$F(x)$ and $G(y)$ are distribution functions. Define the $\tau$th quantile for cdf $F(x)$, $G(y)$ as $$\xi_\tau\equiv F^{1}(\tau)=\inf\{x:F(x)\ge \tau\}$$ and $$\eta_\tau\equiv G^{1}(\tau)=\inf\{y:G(y)\ge \tau\}.$$ Does the linear combination of the quantile, say, $\alpha F^{1}(\tau)+\beta G^{1}(\tau)$ still a quantile of some cdf? That is, does there exist a cdf $H(z)$ s.t. $$H^{1}(\tau)=\alpha F^{1}(\tau)+\beta G^{1}(\tau)?$$ Do we need some constraints about $\alpha$ and $\beta$, say, $\alpha,\beta>0$ and $\alpha+\beta=1$?
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$\begingroup$ Those conditions are the sufficient (and morally the correct conditions). $\endgroup$ – Anthony Quas Jan 1 at 21:29

$\begingroup$ @AnthonyQuas I take your comment to mean that in "nontypical" cases they may not be necessary. Can you give an example? $\endgroup$ – kodlu Jan 1 at 21:55

1$\begingroup$ No, we do not need $\alpha+\beta=1$. Eg: let $x$ be distributed uniformly on [0,1], let $y=2x^2$, then $\alpha=3,\ \beta =5$ gives the quantile function for $3x + 10x^2$. $\endgroup$ – Matt F. Jan 2 at 3:54

2$\begingroup$ Indeed, doesn't any rightcontinuous nondecreasing function $g$ with domain $[0,1]$ define a quantile function? (Just check that $g(U)$, for $U$ a standard uniform random variable defines a random variable with cdf $g^{1}$, the (right) generalized inverse function of $g$)? $\endgroup$ – Stephan Sturm Jan 2 at 6:48

1$\begingroup$ @StephanSturm, almost, but the domain should be $(0,1)$ to allow for unbounded distributions. $\endgroup$ – Matt F. Jan 2 at 13:16