# Does the linear combination of the quantile $\alpha F^{-1}(\tau)+\beta G^{-1}(\tau)$ still a quantile

$$F(x)$$ 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 $$\alpha+\beta=1$$?

• Those conditions are the sufficient (and morally the correct conditions). – Anthony Quas Jan 1 at 21:29
• @AnthonyQuas I take your comment to mean that in "nontypical" cases they may not be necessary. Can you give an example? – kodlu Jan 1 at 21:55
• 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 $3x + 10x^2$. – Matt F. Jan 2 at 3:54
• Indeed, doesn't any right-continuous 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 $g$)? – Stephan Sturm Jan 2 at 6:48
• @StephanSturm, almost, but the domain should be $(0,1)$ to allow for unbounded distributions. – Matt F. Jan 2 at 13:16