Another Take on Averages (Tuklas Vol. 15, No. 3 - November 23, 2013)

ANOTHER TAKE ON AVERAGES

One day, Carlo and Hanz decided to perform an experiment. They each had an opaque box in which they placed a red ball and a white ball. They would then draw from their respective boxes, record the color of the drawn ball, then place the ball back. They can draw as many times as they prefer. The goal is to have the highest percentage of red balls drawn.

Carlo's strategy was to draw significantly more than Hanz, and it paid off. In their first session, Carlo was able to draw the red ball 7 out of 19 times to get a percentage of 0.368. Hanz only made 9 attempts and drew the red ball 4 times to get a percentage of 0.364.

For the second session, Hanz made sure he will draw more times than Carlo, but things would still not go Hanz's way. Carlo drew the red ball 17 times out of 39 (43.6%), while Hanz drew 32 out of 74 (43.2%).

Hanz almost admitted defeat, but the statistician in him found hope by taking the aggregate draws as follows:

PLAYER	1ST SESSION		2ND SESSION		TOTAL
Carlo	7/19	0.3684	17/39	0.4360	24/58	0.4138
Hanz	4/11	0.3636	32/74	0.4324	36/85	0.4235

Using the total data, Hanz argued that he won.

We now ask two questions. First, who ACTUALLY won? And more importantly, how did this happen?

This story is an example of a phenomenon called the Simpson's Paradox. This states that if a group of data were separated into several categories, the population which exhibits the most successes may not exhibit less in each of the categories. What this means is that when comparing two groups of data, the successes of one group can still be ahead when we look at the aggregate, even if that group is behind percentage-wise in all the specific categories. This phenomenon had been observed as early as the 1920's through medical and sociostatistical data. However, Edward H. Simpson was one of the first who explicitly pointed it out in 1951, and it was not until 1972 that a paper formalized its definition.

A CLASSIC EXAMPLE

Before we go deeper into the situations that make up the paradox, let us introduce one of the most classic examples. A popular scenario where Simpson's Paradox occurs is through a sport -- baseball. Although this was not the first occurrence of the paradox on record, it gained the interest of statisticians and baseball junkies alike, especially those who utilized Sabermetrics (the Math of Baseball). These people, called sabermetricians, use baseball data in order to analyze the potential of a new player, team chemistry, even the use of performance-enhancing drugs. Sabermetricians may find this particular set of data interesting, since the presence of this phenomenon will make them rethink their traditional ways of looking into data.

For 1995-1996, the batting averages of Derek Jeter and David Justice (both of whom were very popular at the time) were compared, and the summary is in the next table.

PLAYER	1995		1996		TOTAL
Jeter	12/48	0.250	183/582	0.314	195/630	0.310
Justice	4/11	.253	32/74	0.321	149/551	0.270

Similar to the Carlo-Hanz experiment, Justice had better averages if we look at the seasons independently. Yet when we take the aggregate, Jeter is the actual winner. Because of this, baseball junkies should consider the yearly averages or even the aggregate (more commonly called the "career" statistic) separately, and not take one or the other alone.

REASONING

So after the two scenarios, you might be thinking, "Is there something wrong with the computation of the data?", or "Is there something that can be done to make the data make sense?"

To answer these questions, we go back to the data. One of the main reasons for the paradox is the discrepancy in the trials. To see this, let us try to visualize what is happening. We assign to the y-axis the number of successful trials per category, and to the x-axis the total number of trials per category. The success rate per category per group can be thought of as the ratio \(m = \frac{p}{q}\), with \(p\) successes out of \(q\) trials done by a particular group. On the Cartesian plane, this can be represented as a vector with slope equal to \(m\). We assign trials \(a\) and \(b\) (in black) to Group A and trials \(c\) and \(d\) to Group B (in blue). Trials \(a\) and \(c\) belong to the same category, as do trials \(b\) and \(d\). The total success rate for Groups A and B can then be represented by the slope of the vectors \(a+b\) and \(c+d\), respectively (see figure below).

Let us now compare the slopes. Note that \(m_a > m_c\) and \(m_b > m_d\). That is, Group A performed better than Group B under each individual category. But upon forming \(a+b\) and \(c+d\) (using the Parallelogram Law), we can see that \(m_{a+b} < m_{c+d}\), despite the fact that \(a+b\) is longer than \(c+d\). This indicates that the trials themselves are not considered, only the ratios.

