Uncategorized – Page 5 – Numerical Methods Guy

Abuses of regression

There are three common abuses of regression analysis.

Extrapolation
Generalization
Causation.

Extrapolation

If you were dealing in the stock market or even interested in it, we remember the stock market crash of March 2000. During 1997-1999, many investors thought they would double their money every year, started buying fancy cars and houses on credit, and living the high life. Little did they know that the whole market was hyped on speculation and little economic sense? Enron and MCI financial fiascos were soon to follow.

Let us look if we could have safely extrapolated NASDAQ index from past years. Below is the table of NASDAQ index, S as a function of end of year number, t (Year 1 is the end of year 1994, and Year 6 is the end of year 1999).

Table 1 NASDAQ index as a function of year number.

Year Number (t)	NASDAQ Index (S)
1 (1994)	752
2 (1995)	1052
3 (1996)	1291
4 (1997)	1570
5 (1998)	2193
6 (1999)	4069

A relationship S = a₀+a₁t+a₂t²between the NASDAQ index, S and the year number, t is developed using least square regression and is found to be

S=168.14t² – 597.35t + 1361.8

The data is given for Years 1 thru 6 and it is desired to calculate the value for t>=6. This is extrapolation outside the model data. The error inherent in this model is shown in Table 2. Look at the Year 7 and 8 that was not included in the regression data – the error between the predicted and actual values is 119% and 277%, respectively.

Table 2 NASDAQ index as a function of year number.

Year Number (t)	NASDAQ Index (S)	Predicted Index	Absolute Relative True Error (%)
1 (1994)	752	933	24
2 (1995)	1052	840	20
3 (1996)	1291	1082	16
4 (1997)	1570	1663	6
5 (1998)	2193	2578	18
6 (1999)	4069	3831	6
7 (2000)	2471	5419	119
8 (2001)	1951	7344	277

This illustration is not exaggerated and it is important that a careful use of any given model equations is always called for. At all times, it is imperative to infer the domain of independent variables for which a given equation is valid.

Generalization

Generalization could arise when unsupported or overexaggerated claims are made. It is not often possible to measure all predictor variables relevant in a study. For example, a study carried out about the behavior of men might have inadvertently restricted the survey to Caucasian men. Shall we then generalize the result as the attributes of all men irrespective of race? Such use of regression equation is an abuse since the limitations imposed by the data restrict the use of the prediction equations to Caucasian men.

Misidentification

Finally, misidentification of causation is a classic abuse of regression analysis equations. Regression analysis can only aid in the confirmation or refutation of a causal model ‑ the model must however have a theoretical basis. In a chemical reacting system in which two species react to form a product, the amount of product formed or amount of reacting species vary with time. Although a regression equation of species concentration and time can be obtained, one cannot attribute time as the causal agent for the varying species concentration. Regression analysis cannot prove causality; rather they can only substantiate or contradict causal assumptions. Anything outside this is an abuse of the use of regression analysis method.

This post used textbook notes written by the author and Egwu Kalu, Professor of Chemical and Biomedical Engineering, FAMU, Tallahassee, FL.

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How do you know that the least squares regression line is unique and corresponds to a minimum

We already know that using the criterion of either

minimizing sum of residuals OR
minimizing sum of the absolute value of residuals

is BAD as either of the criteria do not give a unique line. Visit these notes for an example where these criteria are shown to be inadequate.

So we use minimizing the sum of the squares of the residuals as the criterion. How can we show that this criterion gives a unique line?

The proof is given below as image files because the proof is equation intensive. I made a better resolution pdf file also.

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Finding the optimum polynomial order to use for regression

Many a times, you may not have the privilege or knowledge of the physics of the problem to dictate the type of regression model. You may want to fit the data to a polynomial. But then how do you choose what order of polynomial to use.

Do you choose based on the polynomial order for which the sum of the squares of the residuals, Sr is a minimum? If that were the case, we can always get Sr=0 if the polynomial order chosen is one less than the number of data points. In fact, it would be an exact match.

