Q.6.29 - A First Course in Probability

Question

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

Step-by-Step Solution

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 1 : 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.

Step 2 : 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,

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

Now, let the event $E=\left\{x_i\le X_{\left(r\right)}\le x_i+{\delta }_{x_i}\cap x_j\le X_{\left(r\right)}\le x_j+{\delta }_{x_j}\right\}$ 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 $p_1=P\left[X_i\le x_i\right]=F\left(x_i\right)$

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

Step 4 : Explanation :

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

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

Hence proved.