Let $Y_{1} < Y_{2} < ... < Y_{n}$ be the order statistics of a random sample from a distribution with pdf $f(x) = e^{-x} , 0 < x < \infty$, zero elsewhere. Determine the limiting distribution of $Z n = ( Y_{n} – ln(n) )$
Course textbook is Introduction to Mathematical Statistics, Seventh Edition, by Robert V. Hogg, Joseph W. McKean, Allen T. Craig., this is from ch. 5.2.
I understand that I need to do something like $F_{Z_{n}} = P(Y_{n} - ln(n)\leq z) = P(Y_{n} \leq z + ln(n))$, but I can't figure out what to do after this step.
First, we obtain the CDF of $X$. $$ \begin{align} F_X(x)&=\int_0^xf(t)\ dt\\ \text{Pr}[X\le x]&=\int_0^x e^{-t}\ dt\\ &=-\left.e^{-t}\right|_{t=0}^x\\ &=1-e^{-x}. \end{align} $$
Note that from transformation $Z_n = Y_n - \ln\ n$, we have $Y_1<Y_2<\cdots<Y_n\le Z_n$ implies $X_i\le Z_n$, for $i=1, 2,\cdots, n$. Therefore, the fact that $X_i$s are i.i.d. implies $$ \begin{align} F_{Y_n}(y)&=\text{Pr}[Y_n\le y]\\ &=\text{Pr}[Y_n\le z+\ln\ n]\\ &=\left(\text{Pr}[X_i\le z+\ln\ n]\right)^n\\ &=\left(1-e^{-(z+\ln n)}\right)^n\\ &=\left(1-e^{-z}e^{-\ln n)}\right)^n\\ &=\left(1-e^{-z}\cdot\frac{1}{n}\right)^n\\ &=\left(1-\frac{e^{-z}}{n}\right)^n.\\ \end{align} $$ Thus, as $n\to\infty$, the limiting distribution of $Z_n = Y_n - \ln\ n$ is $$ \begin{align} \lim_{n\to\infty}F_{Y_n}(y_n)&=\lim_{n\to\infty}\left(1-\frac{e^{-z}}{n}\right)^n\\ &=\Large\color{blue}{\exp\left(-e^{-z}\right)}. \end{align} $$
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$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$