So what do we do? We choose the degree of polynomial for which the variance as computed by

Sr(m)/(n-m-1)

is a minimum or when there is no significant decrease in its value as the degree of polynomial is increased. In the above formula,

Sr(m) = sum of the square of the residuals for the mth order polynomial

n= number of data points

m=order of polynomial (so m+1 is the number of constants of the model)

Let’s look at an example where the coefficient of thermal expansion is given for a typical steel as a function of temperature. We want to relate the two using polynomial regression.

Temperature	Instantaneous Thermal Expansion
^oF	1E-06 in/(in ^oF)
80	6.47
40	6.24
0	6.00
-40	5.72
-80	5.43
-120	5.09
-160	4.72
-200	4.30
-240	3.83
-280	3.33
-320	2.76

If a first order polynomial is chosen, we get

$latex alpha=0.009147T+5.999$, with Sr=0.3138.

If a second order polynomial is chosen, we get

$latex alpha=-0.00001189T^2+0.006292T+6.015$ with Sr=0.003047.

Below is the table for the order of polynomial, the Sr value and the variance value, Sr(m)/(n-m-1)

Order of polynomial, m	Sr(m)	Sr(m)/(n-m-1)
1	0.3138	0.03486
2	0.003047	0.0003808
3	0.0001916	0.000027371
4	0.0001566	0.0000261
5	0.0001541	0.00003082
6	0.0001300	0.000325

So what order of polynomial would you choose?

From the above table, and the figure below, it looks like the second or third order polynomial would be a good choice as very little change is taking place in the value of the variance after m=2.

Optimum order of polynomial for regression

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Data for aluminum cylinder in iced water experiment

A colleague asked me what if he did not have time or resources to do the experiments that have been developed at University of South Florida (USF) for numerical methods. He asked if I could share the data taken at USF.

Why not – here is the data for the experiment where an aluminum cylinder is placed in iced water. This link also has the exercises that the students were asked to do.

The temperature vs time data is as follows: (0,23.3), (5,16.3), (10,13), (15,11.8), (20,11), (25,10.7), (30,9.6), (35,8.9), (40,8.4). Time is in seconds and temperature in Celcius. Other data needed is

Ambient temperature of iced water = 1.1^oC

Diameter of cylinder = 44.57 mm

Length of cylinder = 105.47 mm

Density of aluminum = 2700 kg/m³

Specific heat of aluminum = 901 J/(kg-^oC)

Thermal conductivity of aluminum = 240 W/(m-K)

Table 1. Coefficient of thermal expansion vs. temperature for aluminum (Data taken from http://www.llnl.gov/tid/lof/documents/pdf/322526.pdf by using mid values of temperatures at which CTE is reported)

Temperature (^oC)	Coefficient of thermal expansion (μm/m/^oC)
-10	58
12.5	59
37.5	60
62.5	62
87.5	66
112.5	71

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In regression, when is coefficient of determination zero

The coefficient of determination is a measure of how much of the original uncertainty in the data is explained by the regression model.

The coefficient of determination, $latex r^2$ is defined as

$latex r^2$=$latex \frac{S_t-S_r}{S_r}$

where

$latex S_t$ = sum of the square of the differences between the y values and the average value of y

$latex S_r$ = sum of the square of the residuals, the residual being the difference between the observed and predicted values from the regression curve.

The coefficient of determination varies between 0 and 1. The value of the coefficient of determination of zero means that no benefit is gained by doing regression. When can that be?

One case comes to mind right away – what if you have only one data point. For example, if I have only one student in my class and the class average is 80, I know just from the average of the class that the student’s score is 80. By regressing student score to the number of hours studied or to his GPA or to his gender would not be of any benefit. In this case, the value of the coefficient of determination is zero.

What if we have more than one data point? Is it possible to get the coefficient of determination to be zero?

The answer is yes. Look at the following data pairs (1,3), (3,-2), (5,4), (7,-5), (9,4.2), (11,3), (2,4). If one regresses this data to a general straight line

y=a+bx,

one gets the regression line to be

y=1.6

When is rsquared zero?

