Statistical Modeling and Analysis
Chapter 15
Modeling of Data
Statistics of Data
•
Mean (or average):
•
Variance:
•
Median: a value
x
j
such that half of the
data are bigger than it, and half of data
smaller than it.
Higher Moments
Gaussian Distribution
Least Squares
•
Given
N
data points (
x
i
,
y
i
), i = 1, …,
N
, find
the fitting parameters
a
j
, j = 1, 2, …,
M
of a
function
f
(
x
)
= y
(
x; a
1
,a
2
,…,a
M
)
such that
is minimized over the parameters
a
j
.
Why Least Squares
•
Given the parameters, what is the
probability that the observed data
occurred?
•
Assuming independent, Gaussian
distribution, that is:
Chi-Square Fitting
•
Minimize the quantity:
•
If each term is an independent Gaussian,
2
follows so-called
2
distribution. Given the
value
2
above, we can compute
Q
=
Prob(random variable chi
2
>
2
)
•
If
Q
< 0.001 or
Q
> .999, the model may be
rejected.
Meaning of Goodness-of-Fit Q
2
Observed
value of
2
Area = Q
If the statistic
2
indeed
follows this distribution,
the probability that chi-
square value is the
currently computed value
2
, or greater, equals the
hashed area Q.
It is quite unlikely if Q is
very small or very close to
1. If so, we reject the
model.
Number of degrees of
freedom
=
N
–
M
.
0
Fitting to Straight Line
(with known error bars)
x
y
fitting to
y
=
a
+
bx
Given (
x
i
,
y
i
±
σ
i
)
Find interception
a
and slope
b
such that
the chi-square merit function
is minimized.
Goodness-of-fit is
Q
=gammq((
N
-2)/2,
2
/2). If
Q
> 0.1, the fitting is good, if
Q
≈ 0.001, may be OK, but if
Q
< 0.001, fitting is
questionable.
If
Q
> 0.999, fitting is too
good to be true.
Linear Regression Model
x
y
fitting to
y
=
a
+
bx
ε
Data do not
follow exactly
the straight line.
The basic
assumption in
linear
regression
(least squares
fit) is that the
deviations
ε
are
independent
gaussian
random noise.
Error in
y
, but no
error in
x
.
Solution of Straight Line Fit
Error Propagation
•
Let
z
=
f
(
y
1
,
y
2
,…,
y
N
) be a function of
independent
random variables
y
i
.
Assuming the variances are small, we
have
•
Variance of
z
is related to variances of
y
i
by
Error Estimates on
a
and
b
•
Using error propagation formula, viewing
a
as a function of
y
i
, we have
•
Thus
•
Similarly
What if error in
y
i
is unknown?
•
The goodness-of-fit
Q
can no longer be
computed
•
Assuming all data have same
σ
:
•
Error in
a
and
b
can still be estimated,
using
σ
i
=
σ
(but less reliably)
M
is number of basis functions,
M
=2 for straight line fit.
General Linear Least-Squares
•
Fit to a linear combination of arbitrary
functions:
•
E.g., polynomial fit
X
k
(
x
)=
x
k-1
, or harmonic
series
X
k
(
x
)=sin(
kx
), etc
•
The basis functions
X
k
(
x
) can be nonlinear
Merit Function & Design Matrix
•
Find
a
k
that minimize
•
Define
•
The problem can be stated as
Let
a
be a
column vector:
Normal Equation & Covariance
•
The solution to min ||
b
-
Aa
||
2
is
A
T
Aa
=
A
T
b
•
Let
C
= (
A
T
A
)
-1
, then
a
=
CA
T
b
•
We can view data
y
i
as a random variable
due to random error,
y
i
=y(x)+
ε
i
. <
ε
i
>=0,
<
ε
i
ε
j
>=
σ
i
2
ij
. Thus
a
is also a random
variable.
Covariance of
a
is precisely
C
•
<
aa
T
>-<
a
><
a
T
> =
C
•
Estimate of the fitting coefficient
is
Singular Value Decomposition
•
We can factor arbitrary complex matrix as
A = U
Σ
V
†
N
M
N
N
N
M
M
M
U and V are unitary, i.e., UU
†
=1, VV
†
=1
Σ
is diagonal (but need not square), real and
positive,
w
j
≥ 0.
Solve Least-Squares by SVD
•
From normal equation, we have
Or
Omitting terms with very
small
w
gives robust method.
N
o
n
l
i
n
e
a
r
M
o
d
e
l
s
y
=
y
(
x
;
a
)
•
2
i
s
a
n
o
n
l
i
n
e
a
r
f
u
n
c
t
i
o
n
o
f
a
.
