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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 <x^{\ast}]-\lambda g(x^{\ast})=0 \end{align}

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 <x^{\ast}] + \sum_{j=1}^n \textbf{1}_{X_j \in [x^{\ast},+\infty)}l_{i,j}X_j -\sum_{j=1}^n \textbf{1}_{X_j \in [x^{\ast},+\infty)}(l_{i,j-1}+l_{i,j+1})g(X_j) - X_i-\textbf{1}_{X_{i-1} \in [x^{\ast},+\infty)}g(X_{i-1})-\textbf{1}_{X_{i+1} \in [x^{\ast},+\infty)}g(X_{i+1})\geq -a \qquad i =1,\ldots,n \end{align}

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 <x^{\ast}] + \sum_{j=1}^n \textbf{1}_{X_j \in [x^{\ast},+\infty)}l_{i,j}X_j \end{align}

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?

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