Convolution Finite vs Infinite Support

Question

It is known that the convolution of two Gaussian function is also a scaled Gaussian function. This convolution is taken from $–\infty$ to $\infty$ since the Gaussian function has infinite support.

Convolutions are typically used as linear filters to smooth a function observed over some data points. An example is a data set where $x=[0,1,2,3,4,5], y=[10,13,16,14,30]$.

Typically people smooth out these data through a convolution with a kernel (Gaussian Kernel) using an infinite integral. Why don’t they only integrate over the space the data points are observed in. Ex integrate over $[0,5]$.

I'm not sure I fully understand the question. One can freely extend the data to be zero outside of where it was observed and then integrate from $-\infty$ to $\infty$, getting the same result as you would if you had merely integrated where the data was observed. What is the drawback to doing this? — Ian, Apr 25 '16 at 17:54
@Ian Thus if I smooth with a gaussian kernel over -Inf to Inf I will get the same result as smoothing from [0,5] ? — raK1, Apr 25 '16 at 18:32
You need to be careful how exactly you write things out, but if the support of $f$ is in $[0,5]$ and $g$ is a Gaussian density then $\int_{-\infty}^\infty f(y) g(x-y) dy = \int_0^5 f(y) g(x-y) dy$. In this particular sense the answer to your question is "yes". — Ian, Apr 25 '16 at 18:40
@Ian can you pelase provide me an explanation with why the equation in your last comment holds given that g is a Gaussian density,Thanks — raK1, Apr 25 '16 at 20:34
It is simply because $f$ is zero except for $y$ between $0$ and $5$. — Ian, Apr 25 '16 at 20:36
@Ian my last question is what if f is also a Gaussian density, however my observations of x are only in the [0,5]. I am thinking of this diffrently since the support of f is infinite, however I am only have observation in the x=[0,5]. Will teh same equation in your previous comment hold ? thank you a lot — raK1, Apr 25 '16 at 21:01
You'll have to be more precise about what you mean by that. $f$ is either a Gaussian density with infinite support or it's not. You may be talking about some kind of distributional approximation by data or something, but then you are not really convolving a Gaussian density with another Gaussian density. — Ian, Apr 25 '16 at 21:19
@Ian yes exactly, I am trying to smooth out data using gaussain convolution. But for instance assume that the data can only take positive values then can I convolve it with and infinite integral ? — raK1, Apr 25 '16 at 21:23
You can use an infinite integral as you wish; if you had a $L^1$ function then you'll get a smooth result. But you should be careful by how you turn "data" into a $L^1$ function. — Ian, Apr 25 '16 at 21:25
@Ian I am sorry for the repetitive questions. So the convolution of 2 gaussians a gaussain, what if I say that the support of both my gaussians f and g is [0,5] and the perform a finite integration will I get the same result in the [0,5] area if I integrate over infinity — raK1, Apr 25 '16 at 21:39
Yes, but the result won't be of the same type that you started with. — Ian, Apr 25 '16 at 21:40
@Ian, i dont iunderstand what do you mean by differnt type ? is that that the finite and infinit econvolution will have different results in the space outside [0,5] and similair in the space in [0,5] — raK1, Apr 25 '16 at 21:41
Your convolution of two Gaussians in $[0,5]$ won't be a Gaussian in $[0,5]$. The easier way to see this is to look at indicator functions: the convolution of the indicator function of $[0,1]$ with itself is in fact a "hat function" on $[0,2]$. — Ian, Apr 25 '16 at 21:47
@ian How can the finite convolution not be Gaussian and the result of both finite and infinite convolutions is the the same in the [0,5] support ? — raK1, Apr 25 '16 at 21:48
Try it. You should find that the convolution of a Gaussian truncated to $[0,5]$ with itself is supported in $[0,10]$. I'm not sure it will have the Gaussian form, either. — Ian, Apr 25 '16 at 21:52

Convolution Finite vs Infinite Support

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