# Scale invariant Image Moments - not scale variant?

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 $\eta$?

$$\mu_{20} = 4 \cdot 1^2 + 4 \cdot 1^2 + 4 \cdot (-1)^2 + 4 \cdot (-1)^2 = 16$$
$$\mu_{00} = 4 \cdot 4 = 16$$
$$\eta_{20} = \frac{16}{16^2} = 0.0625$$