Proof.knit

Example 1 (Maximum of a Sample from a Uniform Distribution) Let $${X}_{1},{X}_{2},...,{X}_{n}$$ be a random sample from a uniform$$\left( 0 \, , \, \gamma \right)$$ distribution. Then, $${X}_{\left( n \right)}\stackrel{p}{\to} \gamma$$, where $${X}_{\left( n \right)} = \mbox{max}\left\{ {X}_{1},{X}_{2},...,{X}_{n} \right\}$$.

Proof. Let $$T = \gamma$$, whereby, $$T$$ is degenerate and, therefore, its cumulative distribution function (CDF) is given by $\begin{equation} F\left( t \right) = \left\lbrace \begin{array}{ll} 0, & t < \gamma\\ 1, & t \ge \gamma. \end{array} \right. \end{equation}$ Taking $${X}_{1},{X}_{2}, \dots ,{X}_{n}$$ as a random sample from a uniform$$\left( 0 \, , \, \gamma \right)$$ distribution into account, $${X}_{i} \sim U \left( 0 \, , \, \gamma \right)$$, for $$i = 1,2, \dots , n$$ and, by that, its probability distribution function (PDF) is $\begin{equation} {f}_{X}\left( x \right) = \left\lbrace \begin{array}{ll} \frac{1}{\gamma}, & 0 < x < \gamma\\ 0, & \mbox{elsewhere} \end{array} \right. \end{equation}$ and its CDF is $\begin{equation} {F}_{X}\left( x \right) = \left\lbrace \begin{array}{ll} 0, & x \leq 0\\ \frac{x}{\gamma}, & 0 < x < \gamma\\ 1, & x \ge \gamma \end{array} \right. \end{equation}$ Thence, the PDF and CDF of $${X}_{\left( n \right)}$$ are, respectively, $\begin{equation} {f}_{{X}_{\left( n \right)}}\left( t \right) = \frac{n}{\gamma} \left( \frac{t}{\gamma} \right)^{n-1},\quad 0 < t < \gamma \end{equation}$ and $\begin{equation} {F}_{{X}_{\left( n \right)}}\left( t \right) = \left\lbrace \begin{array}{ll} 0, & t \leq 0\\ \left( \frac{{t}}{{\gamma}} \right)^{n}, & 0 < t < \gamma\\ 1, & t \ge \gamma. \end{array} \right. \tag{1} \end{equation}$ From Expression (1), if $$0 < t < \gamma$$, then $$0 < \frac{t}{\gamma} < 1$$, in such way, $$\lim _{n \, \to \, \infty}{\left( \frac{{t}}{{\gamma}} \right)^{n}} = 0$$. If $$t = \gamma$$, then $$\frac{t}{\gamma} = 1$$, in such way, $$\lim _{n \, \to \, \infty}{\left( \frac{{t}}{{\gamma}} \right)^{n}} = 1$$. As a result, $$\lim _{n \, \to \, \infty}{{F}_{{X}_{\left( n \right)}}\left( t \right)} = F\left( t \right)$$, thus, $${X}_{\left( n \right)}\stackrel{d}{\to} \gamma$$ and, in consequence, $${X}_{\left( n \right)}\stackrel{p}{\to} \gamma$$. $$\square$$

Example 2 (Minimum of a Sample from a Shifted Exponential Distribution) Let $${X}_{1},{X}_{2}, \dots ,{X}_{n}$$ be a random sample from an exponential distribution that is shifted $$\gamma > 0$$ units, with $$\lambda = 1$$. Then, $${X}_{\left( 1 \right)}\stackrel{p}{\to} \gamma$$, where $${X}_{\left( 1 \right)} = \min\left\{ {X}_{1},{X}_{2}, \dots ,{X}_{n} \right\}$$.

Proof. Let $$Q = \gamma$$, whereby, $$Q$$ is degenerate and, therefore, its CDF is given by $\begin{equation} F\left( q \right) = \left\lbrace \begin{array}{ll} 0, & q < \gamma\\ 1, & q \ge \gamma \end{array} \right., \quad \gamma > 0. \end{equation}$ Since $${X}_{i}$$ ($$i = 1,2, \dots , n$$) is distributed as a shifted exponential$$\left( 1 \, , \, \gamma \right)$$, then, its PDF is $\begin{equation} {f}_{X}\left( x \right) = \left\lbrace \begin{array}{ll} {e}^{- \, \left( x \, - \, \gamma \right)}, & x \ge \gamma\\ 0, & x < \gamma \end{array} \right., \quad \gamma > 0. \end{equation}$ and its CDF is $\begin{equation} {F}_{X}\left( x \right) = \left\lbrace \begin{array}{ll} 1 \, - \, {e}^{- \, \left( x \, - \, \gamma \right)}, & x \ge \gamma\\ 0, & x < \gamma \end{array} \right., \quad \gamma > 0. \end{equation}$ Thence, the PDF and CDF of $${X}_{\left( 1 \right)}$$ are, respectively, $\begin{equation} {f}_{{X}_{\left( 1 \right)}}\left( q \right) = n \, {e}^{-n \, \left( q \, - \, \gamma \right)}, \quad q \ge \gamma \quad \mbox{and} \quad \gamma > 0 \end{equation}$ and $\begin{equation} {F}_{{X}_{\left( 1 \right)}}\left( q \right) = \left\lbrace \begin{array}{ll} 0, & q < \gamma\\ 1 \, - \, {e}^{-n \, \left( q \, - \, \gamma \right)} , & q \ge \gamma \end{array} \right. , \quad \gamma > 0. \end{equation}$ Then $\begin{equation} \lim _{n \, \to \, \infty}{{F}_{{X}_{\left( 1 \right)}}\left( q \right)} = \lim _{n \, \to \, \infty}{\begin{cases} 0, & q < \gamma\\ 1 \, - \, {e}^{-n \, \left( q \, - \, \gamma \right)}, & q \ge \gamma \end{cases}} = \begin{cases} 0, & q < \gamma\\ 1 , & q \ge \gamma \end{cases} = F\left( q \right). \end{equation}$ Thus, $${X}_{\left( 1 \right)}\stackrel{d}{\to} \gamma$$ and, in consequence, $${X}_{\left( 1 \right)}\stackrel{p}{\to} \gamma$$. $$\square$$