# Find the limiting distribution of $Z_n=n[1-F(Y_n)]$

Let $Y_n$ denote the maximum (the last order statistic) of a random sample of size $n$ from a distribution of the continuous type that has c.d.f $F(x)$ and pdf $f(x) = F'(x)$. Find the limiting distribution of $Z_n=n[1-F(Y_n)]$

I dont have the solution to this but the answer my teacher gave us is that the limiting distribution of $Z_n$ should be exponential with mean $1$. If you define $Z_n=n[1-Y_n]$ then that limiting distribution is exponential with mean $1$ but I dont know how to work with $F(Y_n)$

I'm going to denote your "maximum" $Y_n$ as $Y_{(n)}$. Observe that \begin{align*} F_{Z_n}(z) &= P(Z_n \le z) \\ &= P([1 - F_{}(Y_{(n)})] \le z/n ) \\ &= P (F_{}(Y_{(n)}) \ge 1-z/n) \\ &= P(Y_{(n)} \ge F^{-1}(1-z/n)) \\ &= 1- P(Y_{(n)} \le F^{-1}(1-z/n)) \quad \text{(continuous dist.)} \\ \end{align*}

Suppose our random sample was $Y_1,\dots,Y_n$ from a distribution with cdf $F$. Then $$P(Y_{(n)} \le c) = P((Y_1 \le c) \ \cap \cdots \cap (Y_n \le c)) \stackrel{ind.}{=} P(Y_1 \le c) \cdots P(Y_n \le c) \stackrel{i.d.}{=} [P(Y_1 \le c)]^n = [F(c)]^n$$

Returning and using this in the main problem, we have $$F_{Z_n}(z) = 1 - [F \left( F^{-1}(1-z/n) \right)]^n = 1- [1-z/n]^n \xrightarrow{n \to \infty} 1- e^{- z}.$$ which we recognize as Exponential rate $=1$.

This is essentially the same idea given by @user365239, but stated a little differently.

Let $X_1,\dots,X_n$ be IID with DF $F$. Let $Y_n =\max_{j=1\dots n} X_j$.

Let $Z_j = n(1-F(X_j))$. Then $Z_1,\dots,Z_n$ are IID, and observe that

$$n(1-F(Y_n)) = \min_{j=1\dots n}Z_j.$$

Also, since $F(X_j)\sim U[0,1]$, we also have $1-F(X_j)\sim U[0,1]$, and so $Z_j\sim U[0,n]$.

From this we obtain:

$$P( \min_{j=1...n} Z_n > x) = P( U[0,n]>x)^n = (1-x/n)^n \to e^{-x},$$

where the second equality holds when $n>x$.