Jasper den Ouden: 2020-08-14

### 0.1   The model

Real image k −> Rk where k indexes values. The seen value is defined as:

Vl =
 ∑ kl
fkl Rk

Where fkl is presumed known the factor how much k pixel of R contributes to l pixel of V.

To be clearer; the model has Rk which is an image we expect there to be out there. k indexes that image, in a more specific example later, it will be a higher resolution raster image with k=(i,j) just two image coordinates.

Also in the later specific example Vl is the real value as measured by a pixel of a camera, and vl the measured value, off due to random variation. l=(i,j,t) with i and j image coordinates and t the time, or frame number. The camera has a lower resolution than the real image in the model.

I was mistaken that weighed averages are of images scaled and translated over where the perfect choice under the assumptions. See below about that.

The actual measuring is also part of the model,Vl is the value, and vl the actually measured number. Yielding independent gaussian solutions, which means you can multiply them for the combined probabilty;

P(k−>Rk) ∝
 ∏ l
exp
 (Vl − vl)2 2σl2
= exp
 1 2
 ∑ l
 (Vl − vl)2 2σl2

### 0.2   Processing

Can easily apply the maximum likelyhood to that sum; we want the all the real image values of index n

0 =
 d dRn
 ∑ l
 (Vl − vl)2 2σl2
=  2
 ∑ l
 1 σl2
(Vl − vl)dVl/dRn = 2
 ∑ l
 fnl σl2
(Vl − vl)

Rearrange the sum to both sides:

 ∑ l

 fnlvl σl2
=
 ∑ k
Rk
 ∑ l
 fnlfkl σl2

It is a linear equation to solve to get Rk,

### 0.3   My mistake about disjoint areas for Vl

I initially thought disjoint areas on l didn’t follow the linear equation. Here is how i defined it.(once i went to the math of it)

 fkl = 1 if k ∈ D(l)
 fkl = 0 otherwise

And that D(l) is disjoint; l1≠ l2 => D(l1) ∩ D(l2) = ∅ in the sum:

 ∑ l

 fnlvl σl2
=
 ∑ k
Rk
 ∑ l
 fnlfkl σl2

The factors are only nonzero if both n and k are in the same disjoint area, and otherwise they’re one. I.e. we can sum over the area instead;

 ∑ l: n ∈ D(l)

 vl σl2
=
 ∑ k ∈ D(l2)
Rk
 ∑ l: n ∈ D(l)

 1 σl2
with  l2  the one value with  n ∈ D(l2)

So my mistake: basically i thought you could select just k=n there. I.e. that if k, nD(l) then k=n but that’s obviously not true, but not when doing in my head, apparently. (if it is just Rn instead of the sum, it’s the weighed average)

This document was translated from LATEX by HEVEA.