C
l
o
s
e
t
o
m
i
n
i
m
u
m
,
w
e
h
a
v
e
(
T
a
y
l
o
r
e
x
p
a
n
s
i
o
n
)
Solution Methods
•
Know gradient only, Steepest descent:
•
Know both gradient and Hessian matrix:
•
Define
Levenberg-Marquardt Method
•
Smoothly interpolate between the two
methods by a control parameter
.
=0,
use more precise Hessian;
very large,
use steepest descent.
•
D
e
f
i
n
e
n
e
w
m
a
t
r
i
x
A
'
w
i
t
h
e
l
e
m
e
n
t
s
:
Levenberg-Marquardt Algorithm
•
S
t
a
r
t
w
i
t
h
a
n
i
n
i
t
i
a
l
g
u
e
s
s
o
f
a
•
C
o
m
p
u
t
e
2
(
a
)
•
Pick a modest value for
, say
=0.001
•
(
†
)
S
o
l
v
e
A
'
a
=
β
,
e
v
a
l
u
a
t
e
2
(
a
+
a
)
•
If
2
increase, increase
by a factor of 10
and go back to (
†)
•
I
f
2
d
e
c
r
e
a
s
e
,
d
e
c
r
e
a
s
e
b
y
a
f
a
c
t
o
r
o
f
1
0
,
u
p
d
a
t
e
a
a
+
a
,
a
n
d
g
o
b
a
c
k
t
o
(
†
)
Problem Set 9 (3 Nov)
Three things to provide (i) estimate of the fitting parameters, (ii) errors on them, (iii) a statistical measure of goodness-of-fit.
Exploring statistical concepts such as mean, variance, skewness, kurtosis, Gaussian distribution, least squares fitting, chi-square fitting, and goodness-of-fit in data analysis. Learn about fitting parameters, probability computation, and interpretation of model goodness.
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Presentation Transcript
Chapter 15 Modeling of Data
Statistics of Data N 1 N = Mean (or average): x x j = 1 j N 1 ( ) 2 Variance: = = 2 Var( , , ) x x x x 1 N j 1 N = 1 j Median: a value xjsuch that half of the data are bigger than it, and half of data smaller than it. is called standard deviation. Error estimate ? ?/ ?
Higher Moments 3 x x N 1 N j = Skew( , , ) x x 1 N = 1 j 4 x x N 1 N j = Kurt( , , ) 3 x x 1 N = 1 j
Gaussian Distribution 2 ( ) x a 1 = 2 ( ; , ) N x a e 2 2 = x a = = = 2 Var( ) Skew( ) Kurt( ) x 0 x 0 x
Least Squares Given N data points (xi,yi), i = 1, , N, find the fitting parameters aj, j = 1, 2, , M of a function f(x) = y(x; a1,a2, ,aM) such that 1 1 i = N 2 ( ; , i y x a , ) y a i M is minimized over the parameters aj.
Why Least Squares Given the parameters, what is the probability that the observed data occurred? Assuming independent, Gaussian distribution, that is: 2 N ( ) y x 1 2 y exp i i P y i = 1 i i
Chi-Square Fitting Minimize the quantity: 2 ( ; , i y x a , ) y a N 2 1 i M = 1 i i If each term is an independent Gaussian, 2 follows so-called 2 distribution. Given the value 2 above, we can compute Q = Prob(random variable chi2 > 2) If Q < 0.001 or Q > .999, the model may be rejected.
