I have a problem that I am working on. I have thought of two approaches, but they are giving me different answers. I wanted to know which (or if each) approach was wrong and why.
Prompt: $X_1, X_2, ...$ are $\overset{iid}{\sim}Expo(1)$. Let $X_{n:n}$ be an order statistic. Define $Y_n = X_{n:n} - ln(n)$. Show that $Y_n$ converges in law to a limit distribution as $n$ tends to infinity.
Approach 1: Use properties of order statistics to find a distribution for $X_{n:n}$, then transfrom $X_{n:n}$ to $Y_n$
$$f_{X_{n:n}}(X_{n:n}=x) = n \left [ 1 - exp(-x)\right ]^{n-1}exp(-x)$$
Which gives us,
$$F_{Y_n}(Y_n=y) = \left [ 1-exp(-y)\right]^{n-1}exp(-y)$$
Approach 2: Use properties of order statistics and probabilities to derive $F_{Y_n}(y)$
$$F_{Y_n}(y) = P(Y_n \leq Y) = P(X_{n:n} - ln(n)\leq y)$$
$$=P(X_{n:n} \leq y + ln(n))$$
$$=\left [ P(X_1 \leq y +ln(n)) \right ]^n$$
$$=\left [ 1 - exp(-y -ln(n))\right ]^n$$
$$=\left [ 1 - \frac{exp(-y)}{n}\right ]^n$$
$$\Rightarrow \underset{n \rightarrow \infty}{lim} e^{-e^{-y}}$$
Thoughts appreciated!