Lecture 8 Advanced Topics in Least Squares - Part Two -

Lecture 8

Advanced Topics in

Least Squares

- Part Two -

Concerning the Homework

the dirty little secret of data analysis

You often spend more time futzing with reading files that are

in inscrutable formats

than the intellectually-interesting side of data analysis

Sample MatLab Code

cs = importdata('test.txt','=');Ns = length(cs);mag = zeros(Ns,1);Nm=0;for i = [1:Ns] s=char(cs(i)); smag=s(48:50); stype=s(51:52); if( stype == 'Mb' ) Nm=Nm+1; mag(Nm,1)=str2num(smag); endend

mag = mag(1:Nm);

a routine to read a text file

choose non-occurring delimiter to force complete line into one cell

returns “cellstr” data type: array of strings

convert “cellstr” element to string

convert string to number

What can go wrong in least-squares

m = [GTG]-1 GT d

the matrix [GTG]-1 is singular

m =

d1

d2

d3

…

dN

1 x1

1 x2

1 x3

…

1 xN

EXAMPLE - a straight line fit

N i xi

i xi i xi2

GTG =

det(GTG) = N i xi2 – [i xi]2

[GTG]-1 singular when determinant is zero

N=1, only one measurement (x,d)

N i xi2 – [i xi]2 = x2 - x2 = 0

you can’t fit a straight line to only one point

N1, all data measured at the same x

N i xi2 – [i xi]2 = N2 x2 – N2 x2 = 0

measuring the same point over and over doesn’t help

det(GTG) = N i xi2 – [i xi]2 = 0

m =

s1

s2

…

dNd-1

dNd-1

1 1

1 1…

1 -1

1 -1

another example – sums and differences

Ns+Nd Ns-Nd

Ns-Nd Ns+Nd

GTG =

det(GTG) = 0 = [Ns+Nd]2 - [Ns-Nd]2 =

[Ns2+Nd

2 +2NsNd] - [Ns2+Nd

2 -2NsNd] =

4NsNd

Ns sums, si, and Nd differences, di, of two unknowns m1 and m2

zero when Ns=0 or Nd=0, that is, only measurements of one kind

This sort of ‘missing measurement’might be difficult to recognize in a

complicated problem

but it happens all the time …

Example - Tomography

in this method, you try to plaster the subject with X-ray beams made at every possible position and direction, but you can easily wind up missing some small region …

no data coverage here

What to do ?

Introduce prior information

assumptions about the behavior of the unknowns

that ‘fill in’ the data gaps

Examples of Prior Information

The unknowns:

are close to some already-known valuethe density of the mantle is close to 3000 kg/m3

vary smoothly with time or with geographical position

ocean currents have length scales of 10’s of km

obey some physical law embodied in a PDEwater is incompressible andthus its velocity satisfies div(v)=0

Are you only fooling yourself ?

It depends …

are your assumptions good ones?

Application of theMaximum Likelihood Method

to this problem

so, let’s have a foray into the world of probability

Overall Strategy

1. Represent the observed data as a probability distribution

2. Represent prior information as a probability distribution

3. Represent the relationship between data and model parameters as a probability distribution

4. Combine the three distributions in a way that embodies combining the information that they contain

5. Apply maximum likelihood to the combined distribution

How to combine distributions in a way that embodies combining the information that they contain …

Short answer: multiply them

But let’s step through a more well-reasoned analysis of why we should do that …

x

p1(x)

x

p2(x)

x

pT(x)

how to quantify the information in a distribution p(x)

Information compared to what?

Compared to a distribution pN(x) that represents the state of complete ignorance

Example: pN(x) = a uniform distribution

The information content should be a scalar quantity, Q

Q = ln[ p(x)/pN(x) ] p(x) dx

Q is the expected value of ln[ p(x)/pN(x) ]

Properties:

Q=0 when p(x) = pN(x)

Q0 always (since limitp0 of pln(p)=0)

Q is invariant under a change of variables xy

Combining distributions pA(x) and pB(x)

Desired properties of the combination:

pA(x) combined with pB(x) is the same as pB(x) combined with pA(x)

pA(x) combined [ pB(x) combined with pC(x)]is the same as [ pA(x) combined pB(x) ] combined with pC(x)

Q of [ pA(x) combined with pN(x) ] QA

pAB(x) = pA(x) pB(x) / pN(x)

When pN(x) is the uniform distribution …

… combining is just multiplying.

But note that for “points on the surface of a sphere’, the null distribution, p(,), is latitude and is longitude, where would not be uniform, but rather proportional to sin().

Overall Strategy

1. Represent the observed data as a Normal probability distribution

pA(d) exp{ -½ (d-dobs)T Cd-1 (d-dobs) }

In the absence of any other information, the best estimate of the mean of the data is the observed data itself.

Prior covariance of the data.

I don’t feel like typing the normalization

Overall Strategy

2. Represent prior information as a Normal probability distribution

pA(m) exp{ -½ (m-mA)T Cm-1 (m-mA) }

Prior estimate of the model, your best guess as to what it would be, in the absence of any observations.

