This document provides an overview of least-squares regression techniques including: - Simple linear regression to fit a line to data - Polynomial regression to fit higher order curves - Multiple regression to fit surfaces using two or more variables It discusses calculating regression coefficients, quantifying errors, and performing statistical analysis of the regression results including determining confidence intervals. Examples are provided to demonstrate applying these techniques.

1. Part 5a:
LEAST-SQUARES REGRESSION
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Simple Linear Regression
Polynomial Regression
Multiple Regression
Statistical Analysis of L-S Theory
Non-Linear Regression

2. Introduction:
Consider the falling object in air problem:
t0
m
v0
t1
m
“Best fit”
v1
v
t
tn
m
vn
 (t) values considered to be error-free.
 Every measurement of (v) contain some error.
 Assume error in (v) are normally distributed
(random error).
 Find “best fit” curve to represent v(t)

3. Simple Linear Regression
 Consider a set of n scattered data
 Find a line that “best fits” the scattered data
y
a0
a0= intercept
a1= slope
a1 x
 There are a number of ways to define the “best fit” line. However
we want to find one that is unique, i.e., for a particular set of data.
 A uniquely defined best-fit line can be found by minimizing the
sum of the square of the residuals from each data point:
n
Sr
n
( ymeas
i 1
y fit ) 2
( yi
a0 a1 xi ) 2
i 1
Find a0 and a1 that minimizes Sr (least-square)
sum of the square
of the residuals
(or spread)

4. To minimize Sr (a0 , a1), differentiate and set to zero:
n
Sr
a0
2
( yi
a0
a1 xi )
0
[( yi
a0
a1 xi ) xi ] 0
i 1
n
Sr
a1
2
i 1
or
0
na0
xi a0
yi
a0
xi a1
xi2 a1
a1xi
yi
0
yi xi
a0 xi
a1 xi2
Normal equations for
simple linear L-S regression
xi yi
Need to solve these simultaneous equations for the unknowns a0
and a1

5. Solution for a1 and a0 gives
n
a1
xi yi
n
xi
2
i
x
xi
yi
2
and
yi
a0
n
xi
a1
n
y a1 x
EX: Find linear fit for the set of measurements:
x
1
0.5
2
2.5
3
2.0
4
4.0
5
3.5
6
6.0
7
y
y
5.5
0.0714 0.839x
n 7
xi
28
x 4
xi yi 119.5
a1
a0
yi
y
3.4286
xi2 140
7(119 .5) 2(24 )
7(140 ) (28 ) 2
3.4286
24
0.839 (4)
0.839
0.0714

6. Quantification of Error:
Sum of the square of the
residuals for the mean
2
n
St
sum of the square of the
residuals for the linear
regression
( yi
y)
Sr
( yi
a0 a1 xi )
i 1
i 1
standard
deviation
sy
2
n
St
n 1
standard error of
the L-S estimate
sy/ x
Sr
n 2
All these
approaches are
based on the
assumptions:
x > error-free
y > normal error

7. “Coefficient of determination” is defined as
r
St
2
Sr
r
St
Sr =0 (r=1)
Sr =St (r=0)
“correlation coefficient”
Perfect fit
No improvement by fitting the line
Alternative formulation for the correlation coefficient
n
r
n
2
i
x
xi yi
xi
xi
2
n
yi
2
i
y
yi
Note: r 1 does not always necessarily mean
that the fit is “good”. You should always plot
the data along with the regression curve to
see the goodness of the fit.
2
Four set of data with same r=0.816

8. Linearization of non-linear relationships:
 Many engineering applications involve non-linear
relationships, e.g., exponential, power law, or saturated growth
rate.
exponential
power-law
saturated growth-rate
y
a1e b1x
y
a2 x b2
y
a3
x
b3
x
 These relationships can be linearized by some mathematical
operations:
ln y
ln a1 b1 x
log y
b2 log x log a2
1
y
b3 1
a3 x
 Linear L-S fit can be applied to find the coefficients.
1
a3

9. EX: Fit a power law relationship to the following dataset:
x
y
1
0.5
2
1.7
3
3.4
4
5.7
5
8.4
Power law model
y
log y
a2 x b2
b2 log x log a2
(find a2 and b2)
Calculate logarithm of both data:
log x
log y
0
-0.301
0.301
0.226
0.477
0.534
0.602
0.753
0.699
0.922
Applying simple linear regression gives;
slope=1.75 and intercept=-0.300
b2
1.75 log a2
y
0.5 x1.75
0.300
a2
0.5

