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What's the difference between convolution and crosscorrelation?

  1. May 1, 2013 #1
    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 [itex]X[/itex] and [itex]Y[/itex] independent integer valued random variables with respective distributions [itex]p=(p_n)_n[/itex] and [itex]q=(q_n)_n[/itex].

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

    To sum up, [itex]\ast[/itex]acts as an addition while [itex]\circ[/itex] acts as a difference."
    http://math.stackexchange.com/quest...onvolution-and-crosscorrelation/353309#353309
     
  2. jcsd
  3. May 1, 2013 #2

    MarneMath

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    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.
     
  4. May 1, 2013 #3
    " 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.
     
  5. May 1, 2013 #4

    MarneMath

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    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
  6. May 2, 2013 #5
    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?
     
  7. May 2, 2013 #6

    MarneMath

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    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.
     
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