Prior covariance of the model quantifies how good you think your prior estimate is …

example

one observationdobs = 0.8 ± 0.4

one model parameter withmA=1.0 ± 1.25

mA=1

dobs =

0.8

0 2

20

pA(d) pA(m)

Overall Strategy

3. Represent the relationship between data and model parameters as a probability distribution

pT(d,m) exp{ -½ (d-Gm)T CG-1 (d-Gm) }

Prior covariance of the theory quantifies how good you think your linear theory is.

linear theory, Gm=d, relating data, d, to model parameters, m.

example

theory: d=m

but only accurate to ± 0.2

mA=1

d obs=

0.8

0 2

20

pT(d,m)

Overall Strategy

4. Combine the three distributions in a way that embodies combining the information that they contain

p (m,d) = pA(d) pA(m) pT(m,d)

exp{ -½ [

(d-dobs)T Cd-1 (d-dobs) +

(m-mA)T Cm-1 (m-mA) +

(d-Gm)T CG-1 (d-Gm) ]}

a bit of a mess, but it can be simplified ,,,

0 2

20

p(d,m)=pA(d) pA(m) pT(d,m)

Overall Strategy

5. Apply maximum likelihood to the combined distribution, p(d,m) = pA(d) pA(m) pT(m,d)

There are two distinct ways we could do this:

Find the (d,m) combinations that maximized the joint probability distribution, p(d,m)

Find the m that maximized the individual probability distribution, p(m) = p(d,m) dd

These do not necessarily give the same value for m

mest

dpre

0 2

20

p(d,m)

maximum of p(d,m)=pA(d) pA(m) pT(d,m)

Maximum likelihood point

maximum of p(m) = p(d,m)dd

m

p(m)

mest

Maximum likelihood point

special case of an exact theory

in the limit CG0

exp{ -½ (d-Gm)T CG-1 (d-Gm) } (d-Gm)

and p(m) = p(d,m) dd

= pA(m) pA(d) (d-Gm) dd

= pA(m) pA(d=Gm)

so for normal distributions p(m) = exp{ -½ [

(Gm-dobs)T Cd-1 (Gm-dobs) + (m-mA)T Cm

-1 (m-mA) ]}

Dirac delta function, with property f(x) (x-y) dx = f(y)

special case of an exact theory

maximizing p(m) is equivalent to minimizing

(Gm-dobs)TCd-1(Gm-dobs) + (m-mA)TCm

-1(m-mA)

weighted “prediction error” weighted “distance of the model from its prior value”

+

solutioncalculated via the usual messy minimization process

mest = mA + M [ dobs – GmA]

where M = [GTCd-1G + Cm

-1]-1 GT Cd-1

Don’t Memorize, but be prepared to use

(e.g. in homework).

interesting interpretation

mest - mA = M [ dobs – GmA]

estimated model minus its prior

observed data minus the prediction of the prior model

linear connection between the two

special case of no prior informationCm

M = [GTCd-1G + Cm

-1]-1 GT Cd-1[GTCd

-1G]-1 GT Cd-1

mest = mA + [GTCd-1G]-1 GT Cd

-1 [ dobs – GmA]

= mA+[GTCd-1G]-1GTCd

-1dobs–[GTCd-1G]-1GTCd

-1GmA

= mA+[GTCd-1G]-1GTCd

-1dobs–mA

= [GTCd-1G]-1GTCd

-1dobs

recovers weighted least squares

special case of infinitely accurate prior information Cm 0

M = [GTCd-1G + Cm

-1]-1 GT Cd-1 0

mest = mA + 0

= mA

recovers prior value of m

special uncorrelated case Cm=m

2I and Cd=d2I

M = [GTCd-1G + Cm

-1]-1 GT Cd-1

= [d-2GTG + m

-2I]-1 GT d-2

= [ GTG + (d/m)2I ]-1 GT

this formula is sometimes called “damped least squares”, with “damping factor” =d/m

Damped Least Squaresmakes the process of avoiding

singular matrices associated with insufficient data

trivially easy

you just add 2I to GTG before computing the inverse

GTG GTG + 2I

this process regularizes the matrix, so its inverse always exists

its interpretation is :in the absence of relevant data,

assume the model parameter has its prior value

Are you only fooling yourself ?

It depends …

is the assumption - that you know the prior value - a good one?

Smoothness Constraints

e.g. model is smooth when its second derivative is small

d2mi/dx2 mi-1 - 2mi + mi+1

(assuming the data are organized according to one spatial variable)

matrix D approximates second derivative

1 -2 1 0 0 0 …

0 1 -2 1 0 0 …

…

0 0 0 … 1 -2 1

D =

d2m/dx2 Dm

Choosing a smooth solution is thus equivalent to minimizing

(Dm)T (Dm) = mT (DTD) m

comparing to the

(m-mA)TCm-1(m-mA)

minimization implied by the general solution

mest = mA + M [ dobs – GmA]where M = [GTCd

-1G + Cm-1]-1 GT Cd

-1

indicates that we should make the choicesmA = 0

Cm-1 = (DTD)

To implement smoothness