10. Polynomial Regression
 In some cases, we may want to fit our data to a curve rather than a
line. We can then apply polynomial regression (In fact, linear
regression is nothing but an n=1 polynomial regression).
Data to fit to a second order polynomial:
y
a0
a1 x a2 x 2
Sum of the square of the residuals (spread)
2
n
Sr
( yi ,obs
yi , fit )
i 1
2
n
( yi a0 a1 xi
a2 xi2 )
i 1
To minimize Sr(a0, a1, a2), take derivatives and equate to zero:
Sr
a0
n
2
( yi
i 1
a0
a1 xi
a2 xi2 )
0

11. Sr
a1
Sr
a2
n
xi ( yi
a0
a1 xi
a2 xi2 )
0
xi2 ( yi
2
a0
a1 xi
a2 xi2 )
0
i 1
n
2
i 1
Three linear equations with three unknowns a0, a1, a2 :
(n)a0
xi a1
xi2 a2
yi
xi a0
xi2 a1
xi3 a2
xi yi
“normal equations”
xi2 a0
xi3 a1
xi4 a2
xi2 yi
all summations are i=1..n
 This set of equations can be solved by any linear solution
techniques (e.g., Gauss elimination, LU Dec., Cholesky Dec., etc.)

12.  The approach can be generalized to order (m) polynomial following
the same way. Now, the fit function becomes
y
a1 x a2 x 2 .. am x m
a0
 This will require the solution of an order (m+1) system of linear
equations. The standard error becomes
Because (m+1) degrees of
freedom was lost from data
of (n) due to extraction of
(m+1) coefficients .
Sr
n (m 1)
sy / x
EX 17.5: Fit an 2nd order polynomial to the following data x
i
yi 152.6
m 2
2
i
x
55
xi yi
xi2 yi
3
i
x
225
2488.8
4
i
x
979
2.1
7.7
2
13.6
3
585.6
0
1
xi 15
n 6
yi
27.2
4
40.9
5
61.1

13. System of linear equations:
6
15
55
a0
152.6
15
55
225 a1
585.6
55 225 979 a2
2488.8
We get
a0
2.47857
a1
2.35929
a2
1.86071
Then, the fit function:
y
2.47857
2.35929 x 1.86071 x 2
Standard error:
Sr
n (m 1)
sy / x
where
3.74657
1.12
6 (3)
2
6
Sr
( yi
i 1
2.47857 2.35929xi 1.86071xi2 )
3.74657

14. Multiple Linear Regression
 In some cases, data may
have two or more
independent variables. In
this example, for a
function of two
x 2 variables, the linear
regression gives a planar fit
function.
y ( x1 , x2 )
x1
Function to fit
y
a0
a1 x1 a2 x2
Sum of the square of the residuals (spread)
2
n
Sr
( yi ,obs
i 1
2
n
yi , fit )
( yi a0 a1 x1i a2 x2i )
i 1

15. Minimizing the spread function gives:
n
Sr
a0
2
( yi
a0
a1 x1i
a 2 x2 i )
0
i 1
n
Sr
a1
2
x1i ( yi
a0
a1 x1i
a 2 x2 i )
0
x2 i ( yi
a0
a1 x1i
a 2 x2 i )
0
i 1
n
Sr
a2
2
i 1
The system of equations to be solved:
n
x1i
2
1i
x1i
x
x2 i
x1i x2i
x2 i
a0
yi
x1i x2i
a1
x1i y1i
2
x2 i
a2
x 2 i yi
Normal equations
for multiple linear
regression

16. EX 17.7: Fit a planar surface to the following data
x1
x2
y
0
0
5
2
1
10
2.5
2
9
1
3
0
4
6
3
7
2
27
We first do the following calculations:
y
x1
x2
x1x1
x2x2
x1x2
x1y
x2y
5
0
5
0
0
0
0
0
10
1
10
4
1
2
20
10
9
2
9
6.25
4
5
22.5
18
0
3
0
1
9
3
0
0
3
6
3
16
36
24
12
18
27
2
27
49
4
14
189
54
54
16.5
14
76.25
54
48
243.5
100

17. The system of equations to calculate the fit coefficients:
6
16.5
14 a0
16.5 76.25 48 a1
14
48
54 a2
54
243.5
100
returns
a0
a1
5
The fit function
y
4
a2
3
5 4 x1 3x2
 For the general case of a function of m-variables, the same
strategy can applied. The fit function in this case:
y a0 a1 x1 a2 x2 .. am xm
Standard error:
sy / x
Sr
n (m 1)

18.  A useful application of multiple regression is for fitting a power
law equation of multiple variables of the form:
y
a
a
a0 x1a1 x2 2 .. xmm
Linearization of this equation gives
log y
log a0
a1 log x1 ... am log xm
 The coefficients in the last equation can be calculated using
multiple linear regression, and can be substituted to the original
power law equation.

