What's the difference between convolution and crosscorrelation?

1. May 1, 2013

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

2. May 1, 2013

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.

3. May 1, 2013

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.

4. May 1, 2013

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.

Last edited: May 1, 2013
5. May 2, 2013

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?

6. May 2, 2013

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.