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I came across a problem working with image moments [1]. It is stated that

$\eta_{ij} = \frac{\mu_{ji}}{\mu_{00}^{k}}$

where $k = 1 + \frac{i+j}{2}$

is scale invariant.

However, if I try to reproduce this, it does not appear scale invariant at all.

Consider a simple example:

In a binary image, we calculate $\eta_{20}$ of a 2x2 block of 4 pixels:

☐☐
☐☐

$\mu_{20} = 0.5^2 + 0.5^2 + (-0.5)^2 + (-0.5)^2 = 4 \cdot 0.25 = 1$

$\mu_{00} = 4$

$k = 1 + \frac{2+0}{2} = 1+1 = 2$

$\eta_{20} = \frac{1}{4^2} = \frac{1}{16} = 0.0625$

Now, let's scale this block by the factor two:

☐☐☐☐
☐☐☐☐
☐☐☐☐
☐☐☐☐

$\mu_{20} = 4 \cdot 1.5^2 + 4 \cdot 0.5^2 + 4 \cdot (-0.5)^2 + 4 \cdot (-1.5)^2 = 8 \cdot 2.25 + 8 \cdot 0.25 = 18 + 2 = 20$

$\mu_{00} = 16$

$k = 1 + \frac{2+0}{2} = 1+1 = 2$

$\eta_{20} = \frac{20}{16^2} = \frac{20}{256} = 0.078125$

Why do we have a different result after scaling the object if $\eta_{ij}$ is supposedly scale invariant? Is there any formal proof of the scale invariance of $\eta$?

[1] https://en.wikipedia.org/wiki/Image_moment

Makx
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1 Answers1

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The invariance is exact in the continuous domain. Your problem here is how you scale, which is limited by the discrete grid.

Scaling by a factor 2 of your original 4 squares would result in 4 squares, centered at (-1,-1), (-1,1), (1,-1) and (1,1), each with a weight of 4:

$$ \mu_{20} = 4 \cdot 1^2 + 4 \cdot 1^2 + 4 \cdot (-1)^2 + 4 \cdot (-1)^2 = 16 $$

$$ \mu_{00} = 4 \cdot 4 = 16 $$

$$ \eta_{20} = \frac{16}{16^2} = 0.0625 $$

On the discrete grid, the invariances are approximate.

Cris Luengo
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