Question

Consider a sample \({x_1},{x_2},...,{x_n}\) with neven. Let \({\bar x_L}\) and \({\bar x_U}\) denote the average of the smallest n/2 and the largest n/2 observations, respectively. Show that the mean absolute deviation from the median for this sample satisfies

\(\sum {\left| {{x_i} - \tilde x} \right|} /n = \left( {{{\bar x}_U} - {{\bar x}_L}} \right)/2\)

Then show that if n is odd and the two averages are calculated after excluding the median from each half, replacing non the left with n-1 gives the correct result.(Hint: Break the sum into two parts, the first involving observations less than or equal to the median and the second involving observations greater than or equal to the median.)

Short answer

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

Step-by-step solution

Given information

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.

Derivation of the proof

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.