91Ó°ÊÓ

The pdf for the joint order statistic is given by

\(\begin{aligned}{}{f_{{X_{\left( i \right)}},X\left( j \right)}}\left( {u,v} \right) = \frac{{n!}}{{\left( {i - 1} \right)!\left( {j - 1 - i} \right)!\left( {n - j} \right)!}} \times \\{\left( {{F_x}\left( u \right)} \right)^{i - 1}}{\left( {{F_X}\left( v \right) - {F_x}\left( u \right)} \right)^{j - 1 - i}}{\left( {1 - {F_X}\left( v \right)} \right)^{n - j}}{f_x}\left( u \right){f_x}\left( v \right)\end{aligned}\)

Where\({F_x}\left( u \right)\)is the cdf of random variable X

\({f_x}\left( u \right)\)is the pdf of random variable X

Put i=1 and j=n with\(X = \alpha \)

\({f_\alpha }\left( u \right) = 1\)and\({F_\alpha }\left( u \right) = u\)

Hence we get

\(\begin{aligned}{}{f_{\alpha \left( 1 \right),\alpha \left( n \right)}}\left( {s,s + t} \right) = \frac{{n!}}{{\left( {n - 2} \right)!}}{\left( {s + t - s} \right)^{n - 2}}\\ = n\left( {n - 1} \right){t^{n - 2}}\end{aligned}\)

For a starting point\(s \in \left( {0,1 - t} \right)\)with interval length t.

To find the probability of finding\(\alpha \left( 1 \right)\)and\(\alpha \left( n \right)\)within some interval of t, we need to integrate over all starting positions, S:

\(\begin{aligned}{}\int\limits_0^{1 - t} {{f_{\alpha \left( 1 \right),\alpha \left( n \right)}}\left( {s,s + t} \right)ds} = \int\limits_0^{1 - t} {n\left( {n - 1} \right){t^{n - 2}}ds} \\ = n\left( {n - 1} \right){t^{n - 2}}\int\limits_0^{1 - t} {ds} \\ = n\left( {n - 1} \right){t^{n - 2}}\left( {1 - t} \right)\end{aligned}\)

To calculate the expectation of the range of a random sample of size n is

You can integrate the above equation for\(t \in \left( {0,1} \right)\)or can take the expectation of the difference of the random variable\({\alpha _{\left( n \right)}}\)with\({\alpha _{\left( 1 \right)}}\)

The pdf of the k^th order statistic is:

\({f_{X\left( k \right)}}\left( u \right) = \frac{{n!}}{{\left( {k - 1} \right)!\left( {n - k} \right)!}}{\left( {{F_X}\left( u \right)} \right)^{k - 1}}{\left( {1 - {F_X}\left( u \right)} \right)^{n - k}}{f_x}\left( u \right)\)

Put\(X = \alpha \),\(k = 1\)and\(k = n\)for the n^th-order statistic, we get:

\(\begin{aligned}{l}{f_{\alpha \left( 1 \right)}}\left( u \right) = n{\left( {1 - u} \right)^{n - 1}}\\{f_{\alpha \left( n \right)}}\left( u \right) = n{u^{n - 1}}\end{aligned}\)

Now taking expectations of their difference:

\(E\left( {\alpha \left( n \right) - \alpha \left( 1 \right)} \right) = \int\limits_0^1 {u\left( {n{u^{n - 1}} - n{{\left( {1 - u} \right)}^{n - 1}}} \right)du} \)

\(\frac{n}{{n + 1}} - \frac{n}{{n + 1}} = \frac{{n - 1}}{{n + 1}}\)

Hence, the expectation of the range of a random sample of size n is\(\frac{{n - 1}}{{n + 1}}\)