Let {x_i} be a dataset consisting of N real numbers, x1, …, xN. Prove the function g(m) = ∑_{i=1}^{N} |x_i - m| is minimized when m = median({x_i}).

Steps to Solve

1. Define the Function g(m)

The function ( g(m) ) is defined as:

$$ g(m) = \sum_{i=1}^{n} |x_i - m| $$

where ( x_1, x_2, \ldots, x_n ) are the data points in the dataset.

2. Identify the Median

The median of the dataset is defined as the value that separates the higher half from the lower half of the dataset when arranged in order. For an ordered dataset:

• If ( n ) is odd: the median is ( m = x_{\frac{n+1}{2}} ).
• If ( n ) is even: the median is ( m = \frac{x_{\frac{n}{2}} + x_{\frac{n}{2} + 1}}{2} ).

3. Consider a Value d Different from the Median

Let’s take a value ( d ) that is not equal to the median ( m ). We need to analyze the differences in sums ( g(d) ) and ( g(m) ).

4. Calculate g(d) and g(m)

We compare the sums:

$$ g(d) = \sum_{i=1}^{n} |x_i - d| $$

Recognize that the absolute difference can be represented based on whether ( x_i ) is less than or greater than ( d ).

5. Use the Property of Absolute Values

When calculating ( g(d) ), we can split the dataset into two groups:

• Group 1: Values less than ( d )
• Group 2: Values greater than ( d )

The sum of absolute differences will yield a larger value ( g(d) ) compared to ( g(m) ).

6. Conclude that g(m) is Minimal

We can argue that due to how the median minimizes the total distance of points, the property implies that:

$$ g(m) \leq g(d) $$

for any ( d ) not equal to the median ( m ), proving that the function ( g(m) ) is minimized at ( m ) being the median.