Joint probability distribution of two images

atrus_ovis
Messages
99
Reaction score
0
I'm posting this here, as i feel it's more probability-related than image processing.

I'm reading this lecture pdf.
At end of page 1 , beginning of page 2 it says:
The spatial information obviously required for a registration method is provided by the
definition of a joint probability distribution that depends simultaneously on image A and B.

The conventional expression for it is pA,B (a, b). It is calculated as the number of times out
of the total number of pixels N that a pixel in A contains the value a and the same pixel
that is, the pixel in the same image position, in B contains the value b; this number of pixels
is then divided by the total number of pixels to give the joint probability of a, b.

Is this a bit vague, or am i missing something?
Since we are calculating frequency of the value a in the pixels of image A, when was the information about any position introduced?

Or does it mean that for *any* pixel in image A that contains the value a, check if the pixel(s) in image B at the corresponding position have the value b?

If this is too specific or too dependent on image processing,can you provide some material to build my intuition about joint probability distributions?

Thank you.
 
Physics news on Phys.org
Let's say the images are
A = \begin{pmatrix} 1&1& 2 \\ 2&7&1 \end{pmatrix}
and
B = \begin{pmatrix} 1&2&2 \\ 2&1&2 \end{pmatrix}

My interpretation of that passage is that

p_{A,B}(1,1) = \frac{1}{6}
p_{A,B}(1,2) = \frac{2}{6}
p_{A,B}(1,7) = \frac{0}{6}

p_{A,B}(2,1) = \frac{0}{6}
p_{A,B}(2,2) = \frac{2}{6}
p_{A,B}(2,7) = \frac{0}{6}

p_{A,B}(7,1) = \frac{1}{6}
p_{A,B}(7,2) = \frac{0}{6}
p_{A,B}(7,7) = \frac{0}{6}

Although there are (3)(3) = 9 possible ordered pairs of pixel values to consider, the total match-ups will only be 6.

As to whether we should call this "the" joint distribution of two images - that is merely the terminology that those authors wish to use. There are many many ways to take properties of images and define joint distributions.
 
Namaste & G'day Postulate: A strongly-knit team wins on average over a less knit one Fundamentals: - Two teams face off with 4 players each - A polo team consists of players that each have assigned to them a measure of their ability (called a "Handicap" - 10 is highest, -2 lowest) I attempted to measure close-knitness of a team in terms of standard deviation (SD) of handicaps of the players. Failure: It turns out that, more often than, a team with a higher SD wins. In my language, that...
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...

Similar threads

Back
Top