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 $\eta$?
[1] https://en.wikipedia.org/wiki/Image_moment

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.