19. Generalization of L-S Regression:
 In the most general form, L-S regression can be stated as
y
a0 z0
a1 z1 ... am zm
In general, this form is called
“linear regression” as the
fitting coefficients are
linearly dependant on the fit
function.
functions
z0
x 0 , z1
z0
1 , z1
x1 , ..., z m
x1 , ..., zm
xm
xm
Polynomial regression
Multiple regression
 Other functions can be defined for fitting as well, e.g.,
y
a0
a1 cos t a2 sin t

20. For a particular data point
y
a0 z0
a1 z1 ... am zm
e
data
For n data (in matrix form):
y
z10
Z
Z a
e
y1
y2
...
yn
coefficients
a0
a1
...
am
residuals
z11 ... z1m
...
...
zn 0
y
a
e
Calculated based on the
measured independant
variables
zn1
znm
m: order of the fit function
n: number of data points
Z is generally not a square matrix.
n m 1
e1
e2
...
en

21. Sum of the square of the residuals:
n
2
m
Sr
( yi
i 1
a j z ji )
j 0
To determine the fit coefficients, minimize
S r (a0 , a1 ,.., am )
This is equivalent to the following:
Z
T
Z a
Z
T
y
Normal equations
for the general L-S
regression
 This is the general representation of the normal equations for L-S
regression including simple linear, polynomial, and multiple linear
regression methods.

22. Solution approaches:
Z
T
Z a
Z
T
y
A symmetric and square
matrix of size [m+1 , m+1]
 Elimination methods are best suited for the solution of the above
linear system:
LU Decomposition / Gauss Elimination
Cholesky Decomposition
 Especially, Cholesky decomposition is fast and requires less
storage. Furthermore,
 Cholesky decomposition is very appropriate when the order of
the polynomial fit model (m) is not known beforehand.
Successive higher order models can be efficiently developed.
 Similarly, increasing the number of variables in multiple
regression is very efficient using Cholesky decomposition.

23. Statistical Analysis of L-S Theory
Some definitions:
 If a histogram of the data shows a
bell shape curve, normally
distributed data.
 This has a well-defined statistics
n
yi
y
sy
2
sy
mean
i 1
n
yi
n 1
St
n 1
y
2
Standard
deviation
variance
 For a perfectly normal distribution:
mean±std fall about 68% of the total data.
mean±2std fall about 95% of the total data.
: true mean
: true std

24. Confidence intervals:
 Confidence interval estimates intervals within which the
parameter is expected to fall, with a certain degree of confidence.
 Find L and U values such that
PL
U
1
true mean
significance level
For 95% confidence interval
=0.05
L
U
y
y
sy
n
sy
n
t
t
/ 2,n 1
/ 2,n 2
t-distribution (tabulated in
books); in EXCEL tinv ( ,n)
e.g., for =0.05 and n=20
t /2, n-1=2.086
 T-distribution is used to compramize between a
perfect and an imperfect estimate. For example, if
data is few (small n), t-value becomes larger, hence
giving a more conservative interval of confidence.

25. EX: Some measurements of coefficient of thermal expansion of steel (x10-6 1/°F):
6.495
6.665
6.755
6.565
6.595
6.505
6.625
6.515
6.615
6.435
6.715
6.555
6.635
6.625
6.575
6.395
6.485
6.715
6.655
6.775
6.555
6.655
6.605
6.685
n=8
n=16
n=24
Find the mean and corresponding 95% confidence intervals for the
a) first 8 measurements b) first 16 measurements c) all 24 measurements.
For n=8
L
y
U
y
y
sy
n
sy
n
6.59
t
t
sy
0.089921
t
/ 2,n 1
t0.05 / 2,8
/ 2,n 1
6.59
0.089921
2.364623
8
/ 2,n 2
0.089921
2.364623
8
6.6652
2.364623
6.5148
6.59
1
6.5148
6.6652
For eight measurements,
there is a 95% probability
that true mean falls
between these values.

26. The cases of n=16 and n=24 can be performed in a similar fashion. Hence we
obtain:
n
mean(y)
8
6.5900
16
24
sy
t
L
U
0.089921 2.364623
6.5148
6.6652
6.5794
0.095845 2.131451
6.5283
6.6304
6.6000
0.097133 2.068655
6.5590
6.6410
/2,n-1
Results shows that confidence interval narrows down as the number of
measurements increases (even though sy increases by increasing n!).
For n=24 we have 95% confidence that true mean is between 6.5590 and 6.6410.