In fact, 1.6 is the average value of the given y values. Is this a coincidence? Because the regression line is the average of the y values, $latex S_t=S_r$, implying $latex r^2=0$

QUESTIONS

Given (1,3), (3,-2), (5,4), (7,a), (9,4.2), find the value of a that gives the coefficient of determination, $latex r^2=0$. Hint: Write the expression for $latex S_r$ for the regression line $latex y=mx+c$. We now have three unknowns, m, c and a. The three equations then are $latex \frac{\partial S_r} {\partial m} =0$, $latex \frac{\partial S_r} {\partial c} =0$ and $latex S_t=S_r$.
Show that if n data pairs $latex (x_1,y_1)……(x_n,y_n)$ are regressed to a straight line, and the regression straight line turns out to be a constant line, then the equation of the constant line is always y=average value of the y-values.

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Length of a curve experiment

In a previous post, I mentioned that I have incorporated experiments in my Numerical Methods course. Here I will discuss the second experiment.

Length of the curve experiment In this experiment, we find the length of two curves generated from the same points – one curve is a polynomial interpolant and another one is a spline interpolant.

Motivation behind the experiment: In 1901, Runge conducted a numerical experiment to show that higher order interpolation is a bad idea. It was shown that as you use higher order interpolants to approximate f(x)=1/(1+25x²) in [-1,1], the differences between the original function and the interpolants becomes worse. This concept also becomes the basis why we use splines rather than polynomial interpolation to find smooth paths to travel through several discrete points.

What do students do in the lab: A flexible curve (see Figure) of length 12″ made of lead-core construction with graduations in both millimeters and inches is provided. The student needs to draw a curve similar in shape to the Runge’s curve on the provided graphing paper as shown. It just needs to be similar in shape – the student can make the x-domain shorter and the maximum y-value larger or vice-versa. The student just needs to make sure that there is a one-to-one correspondence of values.

Assigned Exercises: Use MATLAB to solve problems (3 thru 6). Use comments, display commands and fprintf statements, sensible variable names and units to explain your work. Staple all the work in the following sequence.

Signed typed affidavit sheet.
Attach the plot you drew in the class. Choose several points (at least nine – do not need to be symmetric) along the curve, including the end points. Write out the co-ordinates on the graphing paper curve as shown in the figure.
Find the polynomial interpolant that curve fits the data. Output the coefficients of the polynomial.
Find the cubic spline interpolant that curve fits the data. Just show the work in the mfile.
Illustrate and show the individual points, polynomial and cubic spline interpolants on a single plot.
Find the length of the two interpolants – the polynomial and the spline interpolant. Calculate the relative difference between the length of each interpolant and the actual length of the flexible curve.
In 100-200 words, type out your conclusions using a word processor. Any formulas should be shown using an equation editor. Any sketches need to be drawn using a drawing software such as Word Drawing. Any plots can be imported from MATLAB.

Where to buy the items for the experiment:

Flexible curves – I bought these via internet at Art City. The brand name is Alvin Tru-Flex Graduated Flexible Curves. Prices range from $5 to $12. Shipping and handling is extra – approximately $6 plus 6% of the price. You may want to buy several 12″ and 16″ flexible curves. I had to send a query to the vendor when I did not receive them within a couple of weeks. Alternatively, call your local Art Store and see if they have them.
Engineering Graph Paper – Staples or Office Depot. Costs about $12 for a pack for 100-sheet pad.
Pencil – Anywhere – My favorite store is the 24-hour Wal-Mart Superstore. $1 for a dozen.
Scale – Anywhere – My favorite store is the 24-hour Wal-Mart Superstore. $1 per unit.

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A legend used in the movie “The Happening”

Well M. Night Shyamalan may have made another disappointing movie – The Happening, but I somewhat liked it. I would give it a grade of B.

In the movie, John Leguzomo’s character, a math teacher, is distracting his fellow panicking passenger in the Jeep with a mathematical question. The question he asks her is if he gave her a penny on Day 1 of the month, two pennies on Day 2 of the month, four pennies on Day 3 of the month, and so on, how much would money would she have after a month. She shouts $300 or some odd number like that. But, do you know that the amount is actually more than a 10 million dollars (Thanks to a student who mentioned that it was a penny that John offered on the first day, not a dollar – sometimes I do feel generous).