Given all of this, is there any way for us to make sense out of the separate data and try to make the conclusion consistent? To do so, we need to do what we call a "normalization process". For the Hanz-Carlo experiment, we will "normalize" the data by making the percentages consistent with their respective number of trials. That is, we give their trials weights so that the aggregate percentage would be a reflection of the same number of data points. This removes the paradox by removing the inconsistencies that it hinges on. In doing so, we have the following table:

PLAYER	1ST SESSION		2ND SESSION		TOTAL
Carlo	7/19	0.3684	(17/39)*(74/74)	0.4360	39.2564/93	0.4221
Hanz	(4/11)*(19/19)	0.3636	32/74	0.4324	38.9091/93	0.4184

CONCLUSION

On the practical side, the statistical anomaly shown by Simpson's Paradox is a good reminder to be mindful of the computed numbers. For instance, sabermetrics is a young field, and Simpson's Paradox is a good example on how to improve upon current methods.

There is a certain viewpoint from psychology that goes by the mantra "the whole is more than the sum of its parts." Looking at things from this perspective, Simpson's Paradox would be more in line with the viewpoint "the whole is not the same as its parts." That is, we can also say that there are situations wherein the sum or aggregate says something different from the individual parts.

Now that we have a clearer picture of what has happened, we go back to the story of Carlo and Hanz. The question of who won among the two is still open to debate. What about you? Do you believe that the individual sessions have more weight, or that the overall performance is the best indicator?

ABOUT THE AUTHOR:
Victor Andrew Antonio is a Lecturer at the Ateneo de Manila University. He obtained his B.S. in Mathematics at the Ateneo de Manila University in 2012 and is currently taking his M.S. in Mathematics at the same university.

REFERENCES:

[1] Bogomolny, Alexander. Simpson Paradox. Retrieved from http://www.cut-the-knot.org/Curriculum/Algebra/SimpsonParadox.shtml.

[2] Pearl, Judea. Simpson's Paradox: An Anatomy. 1999.
[3] Wagner, Clifford H. Simpson's Paradox in Real Life. The American Statistician, Vol. 36 No. 1 (Feb 1982), 46-48.

OLYMPIAD CORNER

from the Asian Pacific Mathematics Olympiad, 2011

Problem: Let \(a,b,c\) be positive integers. Prove that it is impossible to have all of the three numbers \(a^{2}+b+c,b^{2}+c+a, c^{2}+a+b\) to be perfect squares.

Solution: 

Suppose on the contrary that all three numbers are perfect squares. Since \(a^{2}+b+c\) is a perfect square larger than \(a^{2}\), it follows that \(a^{2}+b+c\geq \left( a+1\right) ^{2}\), which is equivalent to \[ b+c\geq 2a+1.\]Using the same argument, we also obtain \[ c+a\geq 2b+1\]and\[a+b\geq 2c+1.\]Combining the above inequalities, we have \[2\left( a+b+c\right) \geq 2\left( a+b+c\right) +3 \]which results in a contradiction. 

This proves that it is impossible to have all of the three numbers to be perfect squares.

PROBLEMS

10. Two circles have exactly two points \(A\) and \(B\) in common. Find a straight line \(L\) through \(A\) such that the circles cut chords of equal lengths out of \(L\). How many solutions can the problem have?
    
11. For nonnegative real numbers \(a\), \(b\), \(c\) with \(a+b+c=1\), prove that \[ \sqrt{a+\frac{\left(b-c\right)^{2}}{4}}+\sqrt{b}+\sqrt{c}\leq\sqrt{3}. \]
12. If \(2n-1\) is a prime number, then for any group of distinct positive integers \(a_{1}\), \(a_{2}\), \(\dots, a_{n}\) there exist \(i,j\in\left\{ 1,2,\dots,n\right\}\) such that \[ {\displaystyle \frac{a_{i}+a_{j}}{\left(a_{i},a_{j}\right)}}\geq2n-1. \]

We welcome readers to submit solutions to the problems posed below for publication consideration. Solutions may be submitted to the PEM facilitators on the deadline date or online via vantonio1992@gmail.com. Solutions must be preceded by the solver's name, school affiliation and year level. The deadline for submission is 12:30 PM December 7, 2013. 

SOLUTIONS

(for November 16, 2013)

7. Prove that if \(a\) and \(b\) are two sides of a triangle and \(m_{c}\) is the median drawn to the third side, then \[\left|m_{c}\right|\leq\frac{\left|a\right|+\left|b\right|}{2}.\] (Taken from Mathematics as Problem Solving by Alexander Soifer)
   (solved by Jan Kendrick Ong [Chang Kai Shek College] and Farrell Eldrian Wu [MGC New Life Academy]; partial credit for Jayson Dwight S. Catindig [Ateneo de Manila HS])
   
   
   SOLUTION: 
   Let \(O\) be the midpoint of \(AB\). Construct \(D\) on line \(\left|OC\right|\) such that \(\left|OC\right| = \left|OD\right|\).
   
