91影视

Using the definition of probability measure of random vector, we have that

$I=f_{x_1}<......<f_{x_5}{f}_{x_1}\left(x_1\right)......f_{x_5}\left(x_5\right)d{x}_1.....d{x}_5$

The integral in the most inner brackets is equal to I -F (x₄). Hence the integral in second most inner bracket is equal to

${鈭�}_{x_3}^{鈭�}{f}_{x_4}\left(x_4\right).(1-(F\left(x_4\right)d{x}_4$

Using the substitution $u_4=1-F\left(x_4\right)$we have that $d{u}_4=-f{x}_{4}d{x}_4$and boundaries goes from $1-F\left(x_3\right)$to zero. Hence,

${鈭�}_{x_3}^{鈭�}{f}_{x_4}\left(x_4\right).(1-F(x_4\left)\right)d{x}_4={鈭�}_0^{1-F\left(x_3\right)}{u}_{4}d{u}_4=\frac{(1-F{\left(x_3\right))}^2}{2}$

Observe the sequence that has been made here. Sub integration functions are as follows

$$\begin{gathered} 1-F\left(x_4\right),\frac{(1-F{\left(x_3\right))}^2}{2},\frac{(1-F{\left(x_2\right))}^3}{3!},\frac{(1-F{\left(x_1\right))}^4}{4} \\ I={鈭�}_{\mathrm{鈩�}}\frac{(1-F{\left(x_1\right))}^4}{4}f{x}_1{x}_{1}d{x}_1 \end{gathered}$$

and with $u_1=1-F\left(x_1\right)$we end up with

${鈭�}_{\mathrm{鈩�}}\frac{(1-F{\left(x_1\right))}^4}{4}f{x}_1{x}_{1}d{x}_1={鈭�}_0^1\frac{{u^4}_1}{4!}d{u}_1=\frac{1}{5!}$

Since variable $x1,.....x5$ are independent and equally distributed they can order themselves in any possible way with the same probability.