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, \({X}_{\left( 1 \right)}\stackrel{p}{\to} \gamma\).
\(\square\)