Convergence in Distribution and Ordered Statistics 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) &amp; = &amp; P(X\leq x) \\ &amp; = &amp; \int\limits_0^x \frac{1}{\theta}e^{-\frac{t}{\theta}} dt \\ &amp; = &amp; \frac{1}{\theta}\cdot\frac{-\theta}{1}e^{-\frac{t}{\theta}}\Big|_0^x \\ &amp; = &amp; 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) &amp; = &amp; P(\min\{X_1, X_2, X_3, \ldots, X_n\} \geq x) \\ &amp; = &amp; P(X_1 \geq x, \ldots, X_n\geq x) \\ &amp; = &amp; P(X_1\geq x)\cdot P(X_n\geq x) \\ &amp; = &amp; (e^{-\frac{x}{\theta}})^n \\ &amp; = &amp; 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) &amp; = &amp; P(M_n\leq x) \\ &amp; = &amp; P(Y_1 - \theta\ln(n) \leq x) \\ &amp; = &amp; P(Y_1 \leq x + \theta\ln(x)) \\ &amp; =  &amp; 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.
Convergence in Distribution in Order Statistics Let $X_1, X_2, \ldots$ be iid from Exp$(\theta)$ with density function $f(x) = \frac{1}{\theta}e^{-\frac{x}{\theta}}$. (a) 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: I have obtained the CDF of X to be: $$F_x(x) = 1 - e^{-\frac{x}{\theta}}$$ the CDF of $Y_1$ to be: $$F_{Y_1} = 1 = e^{-\frac{xn}{\theta}}$$ the CDF of $M_n$ to be: $$F_{M_n}(x) = \left[1 - \frac{1}{n}e^{-\frac{x}{\theta}}\right]^n$$ with limiting distribution $$e^{-e^{-\frac{x}{\theta}}}$$ the CDF of $T_n$ to be: $$F_{T_n}(x) = 1 - e^{-\frac{x}{\theta}}$$ which is also the limiting distribution. (b) Show that $M_n$ and $T_n$ are asymptotically independent. Answer: I've tried finding the joint PDF for ordered statistics by using the equation $$g_{ij}(y_i, y_j) = \frac{n!}{(i - 1)!(j - i - 1)!(n - j)!}[F(y_i)]^{i - 1}[F(y_j) - F(y_i)]^{j - i - 1}[1 - F(y_j)]^{n - j}f(y_i)f(y_j)$$ however, it is a mess based on my CDFs. Any suggestions on a different approach?
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