1}{w})$ ($w$ is inverse variance or "precision"). $P_j(w)$ is the probability state $j$ will occur (as a function of $w$) and $f(w)$ is the expected outcome.
Intuition: As $w\rightarrow \infty $, the state with the highest $u_j$, call it $j^*$, will occur with probability 1 and $f(w)\rightarrow u_j^*$. Therefore, $f(w)$ is bounded from above. It also seems to be strictly increasing, as the higher the $w$, the higher the probability that $j^*$ will be realized.
