Show that $Y_n=n(1-F(X_{(n)})) $ converges in distribution to an exponential random variable.

Question

related to the topic that I was working on yesterday, I encountered a tougher one.

The assumption is that $X_1, X_2, ... X_n$ are iid with cdf $F(x)$. If $Y_n=n(1-F(X_{(n)}))$ then I am to prove that $Y_n$ converges in distribution to an exponential random variable with the unit mean.

Here is what I have so far.

$$\begin{align} Pr[n(1-F(X_{(n)})) \le y]&=Pr[1-\frac{y}{n} \le F(X_{(n)})]\\ &= 1-Pr[F(X_{(n)})\lt 1-\frac{y}{n}]\\ &= 1-Pr[\left(F(X) \right)^n<1-\frac{y}{n}]\\ &= 1-Pr\left[F(X)<\left(1-\frac{y}{n}\right)^{1/n}\right] \end{align} $$ Here is where I stopped.

I am not confident that what I am doing is right...

I appreciate your help.

Answer 1

Notice that

\begin{align*} \mathbb{P}\left[ n(1 - F(X_{(n)}) \leq y \right] &= 1 - \mathbb{P}\left[ F(X_{(n)}) < 1 - \frac{y}{n} \right]. \end{align*}

Now, without assuming extra condition on the distribution function $F$ of $X_i$'s, we cannot provide a decisive answer.

Example. Let $X_i$ be i.i.d. Bernoulli random variables with the parameter $p \in (0, 1)$, then

$$ F(X_{(n)}) = \begin{cases} 1, & X_i = 1 \text{ for some } i \in \{1, \cdots, n\} \\ 1-p, & X_i = 0 \text{ for all } i \in \{1, \cdots, n\} \end{cases} $$

and thus $n(1 - F(X_{(n)})) \to 0$ almost surely, hence in distribution.

From now on, we assume that $U_i := F(X_i)$ has the uniform distribution over $[0, 1]$. (This is true for any continuous distribution.) Then

\begin{align*} \mathbb{P}\left[ n(1 - F(X_{(n)}) \leq y \right] &= 1 - \mathbb{P}\left[ U_{(n)} < 1 - \frac{y}{n} \right] \\ &= 1 - \left( 1 - \frac{y}{n}\right)^n \\ &\xrightarrow[n\to\infty]{} 1 - e^{-y}. \end{align*}

Therefore the claim follows.