$\left(\chi_{n, \alpha}^{2}-(n-1)\right) / \sqrt{2 n-2}$ converges to the upper $\alpha$ point $z_{\alpha}$ of the standard normal distribution.

Question

$\left(\chi_{n, \alpha}^{2}-(n-1)\right) / \sqrt{2 n-2}$ converges to the upper $\alpha$ point $z_{\alpha}$ of the standard normal distribution where $\chi_{n, \alpha}^{2}$ is the upper $\alpha$ bound of the $\chi_n^2$ distribution.

By the Central Limit Theorem, we have $$ \frac{\chi_{n-1}^{2}-(n-1)}{\sqrt{2 n-2}} \leadsto N(0,1). $$ Thus, $$ P\left(\frac{\chi_{n-1}^{2}-(n-1)}{\sqrt{2 n-2}}\geq z_\alpha\right)\to\alpha. $$ Also, we have $$ P\left(\frac{\chi_{n-1}^{2}-(n-1)}{\sqrt{2 n-2}}\geq \frac{\chi_{n,\,\alpha}^{2}-(n-1)}{\sqrt{2 n-2}}\right)=\alpha. $$ It seems that $z_\alpha$ and $\frac{\chi_{n,\,\alpha}^{2}-(n-1)}{\sqrt{2 n-2}}$ should be close to each other. But how can we prove the convergence holds?

Answer 1

Consider a sequence of CDFs $\{F_n\}$. It is known that $F_n\to F$ (for all continuity points of $F$) if and only if $F_n^{-1}\to F^{-1}$ (for all continuity points of $F^{-1}$). An easy way to show the "only if" part when $F$ is the standard normal cdf is as follows. Let $Z\sim N(0,1)$. Then for $\alpha\in (0,1)$, $$ F(F_n^{-1}(\alpha))=\mathsf{P}(F_n(Z)\le \alpha)\to \mathsf{P}(F(Z)\le \alpha)=F(z_{\alpha}) $$ Since $F^{-1}$ is continuous, $F_n^{-1}(\alpha)\to z_{\alpha}$.