# Upper bound for the probability of violating a set of conditions

Let $$X_1,\ldots,X_n$$ be independent and normally distributed $$\mathcal{N}(\bar{x},\sigma^2)$$ random variables. Let $$g:{\rm R} \to [0,\bar{g}]$$ be a decreasing bounded function. Let $$a$$, $$\lambda$$ and $$l_{i,j}$$ be positive constants, with $$\sum_j l_{i,j}=1$$ for all $$i$$. Let $$x^{\ast}$$ denote the unique solution to

\begin{align} x^{\ast}-{\rm E}[X_i \, | \, X_i

Consider the set of conditions: \begin{align} \sum_{j=1}^n \textbf{1}_{X_j \in (-\infty,x^{\ast})}l_{i,j}{\rm E}[X_j \, | \, X_j

Can someone find an upper bound that does not depend on $$x^{\ast}$$ for the probability that the set of conditions above does not hold (i.e., that at least one of the inequalities does not hold)?

EDIT: If I could show that

\begin{align} \sum_{j=1}^n \textbf{1}_{X_j \in (-\infty,x^{\ast})}l_{i,j}{\rm E}[X_j \, | \, X_j

is concave in $$x^{\ast}$$, or if I could impose that in terms of the primitives, then I would be able to find that upper bound. For instance, if the above expression is concave, then I can find an upper bound for the probability that the inequality does not hold for $$i$$ given by

\begin{align} \max\left\{1-\Phi\left(\dfrac{a-2\bar{g}}{\sigma\left[\sum_{j\neq i}l_{i,j}^2+\left(1-l_{i,i}\right)^2\right]^{\frac{1}{2}}}\right)\,,\,1-\Phi\left(\dfrac{a}{\sigma}\right)\right\} \end{align} where $$\Phi$$ is the standard normal CDF.

Can anyone show that the above expression is concave in $$x^{\ast}$$, or can anyone impose concavity in terms of the primitives?