I want to examine whether the following function, $G(q)$, is concave on $q$ ($q>0$):
$G(q)=(1-\alpha )\int^{\frac{\beta .q}{1-\alpha }}_0{F(x)dx}-\int^{(1+\beta ).q}_0{F(x)dx}$
The assumptions are $0<\alpha ,\beta <1$ and $\frac{\beta }{1-\alpha}\le 1+\beta $ (so, $\frac{\beta.q}{1-\alpha }\le (1+\beta ).q$ hold). Besides, $F(x)$ is CDF for a general random variable $x$ (so, $F(x)$ is non-decreasing and also $F(x)\le 1$ hold). Would you please advise?
For specific distribution functions like uniform and exponential, it is valid. I wondered whether $G(q)$ is concave for any general distribution function. Or is there any property for the distribution function to make it concave?
The function $G$ is not necessarily concave, you need additional hypotheses.
For a continuous positive distribution with pdf $f$, we must have $$ \dfrac{f\!\left(\frac{\beta}{1-\alpha} q\right)}{f((1+\beta)q)} \le (1-\alpha)\frac{(1+\beta)^2}{\beta^2} $$ for every $q > 0$. This is shown by using the fact that a twice differentiable function is concave if and only if its second derivative is non-positive.
Indeed, $$G'(q) = \beta F\!\left(\frac{\beta}{1-\alpha}q\right) - (1+\beta) F\!\left((1+\beta)q\right)$$ and $$G''(q) = \frac{\beta^2}{1-\alpha} f\!\left(\frac{\beta}{1-\alpha}q\right) - (1+\beta)^2 f\!\left((1+\beta)q\right).$$ Now use the hypotheses over $\alpha$ and $\beta$ along with $G''(q) \le 0$ to finish the proof.
As a counter-example, take $\alpha=\beta=1/2$ and the exponential distribution with cdf $F(x) = 1 - e^{-x}.$ Then, $G(q) = \dfrac{1}{2}-q-e^{-\frac{3 q}{2}}+\dfrac{1}{2}e^{-q}$ and $G''(q) = \dfrac{1}{2}e^{-q}-\dfrac{9}{4} e^{-\frac{3 q}{2}}$. However, $G''(q) < 0$ if $q < 2\ln(9/2)$ and $G''(q) > 0$ if $q > 2\ln(9/2)$.