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I have a discrete signal (an image actually), which I am convolving/deconvolving with a zero-mean Gaussian kernel. I would like some proof that these operations do not alter the signal mean. Well, it would suffice in continuous case. Does this have to do something with the result's DC component being the product/fraction of the two signals' DC components? I have searched extensively, but found nothing of relevance. Please help.

Thanks!

EDIT: So, to avoid confusion zero-mean Gaussian is one with zero expected value, meaning it's symmetric to the y axis, and it's not about its DC component.

I've found out that a normalized Gaussian kernel (with its full-domain integral equal to 1) has its DC component as 1, leading to an unchanged signal mean.

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    If your signal has _nonzero_ mean and you feel that the mean of the result of the convolution with a zero-mean kernel equals (signal mean)$\times$(kernel mean) $= 0$, why do you think that it can be proved that the mean is _not changed_ by the convolution?2012-11-18
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    The kernel mean does not equal to its DC component, as the Fourier-transform of the Gaussian is a Gaussian. So the resulting signal doesn't have zero DC component. Nor it has zero mean.2012-11-18
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    To be honest, all the question does not make sense at all. You can not manipulate the wording "zero mean Guassian kernel". Zero mean is used for Random gaussian distributed signals. If you use it for a kernel, it should mean that the average of that kernel should equal zero. Now you say that it is not the case and the average is actually one and it solves your problem. I delete my answer.2012-11-18
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    Sorry for the confusion, I thought it's trivial, since for one: the only Gaussian kernel with zero mean is the one with infinite sigma (and constant zero values), and for two: smoothing kernels are symmetrical.2012-11-18

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