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-- AlexanderFedotov - 2015-03-03

# Linear Least Squares

## 2. Uncorrelated measurements

Let with variances be measurements of functions with the known matrix and unknown parameters .

In linear least square method, one estimates the parameter vector by minimizing over the expression

where the weight matrix of dimension is diagonal and defined as the inverse of the diagonal covariance matrix for : i.e. .
In matrix notation (considering and as columns and respectively), one has

The estimate is the solution of the system of equations

or
In a general case of linear transformation , the covariance matrice for is transformed into that for via . Hence,
With by the definition of , that simplifies to

Note, that

and

## 3. Correlated measurements

Let be uncorrelated measurements as those in the previous section, and ( is an invertible matrix). Then are generally correlated and have the covariance matrix .

With and , one gets

where .

Similarly,

and

Thus, all the formulae for the correlated measurements are similar to those for the uncorrelated , with the only complication being the replacement of a diagonal weight matrix with a non-diagonal one:

## 4. An iterative solution

### 4.1 The procedure

Let indices of parameters be distributed among groups with sizes respectively.

The subset of the parameters corresponding to the group of indices, can be considered as an column
Let denote the set of indices that are complementary to indices in .

Consider the following iterative procedure.

2. Make N steps or, equivalently, iterations for , thus finding vectors by minimizing over at the step :
where is the "point" where as a function of parameters , takes minimum (while the rest parameters are fixed at the values obtained in the previous iteration: )
3. Repeat (4.3) infinitly, defining for    (i.e. ) .
One can expect that

### 4.2 The step

The iteration consists in finding , a set of parameters, that minimizes

Let be an submatrix of built from the columns of with indices belonging to (in other words, it is what remains after removing columns which have indices not in ).
Similarly, let be an submatrix consisting of the columns of with indices belonging to .
Introducing

the (4.2.1) can be written as
which is analogous to eq.(2.2).
Therefore, the solution is given by formulae (2.4):
or
Let us write the last expression via matrices of dimensions
and
such that all the arythmetics is done in rows / columns (or ) while complementary rows / columns contain zeros.

We define matrices and as follows

It is noteworthy that

We also define the matrix (subscript stands for extended) via the matrix :

Then (4.2.5) can be written as
This is a representation of the first line of (4.3.1).
The second line of (4.3.1) can be represented with
Summing up (4.2.10) and (4.2.11) gives
where we denoted

### 4.3 steps

#### 4.3.1 as a function of

Let us define matrices and as

and
Then, from (4.2.12),
Indeed, by induction:
• eq.(4.3.3) holds for :
• and assuming it is true for , leads to

In particular, the eq.(4.3.3) is valid for :

#### 4.3.2 The covariance matrix of

According to the general rule of covariance matrix transformation ( ), eq.(4.3.3) gives rise to

Comparing this with the covariance matrix of the exact solution, eq.(2.6), yields

### 4.4 steps for

With the expression (4.3.6) of the form

we can build the expression for the case in one step as
Substituting (4.3.6) into (4.4.2) gives
Transforming similarly the formula to the one, gives
Then for the case one has
and so on...

This generalizes to

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Topic revision: r11 - 2015-03-25 - AlexanderFedotov

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