Let $X_1, X_2, \ldots$ be iid from Exp$(\theta)$ with density function $f(x) = \frac{1}{\theta}e^{-\frac{x}{\theta}}$. Find the limiting distribution of $M_n = Y_1 - \theta\ln(n)$ and $T_n = nY_n$, where $Y_1$ and $Y_n$ represent the first and last order statistics (among $n$ observations) respectively.
Answer: The CDF of $X$ is: \begin{eqnarray*} F_X(x) & = & P(X\leq x) \\ & = & \int\limits_0^x \frac{1}{\theta}e^{-\frac{t}{\theta}} dt \\ & = & \frac{1}{\theta}\cdot\frac{-\theta}{1}e^{-\frac{t}{\theta}}\Big|_0^x \\ & = & 1 - e^{-\frac{x}{\theta}}. \end{eqnarray*}
or $P(X \geq x) = e^{-\frac{x}{\theta}}.$ Then since $Y_1 = \min\{X_1, X_2, X_3, \ldots, X_n\}$: \begin{eqnarray*} P(Y_1 \geq x) & = & P(\min\{X_1, X_2, X_3, \ldots, X_n\} \geq x) \\ & = & P(X_1 \geq x, \ldots, X_n\geq x) \\ & = & P(X_1\geq x)\cdot P(X_n\geq x) \\ & = & (e^{-\frac{x}{\theta}})^n \\ & = & e^{-\frac{xn}{\theta}}. \end{eqnarray*}
So the CDF of $Y_1$ is: $$F_{Y_1}(x) = P(Y_1 \leq x) = 1 - e^{-\frac{xn}{\theta}}$$
Thus, the CDF of $M_n$ is: \begin{eqnarray*} F_{M_n}(x) & = & P(M_n\leq x) \\ & = & P(Y_1 - \theta\ln(n) \leq x) \\ & = & P(Y_1 \leq x + \theta\ln(x)) \\ & = & 1 - e^{-\frac{n}{\theta}(x + \theta\ln(n)}. \end{eqnarray*}
I do not know how to compute the limiting distribution because I am unsure how to evaluate $\lim\limits_{n\to\infty} F_{M_n}(x)$. Have I completed the mathematics correctly?
I did not attempt $T_n$ portion yet because I believe $M_n$ part is incorrect.