This question is based on a story from India and it goes as follows.

King Shriham of India wanted to reward his grand minister Ben for inventing the game of chess. When asked what reward he wanted, Ben asked for 1 grain of rice on the first square of the board, 2 on the second square of the board, 4 on the third square of the board, 8 on the fourth square of the board, and so on till all the 64 squares were covered. That is, he was doubling the number of grains on each successive square of the board. Although Ben’s request looked less than modest, King Shriham quickly found that the amount of rice that Ben was asking for was humongous.

QUESTIONS:

Write a MATLAB (you can use any other programming language) program for the following using the for or while loop.

Find out how many grains of rice Ben was asking for.
If the mass of a grain of rice is 2 mg, and the world production of rice in recent years has been approximately 600,000,000 tons (1 ton=1000 kg), how many times the modern world production was Ben’s request?
Do the inverse problem – find out how many squares are covered if the the number of grains on the chess board are given to you. For example, how many squares will be covered if the number of grains on the chess board are 16?

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Shortest path for a robot

Imagine a robot that had to go from point to point consecutively (based on x values) on a two dimensional x-y plane. The shortest path in this case would simply be drawing linear splines thru consecutive data. What if the path is demanded to be smooth? Then what!

Well one may use polynomial or quadratic/cubic spline interpolation to develop the path. Which path would be shorter? To find out thru an anecdotal example, click here.

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Do quadratic splines really use all the data points?

There are two reasons that linear splines are seldom used

Each spline uses information only from two consecutive data points
The slope of the splines is discontinuous at the interior data points

The answer to resolving the above concerns are to use higher order splines such as quadratic splines. Read the quadratic spline textbook notes before you go any further. You do want what I have to say to make sense to you.

In quadratic splines, a quadratic polynomial is assumed to go thru consecutive data points. So you cannot just find the three constants of each quadratic polynomial spline by using the information that the spline goes thru two consecutive points (that sets up two equations and three unknowns). Hence, we incorporate that the splines have a continuous slope at the interior points.

So does all this incorporation make the splines to depend on the values of all given data points. It does not seem so.

For example, in quadratic splines you have to assume that the first or last spline is linear. For argument sake, let that be the first spline. If the first spline is linear, then we can find the constants of the linear spline just by the knowledge of the value of the first two data points. So now we know that we can set up three equations for the three unknown constants of the second spline as follows

the slope of the first spline at the 2nd data point and the slope of the second spline at the 2nd point are the same,
the second spline goes thru the 2nd data point
the second spline goes thru the 3rd data point

That itself is enough information to find the three constants of the second spline. We can keep using the same argument for all the other splines.

So the mth spline constants are dependent on data from the data points 1, 2, .., m, m+1 but not beyond that.

Can you now make the call on the kind of dependence or independence you have in the constants of the quadratic splines?

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Extrapolation is inexact and may be dangerous

The NASDAQ was booming – people were dreaming of riches – early retirement and what not. The year was 1999 and NASDAQ was at an all time high of 4069 on the last day of 1999.

Yes, Prince was right, not just about the purple rain, but – “‘Cuz they say two thousand zero zero party over, Oops out of time, So tonight I’m gonna party like it’s 1999 party like 1999.”

But as we know the party did not last too long. The dot com bubble burst and as we know it today (June 2008), the NASDAQ is hovering around 2400.

Year ………………NASDAQ on December 31st

1994………………………… 751

1995 ……………………….1052

1996 ………………………..1291

1997 ………………………..1570

1998 ………………………..2192

1999 ………………………..4069

• End of Year NASDAQ Composite Data taken from www.bigcharts.com

So how about extrapolating the value of NASDAQ to not too far ahead – just to the end of 2000 and 2001. This is what you obtain from using a 5th order interpolant for approximation from the above six values.

End of Year …Actual Value …..5th Order Poly Extrapo……………Abs Rel True Error
2000 ……………..2471…………………. 9128 ………………………………….. 269%
2001………………1950……………….. 20720 ………………………………….. 962%

Do you know what would be the extrapolated value of NASDAQ on June 30, 2008 -a whopping 861391! On June 30, 2008, compare it with the actual value.

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