Image processing - Is mean filter shift-invariant?

Question

Update:
So, according to this article.
For a 2D mean filter,
$$ H[f(x,y)]=\frac{1}{NM}\sum_{k=0}^{N-1}\sum_{p=0}^{M-1}f(x-k,y-p)=g(x,y)\tag{1} $$ $$ g(x+x_0,y+y_0)=\frac{1}{NM}\sum_{k=0}^{N-1}\sum_{p=0}^{M-1}f(x+x_0-k,y+y_0-p)\tag{2} $$ $$ H[f(x+x_0,y+y_0)]=\frac{1}{NM}\sum_{k=0}^{N-1}\sum_{p=0}^{M-1}f(x+x_0-k,y+y_0-p)\tag{3} $$ because (2) equals to (3), a 2D mean filter is shift-invariant.
Is the proof right?


BTW, if f(x,y) is a 2D image, such as

1 2 1
2 3 2
1 2 1

and g(x,y) is (using zero padding)

 8/9 11/9  8/9
11/9 15/9 11/9
 8/9 11/9  8/9

What does these $$ H[f(x+x_0,y+y_0)]\tag{4} $$ and $$ g(x+x_0,y+y_0)\tag{5} $$ mean? (I want to prove it by using a image.)


Origin:
I read this article and this Wikipedia page.
However, I still don't know what it means for (or how to apply to) a mean filter.
Can anyone give me an image example?
(for example, https://dsp.stackexchange.com/a/14435 is easier to understand)

Answer 1

The 1D mean filter defined by the following I/O relation is time/shift invariant;

$$y[n] = T\{x[n]\} = \frac{1}{d_1 + d_2 + 1} \sum_{k=n-d_1}^{n+d_2}{x[k]}$$

For $d_1$ and $d_2$ being positive integers, the filter computes the mean (or the average) of its input $x[n]$ in the range of its summation index. To show that it is shift invariant we shall simply show the truth of the following statement: $y[n-d] = T\{x[n-d]\}$ for and $d$ and any input $x[n]$

$$ y[n-d] = \frac{1}{d_1 + d_2 + 1} \sum_{k=n-d-d_1}^{n-d+d_2}{x[k]} ~~~\overset{?}{=}~~~ T\{x[n-d]\} = \frac{1}{d_1 + d_2 + 1} \sum_{k=n-d_1}^{n+d_2}{x[k-d]}$$

Which is easily shown to be true by making the simple change of variable in the second sum as $k-d=k'$ and then optionally replacing $k'$ with $k$ again: \begin{align} T\{x[n-d]\} &= \frac{1}{d_1 + d_2 + 1} \sum_{k=n-d_1}^{n+d_2}{x[k-d]} ~~~~ &\scriptstyle{\text{replace k-d with k'}}\\ T\{x[n-d]\} &= \frac{1}{d_1 + d_2 + 1} \sum_{k'=n-d-d_1}^{n-d+d_2}{x[k']} &\scriptstyle{\text{replace k' with k again}}\\ T\{x[n-d]\} &= \frac{1}{d_1 + d_2 + 1} \sum_{k=n-d-d_1}^{n-d+d_2}{x[k]} &\scriptstyle{\text{recognise that this is = y[n-d]}}\\ T\{x[n-d]\} &= y[n-d] &\scriptstyle{\text{Hence the given System is LTI}}\\ \end{align}

This is extended to 2D in a straight-forward manner

Answer 2

Suppose your mean filter is: $$ y[n] = \frac{1}{N} \sum_{k=0}^{N-1} x[n-k] \tag{1} $$ then shift invariance means that if an input $x_1[n]$ generates $y_1[n]$ then an input $x_2[n] = x_1[n-n_0]$ results in $y_2[n] = y_1[n - n_0]$

For an 2D mean filter, $$ y[n,m] = \frac{1}{NM} \sum_{k=0}^{N-1} \sum_{p=0}^{M-1} x[n-k, m-p] \tag{2} $$ then shift invariance means that if an input $x_1[n, m]$ generates $y_1[n, m]$ then an input $x_2[n, m] = x_1[n-n_0, m-m_0]$ results in $y_2[n, m] = y_1[n - n_0, m - m_0]$

What part is unclear? All you have to do is apply the definition to the formulae (1) or (2) and see if the conditions are true.

Answer 3

At the risk of sounding picky, I would say that:

• the mean filter, seen as a system, is shift-invariant (or translation-invariant),
• the mean operation is shift equi-variant (its results moves equally with the shift)

The difference between the two ideas is explained at Difference between “equivariant to translation” and “invariant to translation”.

Here, as a system, you are concerned with the following: if the input is delayed, is the result delayed as well? In other words: if you combine temporal values in a mean today, or next year, will it yield the same result (time-invariance)? If you combine spatial values in a mean here, or a few meters away, will it yield the same result (space-invariance)?

The mean filter works this way. To DSP engineers, it is called shift-invariant.