# How does convolution/deconvolution with gaussian affect signal mean

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? –  Dilip Sarwate Nov 18 '12 at 15:06
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. –  user49855 Nov 18 '12 at 15:20
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. –  Seyhmus Güngören Nov 18 '12 at 16:07
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. –  user49855 Nov 18 '12 at 16:21

Suppose $x(t)$ and $y(t)$ are finite-energy signals with with finite "DC" values $\bar{x}=\int_{-\infty}^\infty x(t)\,\mathrm dt$ and $\bar{y}=\int_{-\infty}^\infty y(t)\,\mathrm dt$. Their convolution $z(t)$ is given by $$z(t) = \int_{-\infty}^{\infty} x(t-\tau)y(\tau)\,\mathrm d\tau,$$ and its "DC" value is \begin{align*} \bar{z} &= \int_{-\infty}^{\infty} z(t)\,\mathrm dt\\ &= \int_{-\infty}^{\infty}\left[ \int_{-\infty}^{\infty} x(t-\tau)y(\tau)\,\mathrm d\tau\right]\,\mathrm dt\\ &= \int_{-\infty}^{\infty}y(\tau)\left[ \int_{-\infty}^{\infty} x(t-\tau)\,\mathrm dt\right]\,\mathrm d\tau\\ &= \int_{-\infty}^{\infty}y(\tau)\cdot \bar{x}\,\mathrm d\tau &\text{using a change of variable in the inner integral}\\ &= \bar{x}\cdot\bar{y} \end{align*} If $\bar{y} = 1$, which condition is guaranteed to hold whenever $y(t) = \frac{1}{\sigma\sqrt{2\pi}}\exp(-t^2/2\sigma^2)$ or any other valid probability density function, we have that $\bar{z} = \bar{x}$, that is, the "DC" value is unchanged by the convolution. Note that neither Gaussianity nor being a valid probability density function has anything to do with the matter. $\bar{z}$ would be the same as $\bar{x}$ if $y(t)$ were given by $$y(t) = \begin{cases}\sin(t), & 0 \leq t \leq 3\pi/2,\\ 0. &\text{otherwise,}\end{cases}$$ which is not a probability density function (nor can $cy(t)$ be a probability density function for any choice of $c$) but which does have the property that $\bar{y} = 1$.