Quantile density function: student's t vs normal I'm trying to show that $\frac{\exp\Big(\frac{\Phi^{-1}(x)^2}{2}\Big)}{\Big(1+\frac{\Psi^{-1}(x)^2}{\delta}\Big)^{(\delta-1)/2}}$ goes to infinity as $x$ goes to one, where $\Phi^{-1}$ denotes the inverse CDF of the standard normal, and $\Psi^{-1}$ denotes the inverse CDF of the standard Student's t with $\delta$ degrees of freedom.
If it helps, I've noticed that $\frac{\exp\Big(\frac{\Phi^{-1}(x)^2}{2}\Big)}{\Big(1+\frac{\Psi^{-1}(x)^2}{\delta}\Big)^{(\delta-1)/2}}$ always seems to exceed one, but that $\frac{\exp\Big(\frac{\Phi^{-1}(x)^2}{2}\Big)}{\Big(1+\frac{\Psi^{-1}(x)^2}{\delta}\Big)^{(\delta+1)/2}}$ always seems to fall short of one, where $\exp\Big(\frac{\Phi^{-1}(x)^2}{2}\Big)$ is the quantile density function of the standard normal and $\Big(1+\frac{\Psi^{-1}(x)^2}{\delta}\Big)^{(\delta+1)/2}$ is the quantile density function of the Student's t with $\delta$ degrees of freedom.
Thanks,
Rob
 A: The standard Student's PDF is equivalent to a multiple of $t^{-\delta-1}$ when $t\to+\infty$ hence $1-\Psi(t)$ is equivalent to a multiple of $t^{-\delta}$ when $t\to+\infty$ and $\Psi^{-1}(x)$ is equivalent to a multiple of $(1-x)^{-1/\delta}$ when $x\to1$, $x\lt1$. Thus, the denominator of the ratio $R$ you are interested in is equivalent to a multiple of $(1-x)^{-1+1/\delta}$.
The standard normal PDF is $(2\pi)^{-1/2}\mathrm e^{-t^2/2}$ hence $1-\Phi(t)$ is equivalent to $(2\pi)^{-1/2}t^{-1}\mathrm e^{-t^2/2}$ when $t\to+\infty$ and $\Phi^{-1}(x)$ is equivalent to $t$, where $t$ solves $(2\pi)^{1/2}t\mathrm e^{t^2/2}=(1-x)^{-1}$. Thus, $R$ is equivalent to a multiple of $(1-x)^{1-1/\delta}\mathrm e^{t^2/2}$, which is a multiple of $1/\sqrt{s}$ where $s=(1-x)^{2/\delta}t^2$.
Since $t^2$ is equivalent to $-2\log(1-x)$ and $u^a\log u\to0$ for every positive $a$ when $u\to0$, $u\gt0$, this shows that $s\to0$ when $x\to1$, $x\lt1$, hence $R\to+\infty$.
As a confirmation, the second ratio $R'$ you considered involves $(1-x)^{1+1/\delta}\mathrm e^{t^2/2}$, which is a multiple of $1/\sqrt{s'}$ where $s'=(1-x)^{-2/\delta}t^2$. Since $u^a\log u\to-\infty$ for every negative $a$ when $u\to0$, $u\gt0$, this shows that $s'\to+\infty$ when $x\to1$, $x\lt1$, hence $R'\to0$.
