GG 313 Lecture 17
10/20/05
Chapter 4
Single an Multiple Regression
Line Fit with Errors in Y
Mid-term review
Cruise Plans
Last pre-cruise meeting tomorrow at
1:30 in POST 702 (?).
Who needs a ride? I can take three
from UH.
Departure time is still uncertain, but,
unless you hear from me or Garrett
again before then, we should all be at
the pier by 0730 Monday morning.
Least-square line fitting with errors in y-values.
In the last chapter we developed a method for determining
the least-squares fit to a line from over determined data data where we have some degrees of freedom, and our
data values, n, are larger than the number of values to be
solved for, m.
If we have possible errors in the y-values, we should be
able to obtain an estimate of the uncertainty in our best-fit
line.
Paul changes his notation: In Ch. 3, he used y=a1+a2x, in
Ch. 4 he uses y=a-bx, so be careful.
We set up a chi2 function as follows:
2


yi  ai  bxi
2
 (a,b)  

i

i1 
n
(4.1)
This equation is similar to equation 3.94, the function
 E, except that the observed-predicted values are now
divided by the uncertainty in each y value. Note that
i is NOT the standard deviation, but the individual
possible errors of the yi.
Yi are the observed values and a+bxi is the striaght
line being fit to the data.
What we want to do is find the best values of a and b
AND find the possible errors in a and b.
As before, we want values of a and b that minimize the
value of 2. Thus, we want to find the values where the
slope of 2, the first derivative with respect to a and b,
equals zero.

yi  a  bxi
 0  2
2
a
i
i1
2
n
n 
 2
yi  a  bxi 
 0  2
xi
2
b
i

i1 
(4.2)
We now set up several sums corresponding to the
different terms in eqn. 4.2, similar to what was done in
eqn. 3.100:
S 
1

2
i
, Sx  
xi

2
i
, Sy  
yi

2
i
, Sxx  
xi2

2
i
, S xy  
xi yi
 i2
We can now re-write eqns. 4.2:
aS  bS x  S y
aS x  bS xx  S xy
(4.4)
solving for a and b, and substituting,
S y  bS x
S xy  aS x
a
, b
S
S xx
S y  bS x
S xy 
S x SS  S S
xy
y x
S
b

2
S xx
SS xx  S x
(4.3)
Similarly,
S xy  aS x
Sy 
Sx
2
S y Sxx  Sxy S x  aS x S yS xx  Sxy S x
S xx
a


S
SSxx
SS xx  S x2
letting
  SS xx  S x2
We obtain:

(4.5)
Sy Sxx  Sxy Sx
a

SSxy  Sy Sx
b

(4.6)
This looks different from the equations obtained in
chapter 3 where there was no error in y considered. If the
I are constant, then (4.2) becomes:
n
 2
yi  a  bxi n
 0  2
  yi  a  bxi 
2
a
i
i1
i1
n



yi  a  bxi
 0  2 
xi   yi  a  bxi xi
2
b
i

i1 
i1
2
n
These are essentially the same normal equations as (3.96)
and (3.97), though we can argue about “what if =0”.
We can now solve quite easily for the best-fit line, but is it
a reasonable line? We need a test for whether a and b
are realistic values. For example:
If the points have large errors or are close
together in x, then the line can be very uncertain.
Way back in Chapter 1 (EQN???) we found in the section
on propagation of errors that the variance of a function f,
f2, is given by:
f 
    
yi 
2
2
f
2
i
(4.7)
We use our equations for a and b (eqn. 4.6) as f, and
obtain:

a Sxx  Sx xi

yi
 i2 
b Sxi  Sx

yi
 i2 
(4.8)
Substituting back into (4.7), and after some
simplification, we get:

 Sxx
2
2 Sxx  Sx xi
 a   i  2
 
  i   
2
And:



 S
2
2 Sxi  S x
 b   i  2  
  i   
(4.9)
2
(4.10)
We find the covariance and correlation between a and
b:
a b  Sx
      

yi yi 
Sx
r 
SS xx
2
ab
2
i
(4.11)
(4.12)
The correlation coefficient will be non-zero in most
cases, saying that the intercept is correlated to the
slope. This is not true if we shift the origin to the mean x
 value, in which case r=0.
Having calculated values for a and b, we can now go
back to Eqn (4.1) and calculate a value for 2. We also
get a critical value for 2 for n-2 degrees of freedom and
check to see of our result from (4.1) is smaller or larger. If
(4.1) yields the smaller value, then we can say that the fit
is significant within the  level of confidence.
EXAMPLE: (Hypothetical) Data from 5 Hawaiian seamounts
are given below:
Name
Dist. from Kilauea, km
Age, MY
Laysan
1,818
19.9±0.3
Northampton
1,841
26.6±2.7
Kimmei
3,668
39.9±1.2
Ojin
4,102
55.2±0.7
Suiko
4,860
64.7±1.1
Using the method of chapter 3, and Matlab, let’s find the
best-fit line:
% column 1: dist from Kilauea, km
% column 2: age, MY
% column 3: uncertainty, 1 std dev, MY
n=5; % number of seamounts
hold off; clear all % reset parameters
Haw=[1818 19.9 0.3;...
1841 26.6 2.7;...
3668 39.9 1.2;...
4102 55.2 0.7;...
4860 64.7 1.1]; % data matrix from HVO Black book 1
XXX=Haw(:,1); % x values: dist from Kilauea
YYY=Haw(:,2); % Y-values: ages
Unit=[1 1 1 1 1]; % unit vector for additions
Sy=Unit*YYY; % sum of all y values, see page 63 Wessel notes G
Sx=Unit*XXX; % Sum of all x values
Sxy=Unit*(XXX.*YYY); % sum of x*y
Sxx=Unit*(XXX.*XXX); % sum of x^2
N=[n Sx;Sx Sxx];
% data matrix: Nx=B
B=[Sy;Sxy];
% right side
X=inv(N)*B; % X(1)=intercept, X(2)=slope
BestFit=[Haw(1,1) Haw(n,1);X(2)*Haw(1,1)+X(2)
X(2)*Haw(n,1)+X(2)];
plot(BestFit(1,:),BestFit(2,:))
hold on
plot(XXX,YYY,'+')l
Slope is 0.0133
my/km, or
75 km/my
Or 7.5 cm/yr
FOR HOMEWORK:
Calculate the best fit to the Hawaiian-Emperor data using
the equations we covered in chapter 4, including finding the
critical value, covariance, correlation coef, the uncertainty
in the slope (±1 standard deviation), and whether the
results are significant at 95%.
We will go over this at sea - best finish it before you leave.
The m-file on the previous pages should give you a good
start - and provide a check on your results.