Computing image integral

Question

How do one find the mean, std dev, and gradient from image integral? Given an image such as follows:

Answer 1

What jmch didn't say is, if sqrt( C'+A'-B'-D'/K - (mean*mean) ) is not how you compute the standard deviation from the integral image, then how you do it?

First, let me switch to Python / numpy code, so we get a modicum of notation consistency and expressions are easier to check. Given a sample array X, say:

X = array([random() * 10.0 for i in range(0, 9)])

The uncorrected sample standard deviation of X can be defined as:

std = (sum((X - mean(X)) ** 2) / len(X)) ** 0.5 # 1

Applying the binomial theorem to (X - mean(X)) ** 2 we get:

std = (sum(X ** 2 - X * 2 * mean(X) + mean(X) ** 2) / len(X)) ** 0.5 # 2

Given the identities of the summation operation, we can make:

std = ((sum(X ** 2) - 2 * mean(X) * sum(X) + len(X) * mean(X) ** 2) / len(X)) ** 0.5 # 3

If we make S = sum(X), S2 = sum(X ** 2), M = mean(X) and N = len(X) we get:

std = ((S2 - 2 * M * S + N * M ** 2) / N) ** 0.5 # 4

Now for an image I and two integral images P and P2 calculated from I (where P2 is the integral image for squared pixel values), we know that, given the four edge coordinates A = (i0, j0), B = (i0, j1), C = (i1, j0) and D = (i1, j1), the values of S, S2, M and N can be calculated for the range I[A:D] as:

S = P[A] + P[D] - P[B] - P[C]

S2 = P2[A] + P2[D] - P2[B] - P2[C]

N = (i1 - i0) * (j1 - j0)

M = S / N

Which can then be applied to equation (4) above yielding the standard deviation of the range I[A:D].

Edit: It's not entirely necessary, but given that M = S / N we can apply the following substitutions and simplifications to equation (4):

std = ((S2 - 2 * M * S + N * M ** 2) / N) ** 0.5

std = ((S2 - 2 * (S / N) * S + N * (S / N) ** 2) / N) ** 0.5

std = ((S2 - 2 * ((S ** 2) / N) + (S ** 2 / N)) / N) ** 0.5

std = ((S2 - ((S ** 2) / N)) / N) ** 0.5

std = (S2 / N - (S / N) ** 2) ** 0.5 # 5

Which is quite close to the equation remi gave, actually.