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Let $\mathsf{E}\eta = a<0$ where $\eta$ is an integrable real random value. I need to find $x\geq 0$ such that $g(x):=\mathsf{E}\max(\eta,-x)\leq\frac{2}{3}a.$

Clearly, $g(0)\geq 0$, $g$ non-increasing and $\lim\limits_{x\to\infty}g(x) = a<\frac{2}{3}a.$

I would like to find $x^* = \inf\{x\geq 0: g(x)\leq \frac{2}{3}a\}$. It's ok for me to find it numerically since the solution always exists - but I am interested if there are upper bounds on $x$ in terms of charateristics of $\eta$.

After simple moves I obtain $g(x)\leq\frac{2}{3}a$ iff $ \int\limits_{-\infty}^{-x}t\,\mathrm{d}F_\eta(t)+xF_\eta(-x)\geq \frac{1}{3}a.\tag{1} $ So it's sufficient to take $x$ such that $ x F_\eta(-x)\geq\frac{1}{3}a-\int\limits_{-\infty}^{0}t\,\mathrm{d}F_\eta(t) = \frac{1}{3}a+\mathsf{E}\eta^- $ or $ x{F_\eta(-x)}\geq \frac{1}{3}a+\mathsf{E}\eta^-.\tag{2} $

Edited: I made a misprint with replacing $F_\eta(-x)$ by $F_\eta(x)$ so bound in the first version of the question does not hold. As for now I am not even sure that solution for $(2)$ always exists. I would appreciate if someone can help me with deriving bound on $x^*$.

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    Equation $(1)$ says that $xF_\eta(-x)\ge\frac{1}{3}a-\int_{-\infty}^{-x}t\;\mathrm{d}F_\eta(t)$ but $\frac{1}{3}a-\int_{-\infty}^{-x}t\;\mathrm{d}F_\eta(t)\le\frac{1}{3}a-\int_{-\infty}^{0}t\;\mathrm{d}F_\eta(t)$ because $\int_{-x}^{0}t\;\mathrm{d}F_\eta(t)\le 0$2011-08-16

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