Question

Verify Equation $(6.6)$, which gives the joint density of $X_{\left(i\right)}$ and $X_{\left(j\right)}$.

Short answer

Equation :

$$\begin{gathered} f_{x_{\left(i\right)}{x}_{\left(j\right)}}\left(x_{\left(i\right)},x_{\left(j\right)}\right)=\frac{n!}{(i-1)!(j-i-1)!(n-j)!}\left[F{\left(x_i\right)}^{i-1}\right]{\left[F\left(x_j\right)-F\left(x_i\right)\right]}^{j-i-1} \\ {\left[1-F\left(x_j\right)\right]}^{n-j}f\left(x_j\right)f\left(x_i\right)\forall x_i<x_j \end{gathered}$$proved.

Step-by-step solution

Joint probability distribution :

The related probability distribution on all possible pairings of outputs is the joint probability distribution. For each given number of random variables, the joint distribution may be studied.

Explanation :

Equation $(6.6)$:

$$\begin{gathered} f_{x_{\left(i\right)}{x}_{\left(j\right)}}\left(x_{\left(i\right)},x_{\left(j\right)}\right)=\frac{n!}{(i-1)!(j-i-1)!(n-j)!}\left[F{\left(x_i\right)}^{i-1}\right]{\left[F\left(x_j\right)-F\left(x_i\right)\right]}^{j-i-1} \\ {\left[1-F\left(x_j\right)\right]}^{n-j}f\left(x_j\right)f\left(x_i\right)\forall x_i<x_j \end{gathered}$$

Let the denote the joint probability density function of $X_{\left(i\right)}$and $X_{\left(j\right)}$to prove the equation,

$f_{x_{\left(i\right)}{x}_{\left(j\right)}}\left(x_{\left(i\right)},x_{\left(j\right)}\right)$where $1\le i\le j\le n$, then we have,

role="math" localid="1647420071276" $f_{x_{\left(i\right)}{x}_{\left(j\right)}}\left(x_{\left(i\right)},x_{\left(j\right)}\right)=\underset{\underset{{\delta }_{xj}\to 0}{{\delta }_{xi}\to 0}}{\lim }\frac{P\left[x_i\le X_{\left(r\right)}\le x_i+{\delta }_{x_i}\cap x_j\le X_{\left(r\right)}\le x_j+{\delta }_{x_j}\right]}{{\delta }_{x_i}{\delta }_{x_j}}.....\left(1\right)$

Now, let the event role="math" localid="1647419990583" $E=\left\{x_i\le X_{\left(r\right)}\le x_i+{\delta }_{x_i}\cap x_j\le X_{\left(r\right)}\le x_j+{\delta }_{x_j}\right\}$can be written as :

$1\leftarrow i-1\to \left|1\right|\leftarrow j-i-1\to \left|1\right|\leftarrow n-j\to 1$

Now,

$X_{\left(i\right)}\le x_i$for $i-1$of the $X{\text{'}}_{\left(r\right)}s,$

$x_i<X_{\left(r\right)}\le x_i+{\delta }_{x_i}$for one $X_{\left(r\right)}$

$x_i+{\delta }_{x_i}<X_{\left(r\right)}\le x_j$for $\left(j-i-1\right)$of $X{\text{'}}_{\left(r\right)}s,$

$x_j<X_{\left(r\right)}\le x_j+{\delta }_{x_j}$for one $X_{\left(r\right)}$

And $X_{\left(r\right)}>x_j+{\delta }_{xj}$for $\left(n-j\right)$of the $X{\text{'}}_{\left(r\right)}s,$

Hence, by using multinomial probability law we get the following as :

$P\left(E\right)=P[x_i\le X_{\left(r\right)}\le x_i+{\delta }_{x_i}\cap x_j\le X_{\left(r\right)}\le x_j+{\delta }_{x_j}]=\frac{n!}{(i-1)!1!(j-i-1)!1!(n-j)!}{p_1}^{i-1}{p}_2{p_3}^{j-i-1}{p}_4{p_5}^{n-j}.......\left(2\right)$

where $p_1=P\left[X_i\le x_i\right]=F\left(x_i\right)$

$$\begin{gathered} p_2=P\left[x_i<X_{\left(r\right)}\le x_i+{\delta }_{x_i}\right]=F(x_i+{\delta }_{x_i})-F\left(x_i\right) \\ {p}_3=P\left[x_i+{\delta }_{x_i}<X_{\left(r\right)}\le x_j\right]=F\left(x_j\right)-F(x_i+{\delta }_{x_i}) \\ {p}_4=P\left[x_j<X_{\left(r\right)}\le x_j+{\delta }_{x_j}\right]=F(x_j+{\delta }_{x_j})-F\left(x_j\right) \\ {p}_5=P\left[X_{\left(r\right)}>x_j+{\delta }_{x_j}\right]=1-P\left[X_{\left(r\right)}\le x_j+{\delta }_{x_j}\right] \\ =1-F(x_j+{\delta }_{x_i}) \end{gathered}$$

Explanation :

Thus, substituting $\left(2\right)$in $\left(1\right)$, we get,

localid="1647425314276" $$\begin{gathered} f_{x_{\left(i\right)}{x}_{\left(j\right)}}\left(x_{\left(i\right)},x_{\left(j\right)}\right)=\frac{n!}{(i-1)!(j-i-1)!(n-j)!}\times \underset{{\delta }_{x_i}\to 0}{\lim }\frac{F(x_i+{\delta }_{x_i}-F(x_i\left)\right)}{{\delta }_{x_i}}\times \\ \underset{{\delta }_{x_i}\to 0}{\lim }{\left[F\right(x_j)-F(x_i+{\delta }_{x_i}\left)\right]}^{j-i-1}\times \underset{{\delta }_{x_j}\to 0}{\lim }\left[\frac{F(x_j+{\delta }_{x_j})-F\left(x_j\right)}{{\delta }_{x_j}}\right]\times \underset{{\delta }_{x_j}\to 0}{\lim }{\left[1-F(x_j+{\delta }_{x_j})\right]}^{n-j} \\ =\frac{n!}{(i-1)!(j-i-1)!(n-j)!}\left[F{\left(x_i\right)}^{i-1}\right]f\left(x_i\right){\left[F\left(x_j\right)-F\left(x_i\right)\right]}^{j-i-1}f\left(x_j\right){\left[1-F\left(x_j\right)\right]}^{n-j} \\ {f}_{x_{\left(i\right)}{x}_{\left(j\right)}}\left(x_{\left(i\right)},x_{\left(j\right)}\right)=\frac{n!}{(i-1)!(j-i-1)!(n-j)!}\left[F{\left(x_i\right)}^{i-1}\right]{\left[F\left(x_j\right)-F\left(x_i\right)\right]}^{j-i-1} \\ {\left[1-F\left(x_j\right)\right]}^{n-j}f\left(x_j\right)f\left(x_i\right)\forall x_i<x_j \end{gathered}$$

Hence proved.