# Image averaging for impulse noise

I understand the mathematics behind the use of image averaging for gaussian noise. Why can't it be used for impulse noise ? Would it be because there is no way to prove that impulse noise is additive ? And that it has no mean ?

Also, am I right to say that impulse noise does not affect every pixel in an image, as opposed to gaussian noise ?

Thanks !

• Anyone any comments ? – RuiQi Apr 26 '16 at 13:06

I'm not in the image processing filed, but I would say:
Yes, Gaussian noise means that each of your pixel is assumed to be produced by a stochastic process. Hence you can for instance think of it as a "real" intensity corrupted by an additive normally distributed white noise. Normally distributed means also that if you take $n$ observations of the same pixel you most likely will end up with $n$ wrong measurements. I think this is actually the worst kind of noise since given any pixel, then any other pixel inside a given neighbourhood will be corrupted, hence telling which observation is to trust more is difficult. The way I would go is probably (low pass) filtering.

You can think to the impulsive noise again as an additive noise potentially present in all the pixels, but potentially also localized in only a subset of pixels or observations (either it is or it is not). I personally would not go for averaging or low pass filtering since it will never erase completely the noise.
Assume indeed to have $n$ images of the same scene. Assume that for a given pixel $p$ whose "actual" value is $p=0$ you measure it as $255$ in one image and as $0$ in $n$ images. If you take the mean then you will chose $p = 255/n$ as the actual value. Infact a low pass filter is a process with infinite impulse response and it tends to zero only asymptotically.
In such case for me it looks more effective to take the median, over the $n$ samples. With the median you obtain that you estimate of the actual values of $p$ is exactly $0$.
You can also take the most-frequent value (the mode) and it will perform better than the mean. In general you can also use a median or a mode inside the same image by taking them in a neighbourhood of a given pixel. Infact for example, take a $10\times 10$ sub-image and assume it corresponds to a white background. Assume that 30 over the 100 pixels are black due to the impulsive noise, then you can substitute to each pixel the median over the entire sub-image obtaining a completely white patch. A mean will lead to a 30% grey (76 / 255)

• Hi thanks ! But wouldn't it work if I have many images ? 1000 for example. I am just curious since I can't seem to find anything online that talks about the use of image averaging for impulse noise. – RuiQi Apr 26 '16 at 15:04
• well if you take 1000 images in which the same pixel $p$ is white (255) in 200 and black (0) in 700 you will end up with an estimate of $p$ of $\hat{p} = (255*200 + 0*700)/1000 = 51$ therefore you will have a 80% gray pixel instead of a black pixel. In general it depends on the probability of the impulse to be there. – LJSilver Apr 26 '16 at 15:08
• Ah i get it now. The 'mathematical proof' would be to use probability. Thanks so much ! – RuiQi Apr 26 '16 at 15:12
• well, It depends on what you want to prove. What do you want to prove? – LJSilver Apr 26 '16 at 15:13
• For image with gaussian noise, we can treat it as an addition of the original image with the noise i.e. f(x) + n(x). Image averaging then reduces the noise variance and if I have infinity images then theoretically the noise is eliminated. Can I show the same with impulse noise ? i.e. f(x) + impulse(x). If I can, then theoretically, wouldn't averaging over infinity images eliminate the impulse noise ? – RuiQi Apr 26 '16 at 15:22