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A sample of size n is \({x_1},{x_2},...,{x_n}\)with n even.

\({\bar x_L}\) and \({\bar x_U}\) denote the average of the smallest n/2 and the largest n/2 observations, respectively.

Let\(n = 2d\left( {even} \right)\).

The ascending of n observations is,

\({X_1},{X_2},...,{X_d},{X_{d + 1}},...,{X_{2d}}\).

The sample mean is given as,

\(\begin{aligned}{{\bar X}_L} &= \frac{1}{d}\sum\limits_{i = 1}^d {{X_i}} \\{{\bar X}_U} &= \frac{1}{d}\sum\limits_{i = 1}^d {{X_{d + i}}} \end{aligned}\)

The sample median is given as,

\(\tilde X = {X_{d + 1}}\)

Further calculations are computed as,

\(\begin{aligned}\frac{{n\left( {{{\bar X}_U} - {{\bar X}_L}} \right)}}{2} &= d\left( {{{\bar X}_U} - {{\bar X}_L}} \right)\\ &= \sum\limits_{i = 1}^d {{X_{d + i}}} - \sum\limits_{i = 1}^d {{X_i}} \\ &= \sum\limits_{i = 1}^d {\left( {{X_{d + i}} - {X_i}} \right)} \\ &= \sum\limits_{i = 1}^d {\left( {{X_{d + i}} - \tilde X + \tilde X - {X_i}} \right)} \,\;\;\;\;\;\left( {Adding\;and\;substracting\;median} \right)\\ &= \sum\limits_{i = 1}^d {\left| {{X_{d + i}} - \tilde X} \right|} + \sum\limits_{i = 1}^d {\left| {\tilde X - {X_i}} \right|} \\ &= \sum\limits_{i = 1}^{2d} {\left| {{X_{d + i}} - \tilde X} \right|} + \sum\limits_{i = 1}^d {\left| {{X_i} - \tilde X} \right|} \\ &= \sum\limits_{i = 1}^{2d} {\left| {{X_i} - \tilde X} \right|} \end{aligned}\)

Therefore,

\(\begin{aligned}\frac{{n\left( {{{\bar X}_U} - {{\bar X}_L}} \right)}}{2} = \sum\limits_{i = 1}^{2d} {\left| {{X_i} - \tilde X} \right|} \\ = \sum\limits_{i = 1}^n {\left| {{X_i} - \tilde X} \right|} \end{aligned}\)

Thus,

\(\frac{{\sum\limits_{i = 1}^n {\left| {{X_i} - \tilde X} \right|} }}{n} = \frac{{\left( {{{\bar X}_U} - {{\bar X}_L}} \right)}}{2}\)

For odd number of observations, \(n = 2d + 1\).

The ascending of n observations is,

\({X_1},{X_2},...,{X_d},{X_{d + 1}},...,{X_{2d + 1}}\).

The sample mean is given as,

\(\begin{aligned}{{\bar X}_L} &= \frac{1}{d}\sum\limits_{i = 1}^d {{X_i}} \\{{\bar X}_U} &= \frac{1}{d}\sum\limits_{i = 1}^d {{X_{d + 1 + i}}} \end{aligned}\)

The sample median is given as,

\(\tilde X = {X_{d + 1}}\)

The calculations are as follows,

\(\begin{aligned}\frac{{\left( {n - 1} \right) * \left( {{{\bar X}_U} - {{\bar X}_L}} \right)}}{2} &= d\left( {{{\bar X}_U} - {{\bar X}_L}} \right)\\ &= \sum\limits_{i = 1}^d {{X_{d + 1 + i}}} - \sum\limits_{i = 1}^d {{X_i}} \\ &= \sum\limits_{i = 1}^d {\left( {{X_{d + 1 + i}} - {X_i}} \right)} \\ &= \sum\limits_{i = 1}^d {\left( {{X_{d + 1 + i}} - \tilde X + \tilde X - {X_i}} \right)} \\ &= \sum\limits_{i = 1}^d {\left| {{X_{d + 1 + i}} - \tilde X} \right| + \sum\limits_{I = 1}^d {\left| {\tilde X - {X_i}} \right|} } \\ &= \sum\limits_{i = d + 2}^{2d + 1} {\left| {{X_i} - \tilde X} \right| + \sum\limits_{I = 1}^d {\left| {{X_i} - \tilde X} \right|} } \\ &= \sum\limits_{i = 1}^d {\left| {{X_i} - \tilde X} \right| + \left| {\tilde X - \tilde X} \right|} + \sum\limits_{i = d + 2}^{2d + 1} {\left| {{X_i} - \tilde X} \right|} \\ &= \sum\limits_{i = 1}^{2d + 1} {\left| {{X_i} - \tilde X} \right|} \end{aligned}\)

Therefore,

\(\begin{aligned}\frac{{\left( {n - 1} \right) * \left( {{{\bar X}_U} - {{\bar X}_L}} \right)}}{2} &= \sum\limits_{i = 1}^{2d + 1} {\left| {{X_i} - \tilde X} \right|} \\ &= \sum\limits_{i = 1}^n {\left| {{X_i} - \tilde X} \right|} \\ &= \frac{{\sum\limits_{i = 1}^n {\left| {{X_i} - \tilde X} \right|} }}{{n - 1}}\\ &= \frac{{{{\bar X}_U} - {{\bar X}_L}}}{2}\end{aligned}\)

Hence Proved.