   

   

   
   Then the quadrilateral \(ABCD\) is a parallelogram, and \(\left|BD\right|=\left|CA\right|=\left|b\right|\). In addition, by the Triangle Inequality, \(\left|CD\right|<\left|CB\right|+\left|BD\right|\), which means \[ \begin{eqnarray*} 2\left|m_{c}\right| & < & \left|a\right|+\left|b\right|\\ \left|m_{c}\right| & \leq & \frac{\left|a\right|+\left|b\right|}{2}. \end{eqnarray*} \]
   Now, when \(n\) is even, we can write \(n\) as \(4^km^2\), where \(m\) is odd. Then look at all pairings of divisors. There is an odd number of pairs whose sum is odd. They are \[\left(4^k\left(1\right)\cdot m^2, \dots, 4^k\left(i\right)\cdot \frac{m^2}{i}, \dots, 4^km \cdot m, \dots, 4^k\left(i\right)\cdot \frac{m^2}{i}, \dots, 4^km^2\cdot 1\right).\]So, \(\sigma\left(n\right)\) is odd since we are adding an odd number of odd numbers.
8. Prove that, for all integers \(n\geq2\),\[\frac{1}{1^{2}}+\frac{1}{2^{2}}+\cdots+\frac{1}{n^{2}}\leq2.\]
   (solved by Farrell Eldrian Wu [MGC New Life Academy]; partial credit for Jayson Dwight S. Catindig [Ateneo de Manila HS] and Jan Kendrick Ong [Chang Kai Shek College])
   
   
   SOLUTION: 
   This is equivalent to proving \[ \frac{1}{2^{2}}+\cdots+\frac{1}{n^{2}}\leq1. \] Note that \[ \frac{1}{\left(i-1\right)i}\leq\frac{1}{i^{2}} \] for all \(i\geq2\). This means that \[ \begin{eqnarray*} \frac{1}{2^{2}}+\cdots+\frac{1}{n^{2}} & \leq & \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{\left(n-1\right)n}\\ & = & \left(\frac{1}{1}-\frac{1}{2}\right) + \left(\frac{1}{2}-\frac{1}{3}\right) + \cdots + \left(\frac{1}{n-1}-\frac{1}{n}\right) \\ & = & 1-\frac{1}{n}\\ & = & \frac{n-1}{n}\\ \frac{1}{2^{2}}+\cdots+\frac{1}{n^{2}} & < & 1. \end{eqnarray*} \]
9. Find all solutions of \(2^{n}+7=x^{2}\) where \(n\) and \(x\) are integers. (Taken from International Mathematics: Tournament of the Towns, Questions, and Solutions,Tournaments 6 to 10 (1984 to 1988) by P. J. Taylor)
   (solved by Jayson Dwight S. Catindig [Ateneo de Manila HS] and Farrell Eldrian Wu [MGC New Life Academy]; partial credit for Marielle Macasaet [St. Theresa's College] and Jan Kendrick Ong [Chang Kai Shek College])
   
   
   SOLUTION: 
   Since we are required to make \(x\) an integer, then \(2^{n}\) must be an integer, and so \(n\geq0\). We consider all possible cases when we rewrite the equation modulo \(4\).
   
   If \(n>1\), then \(2^{n}\equiv0\mod4\). On the other hand, if \(n=1\), then \(2^{n}\equiv2\mod4\), and finally, if \(n=0\), then \(2^{n}\equiv1\mod4\). We look at those three possibilities and deduce solutions for \(2^{n}+7\equiv x^{2}\mod4\). Note that the only possible quadratic residues modulo \(4\) are \(0\) and \(1\).
   

   • CASE 1: \(n>1\).
     We start with \[ \begin{eqnarray*} x^{2} & \equiv & \left(2^{n}+7\right)\mod4\\ & \equiv & 3\mod4, \end{eqnarray*} \]which is impossible.
   • CASE 2: \(n=1\). \[ \begin{eqnarray*} x^{2} & \equiv & \left(2^{n}+7\right)\mod4\\ & \equiv & 1\mod4, \end{eqnarray*} \]which is acceptable. Going back to the original equation, it can be seen that \(x=\pm3\).
   • CASE 3: \(n=0\). \[ \begin{eqnarray*} x^{2} & \equiv & \left(2^{n}+7\right)\mod4\\ & \equiv & 0\mod4, \end{eqnarray*} \]which is also acceptable. But this means \(x=\pm2\sqrt{2}\), which are not integers.

   Therefore, the only solution is \(n=1\), \(x=\pm3\).

ERRATA

The right-hand side of Problem #11 was changed from "\(3\)" to "\(\sqrt{3}\)."