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#### Introduction

Recently, I introduced the golden search method – a special way to save computation time by modifying the bisection method with the golden ratio, and I illustrated how to minimize a cusped function with this script. I also wrote an R function to implement this method and an R script on how to apply this method with this example. Today, I will use apply this method to a statistical topic: minimizing the sum of absolute deviations with the median.

While reading Page 148 (Section 6.3) in Michael Trosset’s “An Introduction to Statistical Inference and Its Applications”, I learned 2 basic, simple, yet interesting theorems.

If X is a random variable with a population mean and a population median , then

a) minimizes the function

b) minimizes the function

I won’t prove these theorems in this blog post (perhaps later), but I want to use the golden section search method to show a result similar to b):

c) The sample median, , minimizes the function

.

This is not surprising, of course, since

– is just a function of the random variable

– by the law of large numbers,

Thus, if the median minimizes , then, intuitively, it minimizes . Let’s show this with the golden section search method, and let’s explore any differences that may arise between odd-numbered and even-numbered data sets.

#### Minimizing a Data Set with an Odd Number of Data

Consider this following data set with an odd number of data:

A = {5, 9, 11, 14, ,17}

Here is the plot of the function

for this data set.

This function is clearly cusped at multiple points and has a unique minimum. Thus, the golden section search method is suitable for minimizing it.

To use my R script to find the minimizer, I need to

– define an R function for this above function (I call it sum.of.distances1 – see the R scripts at the bottom of this blog post)

– call both this above function and the function for the golden section search method with the source() command

– feed the 4 required arguments – objective function (sum.of.distances1), the lower and upper bounds (0, 20), and the tolerance (1e-5) – to golden.section.search()

Here is the output after the first iteration:

Iteration # 1 f1 = 23.08204 f2 = 18.36068 f2 < f1 New Upper Bound = 20 New Lower Bound = 7.63932 New Lower Test Point = 12.36068 New Upper Test Point = 15.27864

Here is the output after the final iteration, including the resulting minimizer:

Iteration # 31 f1 = 17 f2 = 17 f2 > f1 New Upper Bound = 11 New Lower Bound = 11 New Upper Test Point = 11 New Lower Test Point = 11 Final Lower Bound = 11 Final Upper Bound = 11 Estimated Minimizer = 11

This is, of course, the median of A = {5, 9, 11, 14, ,17}. You can confirm it in R by using the median() function:

data.vector1 = c(5, 9, 11, 14, 17) median(data.vector1)

#### Minimizing a Data Set with an Even Number of Data

Now, what about a data set with an even number of data?

B = {5, 9, 11, 14, ,17, 18}

Here is the plot of the objective function.

Notice that, between and , the function is horizontal line. This makes intuitive sense; if you think back to learning how to calculate the median of a data set with an even number of data, the median is *NOT* one of the data, but an *interpolated value* between the 2 “middle” data (the average of the 2 “middle” data is usually used). You can confirm this interpolation with the median() function in R.

data.vector2 = c(5, 9, 11, 14, 17, 18) median(data.vector2)

For a data set with an even number of data, interpolation is needed because there is *no unique number* such that there are equal number of data larger and smaller than itself.

This non-uniqueness of the median is reflected in the minimization with the golden section search method. The minimizer varies depending on the initial lower and upper bounds.

– If I use 0 and 20 as the initial lower and upper bounds, here is the output of the final iteration and the resulting minimizer.

Iteration # 31 f1 = 24 f2 = 24 f2 < f1 New Upper Bound = 13.74744 New Lower Bound = 13.74743 New Lower Test Point = 13.74743 New Upper Test Point = 13.74743 Final Lower Bound = 13.74743 Final Upper Bound = 13.74744 Estimated Minimizer = 13.74743

– If I use 1 and 20 as the initial lower and upper bounds instead, here is the corresponding output.

Iteration # 31 f1 = 24 f2 = 24 f2 < f1 New Upper Bound = 13.80178 New Lower Bound = 13.80177 New Lower Test Point = 13.80178 New Upper Test Point = 13.80178 Final Lower Bound = 13.80177 Final Upper Bound = 13.80178 Estimated Minimizer = 13.80178

#### Summary

Thus, the lessons to be taken away from this exercise are:

1) The median minimizes the function

2) There is a unique median for a data set with an odd number of data, but there is no unique median for a data set with an even number of data.

#### Reference

This blog post was inspired by Exercise #2 in Section 7.1 on Page 135 in “A First Course in Statistical Programming with R” by John Braun and Duncan Murdoch.

#### R Script

I saved the 2 objective functions in a file called sum.of.distances.R:

# odd number of data sum.of.distances1 = function(x) { abs(5-x) + abs(9-x) + abs(11-x) + abs(14-x) + abs(17-x) } # even number of data sum.of.distances2 = function(x) { abs(5-x) + abs(9-x) + abs(11-x) + abs(14-x) + abs(17-x) + abs(18-x) }

See my previous post for the function that conducts the golden section search. In a separate script called “minimization.R”, I ran the following:

##### Showing that the median minimizes the sum of absolute deviations ##### By Eric Cai - The Chemical Statistician source('sum.of.distances.R') png('INSERT YOUR DIRECTORY PATH HERE/example data 1.png') curve(sum.of.distances1, from = 0, to = 20, main = 'f(x) = |5-x| + |9-x| + |11-x| + |14-x| + |17-x|', ylab = 'f(x)') dev.off() golden.section.search(sum.of.distances1, 0, 20, 1e-5) data.vector1 = c(5, 9, 11, 14, 17) median(data.vector1) png('INSERT YOUR DIRECTORY PATH HERE/example data 2.png') curve(sum.of.distances2, from = 0, to = 20, main = 'f(x) = |5-x| + |9-x| + |11-x| + |14-x| + |17-x| + |18-x|', ylab = 'f(x)') dev.off() golden.section.search(sum.of.distances2, 0, 20, 1e-5) golden.section.search(sum.of.distances2, 1, 20, 1e-5) data.vector2 = c(5, 9, 11, 14, 17, 18) median(data.vector2)

Filed under: Applied Mathematics, Descriptive Statistics, Numerical Analysis, Plots, R programming, Statistical Computing Tagged: absolute deviations, applied math, applied mathematics, math, mathematics, median, numerical analysis, numerical method, numerical methods, plot, plots, plotting, R, R programming, statistical computing, statistics, sum of absolute deviations

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