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I don't understand answer formulated in ways like this "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 $Y-X$."

Can someone explain this in an easier way? See: What's the difference between convolution and crosscorrelation?

PS: By convolution I meant the type of convolution that is used in image and signal processing. Stuff like this: http://www.songho.ca/dsp/convolution/convolution.html#convolution_2d

Community
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user68610
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  • I have deleted several non-constructive comments. – Zev Chonoles May 01 '13 at 23:51
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    It would help to include what specifically confuses you about Did's answer and the linked website. I think the website's explanation is quite clear, especially the complete example they carry out, so if you need further explanation I need to know what exactly you don't understand. – Alexander Gruber May 05 '13 at 15:19

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You can think of convolution as "flip and shift". Order doesn't matter - convolving A with B is the same as convolving B with A.

Correlation is convolution but without the "flip". Order in this case is important - correlating A with B is not the same as correlating B with A

David Wurtz
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