Meaning of Goodness-of-Fit Q ( ) If the statistic 2 indeed follows this distribution, the probability that chi- square value is the currently computed value 2, or greater, equals the hashed area Q. 2 2 2 ( ) exp /2 P Observed value of 2 It is quite unlikely if Q is very small or very close to 1. If so, we reject the model. Area = Q Number of degrees of freedom = N M. 0 2
Fitting to Straight Line (with known error bars) Given (xi, yi i) Find interception a and slope b such that = y 2 y a bx N 2 ( , ) a b i i = 1 i i the chi-square merit function is minimized. Goodness-of-fit is Q=gammq((N-2)/2, 2/2). If Q > 0.1, the fitting is good, if Q 0.001, may be OK, but if Q < 0.001, fitting is questionable. fitting to y=a+bx If Q > 0.999, fitting is too good to be true. x
Linear Regression Model Error in y, but no error in x. y Data do not follow exactly the straight line. The basic assumption in linear regression (least squares fit) is that the deviations are independent gaussian random noise. fitting to y=a+bx x
Solution of Straight Line Fit = = 2 2 0, 0 a b x y 1, N N N , i i S S S x y 2 i x = 2 i 2 i = = = 1 1 1 i i i 2 i x y N N , i i S S xx xy 2 i 2 i = = 1 1 i i + aS bS S x y S + = 2 x S S , aS bS SS S x xx xy xx S S S S SS xx y x xy xy x y = = , a b
Error Propagation Let z = f(y1,y2, ,yN) be a function of independent random variables yi. Assuming the variances are small, we have ( i i i y y N f = ) z z y y i = 1 i Variance of z is related to variances of yi by 2 2 f i i i y = 2 N f = 1
Error Estimates on a and b Using error propagation formula, viewing a as a function of yi, we have S S x a y = = xx x i 2 i i Thus 2 S S x S N = 2 a 2 i xx x i xx 2 i = 1 i S Similarly = 2 b
What if error in yi is unknown? The goodness-of-fit Q can no longer be computed Assuming all data have same : 1 i = N 2 = 2 M is number of basis functions, M=2 for straight line fit. ( ) /( i y x ) y N M i Error in a and b can still be estimated, using i= (but less reliably)
General Linear Least-Squares Fit to a linear combination of arbitrary functions: M y x a X x = = ( ) ( ) k k 1 k E.g., polynomial fit Xk(x)=xk-1, or harmonic series Xk(x)=sin(kx), etc The basis functions Xk(x) can be nonlinear
Merit Function & Design Matrix Find ak that minimize 2 M ( ) x y a X N i k k i = 2 = 1 k = 1 i i a a 1 Define Let a be a column vector: 2 = a ( ), X x y j i = = i A b ij i a i i M The problem can be stated as 2 2 Min || || b Aa
Normal Equation & Covariance The solution to min ||b-Aa||2 is ATAa=ATb Let C = (ATA)-1, then a = CATb We can view data yi as a random variable due to random error, yi=y(x)+ i. < i>=0, < i j>= i2 ij. Thus a is also a random variable. Covariance of a is precisely C <aaT>-<a><aT> = C Estimate of the fitting coefficientis T j jj j CA b = a C
Singular Value Decomposition We can factor arbitrary complex matrix as A = U V a a a U U U 0 0 w 11 1 11 1 M N 1 * * V V V 11 * 12 1 M 0 N M 21 21 = M M N M N N 0 0 w M * V MM a a U 0 0 1 N NM NN U and V are unitary, i.e., UU =1, VV =1 is diagonal (but need not square), real and positive, wj 0.
Solve Least-Squares by SVD From normal equation, we have = 1 T T ( ) a A A A b = = T T T ( ( ) ) AB AB B A B A 1 1 1 A U V = T but ( ( V ) 1 = U V U V U V T T T T T so ( ) ( ) a b ) ( ) 1 1 = = T T T T T T T T T V U U V V U b V V V U b = = 1 1 T T T T T T T ( ) ( ) V V U b V U b T U b Omitting terms with very small w gives robust method. ( ) j w = a V Or ( ) j = , 1, , w j M j j
Nonlinear Models y=y(x; a) 2 is a nonlinear function of a. Close to minimum, we have (Taylor expansion) 1 2 ( ) = + + 2 3 T ( ) a a a a D a a a a ( ) ( ) ( ) ( ) O min min min min 1 2 + a T T d a D a where 2 2 ( ) a = 2 d D a ( ), a + = D ij a a i j
Solution Methods Know gradient only, Steepest descent: = 2 a a a constant ( ) next cur cur Know both gradient and Hessian matrix: = 1 2 a a D a ( ) min cur cur Define a a ( ; ) i y x a ( ; ) i y x a 1 2 1 N = = 2 , kl 2 i = 1 i k l
Levenberg-Marquardt Method Smoothly interpolate between the two methods by a control parameter . =0, use more precise Hessian; very large, use steepest descent. Define new matrix A' with elements: + = (1 ), if if i i j j ii ij , ij
Levenberg-Marquardt Algorithm Start with an initial guess of a Compute 2(a) Pick a modest value for , say =0.001 ( ) Solve A' a= , evaluate 2(a+ a) If 2 increase, increase by a factor of 10 and go back to ( ) If 2 decrease, decrease by a factor of 10, update a a+ a, and go back to ( )
Problem Set 9 (3 Nov) 1. If we apply the Levenberg-Marquardt method for a linear least-square problem, what are A and ? How many iterations are needed for convergence with a judicious choice of starting condition? 2. Radiative decay data can be fitted to the sum of two exponentials, ?1? ?/?1+ ?2? ?/?2. Apply the Levenberg-Marquardt method for nonlinear least-squares fit implemented in scipy optimize.curve_fit(f, x, y, , method= lm ). Use the data file nfit.dat posted, to determine the fitting parameters.
