What's the difference between convolution and crosscorrelation?

[JonMuchnick]

What's the difference between convolution and crosscorrelation?

I read the answer below, but I don't know enough math to understand it.
Could someone clarify it for me, please?



"The meaning is quite different. To see why in a simple setting, consider $X$ and $Y$ independent integer valued random variables with respective distributions $p=(p_n)_n$ and $q=(q_n)_n$.

The convolution $p\ast q$ is the distribution $s=(s_n)_n$ defined by $s_n=\sum\limits_kp_kq_{n-k}=P[X+Y=n]$ for every $n$. Thus, $p\ast q$ is the distribution of $X+Y$.
The cross-correlation $p\circ q$ is the distribution $c=(c_n)_n$ defined by $c_n=\sum\limits_kp_kq_{n+k}=P[Y-X=n]$ for every $n$. Thus, $p\circ q$ is the distribution of $Y-X$.

To sum up, $\ast$acts as an addition while $\circ$ acts as a difference."
http://math.stackexchange.com/quest...onvolution-and-crosscorrelation/353309#353309

 

[MarneMath]

I generally only have encountered these in time series, so my input will come from there. The convolution is a simple (sometimes) way of modify two signals and producing a third modified signal that is often a filter. Typically you'll have two functions, one that goes on forever, the other that hangs around zero is called the filter. Therefore you can think of this third modified function is a filtered version of the input signal. The advantage of a convolution is that the operation is linear and thus the mathematics is simple.

You can think of a cross-correlation as a modified cross-covariance, except it's being divided by the product of the individual series. There's is a relationship between these two ideas. If you take the difference between the means and divide by the variance and take the convolution, you end up with the cross-correlation coefficient, which is used to test quality of a least-square fit.

I'm sure if this answered your question, but hopefully it points you in the right direction.

 

[JonMuchnick]

" If you take the difference between the means and divide by the variance and take the convolution" How would you do that? Please give an example.

 

[MarneMath]

Um, well you first get the means, then you divide it by the variances, and then apply the definition of the convolution. So, I'm going to ask you some basic questions: You do know how to find the mean, variance and follow the definition of a convolution, right? If not, then perhaps you need to step a few steps back.

 

[JonMuchnick]

I might indeed need to go a few steps back. But does understanding this thing help you to understand why people use a minussign in convolution and why people use convolution in signalprocessing, what the benefit of a flipped signal as a result of the minussign is?

 

[MarneMath]

I think that question is better suited in the electrical engineering forum. Typically, people use a convolution because a convolution has useful mathematical properties that makes handling the two signals much easier. One such property would be the convolution theorem, which I imagine would be extremely useful for an electrical engineer. In time series, you can use a cross-correlation to measure time delay. This also would seem useful for an electrical engineering doing signal process. There are other useful things you can use the cross-correlation in statistical analysis, which is what my first post was mainly getting it. So, if you want a more detail response on how to handle these with regards to signal processing, I would post in the electrical engineering sub-forum.