A moment's question.

Question

Let G be a (absolutely) continuous distribution such that $$\displaystyle{\int_{-\infty}^{\infty}{x^{2}dG(x)}}<\infty$$ or $$\displaystyle{\int_{0}^{1}{\left[G^{-}(t)\right]^{2}dt}}<\infty.$$ Suppose $0< z< 1$ and $\gamma >1$. Can we say that $$\displaystyle{\int_{0}^{z}{t^{\gamma}\left[G^{-}(t)\right]^{2}dt}}+\displaystyle{\int_{z}^{1}{(1-t)^{\gamma}\left[G^{-}(t)\right]^{2}dt}}<\infty\ \ \ ?$$

Answer 1

Since $[G^{-}(t)]^2 \ge 0$, and $t$ in the integrand is in $[0,1]$, we have that $$ t^\gamma [G^{-}(t)]^2 \le [G^{-}(t)]^2 \quad\text{and} \quad (1-t)^\gamma [G^{-}(t)]^2 \le [G^{-}(t)]^2 .$$