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Let $X$ and $Y$ be bounded real-valued random variables. Define $ f(a)=\operatorname{E} \min(aX,Y) $
Is $f$ a quasilinear function of $a$?
That is $f$ is both quasiconvex and quasiconcave.

To answer this, I would pass differentiation under the expectation, but as the function $\min(x,y)$ is not differentiable everywhere, I'm blocked.

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    @DavideGiraudo I mean it's both quasiconcave and quasiconve$x$. Ma$y$be onl$y$ one of these is possible though.2012-06-22

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By definition, a function $u$ is quasi-convex if, for every $a\leqslant c\leqslant b$, $u(c)\leqslant\max(u(a),u(b))$, $u$ is quasi-concave if, for every $a\leqslant c\leqslant b$, $u(c)\geqslant\min(u(a),u(b))$, and $u$ is quasi-linear if $u$ is quasi-convex and quasi-concave.

Note that $\min(a,0)+\min(-a,0)=-|a|$ and $a\mapsto-|a|$ is not quasi-convex. Hence, the function $f$ defined by $f(a)=\mathrm E(\min(aX,Y))$ for some random variables $X$ and $Y$, is not quasi-linear in general.

On the other hand...

If $X\geqslant0$ with full probability, since each function $a\mapsto\min(ax,y)$ with $x\geqslant0$ is nondecreasing, $f$ is nondecreasing. Likewise, if $X\leqslant0$ with full probability, since each function $a\mapsto\min(ax,y)$ with $x\leqslant0$ is nonincreasing, $f$ is nonincreasing. In both cases, $f$ is quasi-linear.

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    Strong as always. Thank you did. It's true that stochastic processes are not revelant. I got confuse in adapting my problem. I will correct the question.2012-06-23