Let $R\_n$ be a random variable with values in $[0,1]$ and $nR\_n$ converges to $\frac{1}{1+C} \chi\_m^2$ in distribution for some constant $C>0$ and $m\in \mathbb{N}$. Is it possible to show that $(1-R\_n)^{-\frac{n-1}{2}} =O\_P(1)$ holds?


I came across this result in the [proof of Theorem 3 of this paper](https://people.eecs.berkeley.edu/%7Ejordan/courses/260-spring10/readings/liang-etal.pdf) but didn't really get it. Some help would be much appreciated.



Random variable is Big O in probability notation
For each natural $n$, let $R\_n$ be random variable with values in $[0,1]$ such that $nR\_n$ converges in distribution, as $n\to\infty$, to a random variable $X$ with a continuous cdf.


Let $T\_n:=(1-R\_n)^{-(n-1)/2}$. Take any real $t>1$. Then for all large enough natural $n$
$$t^{-2/(n-1)}=e^{-2\ln t/(n-1)}<1-\frac{\ln t}n$$
and hence
$$P(T\_n>t)=P(nR\_n>n(1-t^{-2/(n-1)})) \\
\le P(nR\_n>\ln t)\underset{n\to\infty}\longrightarrow P(X>\ln t)
\underset{t\to\infty}\longrightarrow0.$$
So,
$$\lim\_{t\to\infty}\lim\_{n\to\infty}P(T\_n>t)=0;$$
that is (cf. e.g. [equivalence (i)$\iff$(v) in Lemma 1](http://www2.math.uu.se/%7Esvante/papers/sjN6.pdf)), $(1-R\_n)^{-(n-1)/2}=T\_n=O\_P(1)$, as desired.




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The condition that $X$ has a continuous cdf (which actually holds in your setting) is not needed here, but with it the proof gets simpler just a little bit.


