Convergence in distribution to a degenerate distribution This question came up based on a disagreement I had with a TA. This was the specific example:
Let $X_{1},...,X_{n}$ be an iid random sample from a population with pdf $f(x)=3(1-x)^2, 0<x<1$. The $nth$ order statistic is represented as $X_{(n)}$.
Question: Find a constant $v$ such that $n^v(1-X_{(n)})$ converges in distribution.
I believe that the intent of the question was to prompt the invocation of Slutsky's theorem (or maybe not); however, to which distribution the presented snippet was supposed to converge was not specified. So I presented a lazy alternative answer as follows:
As the sample size approaches infinity, $X_{(n)}$ becomes arbitrarily close to 1. Therefore, we can simply set $v=0$ and the statement will converge to a degenerate distribution. In this case, the order statistic converges to 1, so with $v=0$, the statement converges to 0. I later realized that it actually converges to 1 after the transformation, which is reflected in the final degenerate distribution written toward the bottom of the question. I was surprised when the TA said that 0 is not a random variable or a distribution, and my answer made no sense. I second guessed myself and went further:
The cdf of the original distribution is $$F(x)=x^3-3x^2+3x$$ Using this, I derived the pdf of the order statistic,
$$f_{X_{(n)}}(x)=3n(x^2-2x+1)(x^3-3x^2+3x)^{n-1} $$
And the cdf of the order statistic, $$f_{X_{(n)}}(x)=(x^3-3x^2+3x)^n$$
I then used this to graph the cdf of the order statistic with $n=1, n=100, n=1,000,000,000$. Predictably, the graph showed the cdf getting thinner and steeper, until with huge samples it visually looks like a vertical line at $X_{(n)}=1$.
I reiterated my argument, and showed the visualization: it is converging to a degenerate distribution. Again, I was rebuffed. After pestering the TA and asking the instructor for the course, I still have no answer for why my answer was wrong, but I can't keep bothering them about it.
Can someone here tell me if my argument is solid, and if not, tell me specifically what I've done that is in error?
EDIT: Here is the final formalism that I gave to tie my argument together. Following Casella and Berger's definition,

A sequence of random variables, $X_{1}, X_{2},...,$ converges in distribution to  a random variable $X$ if $$\lim_{n \to \infty} F_{X_n}(x) = F_X(x)$$ at all points $x$ where $F_X(x)$ is continuous

Since $$\lim_{n \to \infty} F_{X_(n)}(x) = 1$$
And we can define a degenerate variable $Y$ with the cdf
$$F_Y(x)=\begin{cases}1, & x\ge1\\ 0, & else \\ \end{cases}$$
We can say
$$\lim_{n \to \infty} F_{X_(n)}(x) = F_Y(x) $$
So by definition, $X_{(n)}$ converges in distribution to the degenerate distribution of $Y$.
EDIT: Adding a bounty to this, as it has become important. I did not formally perform the transformation $Y=1-X_(n)$ above, but doing so results in $Y$ having a distribution that still converges to the same degenerate distribution above. I was now told that my answer does not prove convergence in distribution at all. Question is answered if someone can definitively prove either 1) my answer is correct, and show exactly how you would have gone about proving it formally and rigorously, or 2) my answer is wrong, exactly why it is wrong, and prove any correct answer that uses degenerate distributions or degenerate variables if such an answer exists.
 A: Your answer is correct (assuming that you have accurately transcribed the question). The proof:
Let $F_n(c)$ be the cdf of $(1 - X_{(n)})$, where $X_{(n)}$ is the greatest element in a sample of size $n$. Let $F(c)$ be the cdf for the constant 0 distribution.
For $c < 0$, of course $F_n(c) = 0 = F(c)$.
For $c > 1$, of course $F_n(c) = 1 = F(c)$.
For $0 < c \le 1$:
$$
\begin{align}
F_n(c) &= P(1 - X_{(n)} \le c) \\
&= P(X_{(n)} \ge 1-c) \\
&= 1-P(X_1 < 1-c, ..., X_n < 1-c) \\
&= 1 - P(X_1 < 1-c)^n \to 1 = F(c)
\end{align}
$$
And the case $c = 0$ doesn't matter, because $F$ isn't continuous at $0$.
If the people you are arguing with don't realise that convergence in distribution to a constant is a thing, you could point them to e.g. Wikipedia's Proofs of convergence of random variables article.
A: Presumably the real issue here is that the TA does not like your answer because the purpose of scaling when seeking convergence results is to find an asymptotic distribution that is non-degenerate.  However, having said that, your answer is technically correct (the best kind of correct?).  You can easily tighten up your argument by giving the explicit distribution of the quantity of interest and showing the limit of this distribution as $n \rightarrow \infty$.  I will show you how to do this here.
Take $v=0$ and denote the resulting quantity as $Y_n \equiv 1- X_{(n)} = 1 - \max (X_1,...,X_n)$.  Now observe that for all $0 \leqslant y \leqslant 1$ we have:
$$\begin{align}
F_{Y_n}(y) 
\equiv \mathbb{P}(Y_n \leqslant y) 
&= 1 - \mathbb{P}(Y_n > y) \\[12pt]
&= 1 - \mathbb{P}(\max (X_1,...,X_n) < 1-y) \\[6pt]
&= 1 - \prod_{i=1}^n \mathbb{P}(X_i < 1-y) \\[4pt]
&= 1 - F_X(1-y)^n \\[12pt]
&= 1 - (3(1-y)-3(1-y)^2+(1-y)^3)^n \\[12pt]
&= 1 - ((3 - 3y) + (-3 + 6y - 3y^2) + (1 - 3y + 3y^2 - y^3))^n \\[12pt]
&= 1 - (1 - y^3)^n. \\[12pt]
\end{align}$$
Taking the limit (and now considering the broader range $y \in \mathbb{R}$) then gives:
$$\lim_{n \rightarrow \infty} F_{Y_n}(y) = \mathbb{I}(y \geqslant 0),$$
which is the CDF of the point-mass distribution at zero.  So, you are correct that the distribution of $Y_{n}$ converges to a point-mass distribution at zero.  Of course, this is a degenerate distribution, and ideally we would like to get an asymptotic result giving convergence to a non-degenerate distribution.  I recommend you see if you can also derive the latter, which was presumably the intended goal of the exercise.