27. Confidence Interval for L-S regression:
 Using matrix inverse for the solution of (a) is inefficient:
a
Z
T
Z
1
Z
T
y
 However, inverse matrix carries useful statistical information
about the goodness of the fit.
Z
T
Z
1
Inverse matrix
Diagonal terms
coefficients
variances (var) of the fit
Off -diagonal terms
the fit coefficients
covariances (cov) of
2
var(ai 1 ) uii s y / x
cov(ai 1, a j ) ui
2
sy / x
1, j
uij: Elements of the inverse matrix
 These statistics allow calculation of confidence intervals for the
fit coefficients.

28.  Calculating confidence intervals for simple linear regression:
y
a0
a1 x
For the intercept (a0)
L
a0 t
/ 2,n 2
s ( a0 )
U
a0 t
/ 2,n 2
s ( a0 )
For the slope (a1)
L
a1 t
U
a1 t
/ 2,n 2
s (a1 )
/ 2,n 2
s (a1 )
Standard error for the coefficient
(extracted from the inverse matrix)
s(ai )
var(ai )

29. EX 17.8: Compare results of measured versus model data shown below.
a) Plot the measured versus model values.
b) Apply simple linear regression formula to see the adequacy of the measured
versus model data.
c) Recompute regression using matrix approach, estimate standard error of the
estimation and for the fit parameters, and develop confidence intervals.
a)
60
Model
value
8.953
16.405
22.607
27.769
32.065
35.641
38.617
41.095
43.156
44.872
46.301
47.49
48.479
49.303
49.988
50
40
model
Measured
Value
10
16.3
23
27.5
31
35.6
39
41.5
42.9
45
46
45.5
46
49
50
30
20
10
0
0
20
40
60
measured
b) Applying simple linear regression formula gives
y
0.859 1.032x
x: measured
y: model

30. c) For the statistical analysis, first form the following [Z] matrix and (y) vector
1
Z
Then,
10
8.953
1 16.3
.. ..
..
1
16.405
..
y
..
50
Z
..
49.988
T
T
Z a
Z
548.3
a0
552.741
548.3 22191.21 a1
22421.43
15
y
Solution using the matrix inversion
a
a0
a1
0.688414
Z
T
Z
1
Z
0.01701
T
y
552.741
0.85872
0.01701 0.000465 22421.43
1.031592

31. Standard error for the fit function:
Sr
n 2
sy / x
0.863403
Standard error for the coefficients:
s(a0 )
2
u11s y / x
0.688414(0.863403) 2
0.716372
s(a1 )
2
u22 s y / x
0.000465(0.863403) 2
0.018625
For a 95% confidence interval ( =0.05, n=13, Excel returns inv(0.05,13)=2.160368)
a0
a0 t
/ 2, n 2
s(a0 )
0.85872 2.160368(0.716372)
0.85872 1.547627
a1
a1 t
/ 2, n 2
s(a1 ) 1.031592 2.160368(0.018625)
1.031592 0.040237
Desired values of slope=1 and intercept=0 falls in the intervals (hence we can
conclude that a good fit exist between measured and model values).

32. Non-linear Regression
 In some cases we must fit a non-linear model to the data, e.g.,
y
a0 (1 e
a1 x
)
parameters a0 and a1
are not linearly
dependant on y
 Generalized L-S formulation cannot be used for such models.
 Same approach of using sum of square of the residuals are
applied, but the solution is sought iteratively.
Gauss-Newton method:
 A Taylor series expansion is used to (approximately) linearize the
model. Then standard L-S theory can be applied to estimate the
improved estimates of the fit parameters.
In most general form
y
f ( x; a0 , a1 ,..am )

33. Taylor series around the fit parameters
f ( xi ) j
f ( xi ) j
f ( xi )
1
a0
f ( xi ) j
a0
a1
i: i-th data point
j: iteration number
a1
Then
ymeas
f ( xi ) j
y fit
a0
a0
f ( xi ) j
a1
a1
In matrix form:
d
Zj
iteration number
d
a
y1
f ( x1 )
y2
f ( x2 )
...
yn
f ( xn )
Zj
f1
a0
f2
a0
...
fn
a0
f1
a0
f2
a0
...
fn
a0
a
a0
a1

34. Applying the generalized L-S formula
Zj
T
Zj
a
Zj
T
d
 We solve the above system for ( A) for improved values of
parameters:
a0 , j
1
a0 , j
a0
a1, j
1
a1, j
a1
 The procedure is iterated until an acceptable error:
a0 , j
a 0
1
a0 , j
a0 , j
a1, j
a 1
1
1
a1, j
a1, j
1