Student test statistic and self normalizing sum. I study the asymptotic distribution of self normalizing sums which are defined as 
$S_n/V_n$ where 
$S_n=\sum_{i=1}^n X_i$ and $V_n^2 = \sum_{i=1}^n X_i^2$ 
for some i.i.d RV's $X_i$. 
Motivation to study such sums comes from the fact that the classical Student $T_n$ statistic could be expressed as:  
$T_n(X)= \frac{\sum_{i=1}^n X_i}{\sqrt{\frac{n}{n-1}\sum_{i=1}^n (X_i-\overline X)^2}} = \frac{S_n/V_n}{\sqrt{\frac{1}{n-1}(n - (S_n/V_n)^2)}}$ 
From the paper I study (http://arxiv.org/pdf/1204.2074v2.pdf) I know that:

If $T_n$ or $S_n/V_n$ has an asymptotic distribution, then so does the other, and they coincide.

but it do not seem trivial for me. Can someone explain it? 
I'm not sure if it is only showing that the denominator is equal 1 in probability and using the Slutsky-Theorem? 
https://de.wikipedia.org/wiki/Slutsky-Theorem
 A: The authors reference a 1969 paper by Efron. The relevant section in Efron appears to be a reference to what at that time was an unpublished paper by Logan, Mallows, Rice and Shepp (see p.16, last paragraph of Efron).
However, the article you are reading give the actual paper, published in 1973 :Limit distributions of self-normalized sums. They reference back to Efron's 1969 paper, where he uses Fisher's finding that the t-statistic is asymptotically normal under the 0-mean hypothesis as long as the vector of random variates have rotational symmetry.
A: The authors' claim can be proved using the Skorohod representation theorem. To prove one direction: if the ratio $R_n:=S_n/V_n$ converges in distribution to a limit $R$, then there exist variables $R_1^*, R_2^*, \ldots$ and $R^*$ with the same distributions as $R_1, R_2,\ldots$ and $R$ respectively such that $R_n^*\to R^*$ almost surely. The identity
$$
T_n = \frac{R_n}{\sqrt{\frac{1}{n-1}(n - R_n^2)}}\tag1$$
(note you have a typo in your formula) then shows that the correspondingly defined $T_n^*$ converges a.s. to $R^*$, and therefore $T_n$ converges in distribution to $R$.
Conversely, we can rearrange (1) to solve for $R_n$:
$$
R_n = {T_n\over\sqrt{\frac 1n(n-1+T_n^2)}}\tag2
$$
(please check my algebra). In view of (2) it is clear that if $T_n$ converges a.s. to a limit $R$, then so does $R_n$.
