File talk:Comparison convolution correlation.svg

Convolution = Cross-correlation?
it seeems like the convolution example uses a different triangle shaped window from the top row to the lower rows... can anyone correct it?
 * The difference is intentional; in convolution, g(t) is horizontally flipped. That's what distinguishes it from cross-correlation. cm&#610;&#671;ee&#9742;&#9993; 23:23, 17 July 2012 (UTC)

The cross-correlation $$f\star g$$ is incorrect. According to the definition in the article, it should be reversed, making it look exactly like the convolution (this is consistent with the statement in the Properties section that "If f is Hermitian, then $$f\star g = f*g.$$"). It looks like this diagram was made using a different convention from that in the article. --TSchwenn (talk) 23:56, 29 September 2012 (UTC)
 * I've done a bit of checking and still think that the graphic is correct.
 * For convolution, Fig 13-5 in has similar f, g and f * g.
 * I couldn't find proper source for cross-correlation, but this video shows MATLAB giving the same result.
 * This video summarises the differences. cm&#610;&#671;ee&#9094;&#964;a&#671;&#954; 01:22, 19 September 2014 (UTC)

I found a proper source(s) for cross-correlation. For real-valued functions, it is defined as:


 * $$(f \star g)(\tau)\ \stackrel{\mathrm{def}}{=} \int_{-\infty}^{\infty} f(t)\ g(t+\tau)\,dt.$$

It follows that when the $$f$$ function is symmetrical (as in the figure we're talking about), convolution and cross-correlation are identical. The figure is wrong. --Bob K (talk) 22:00, 19 November 2016 (UTC)


 * You've convinced me with the sources and File:Comparison_convolution_james.png. I've updated the graphic and hope it's all right now. Cheers, cm&#610;&#671;ee&#9094;&#964;a&#671;&#954; 19:10, 20 November 2016 (UTC)
 * P.S. Apologies for doubting editors who pointed out my mistake previously.

Asymmetric function to illustrate autocorrelation
IMO it would be more informative if some non-symmetric function (for example g function instead of square f) were used tu illustrate autocorrelation. Marioosz (talk) 17:41, 17 January 2013 (UTC)
 * ✅ cm&#610;&#671;ee&#9094;&#964;a&#671;&#954; 02:03, 19 September 2014 (UTC)

Trying to understand geometric intuition behind correlation’s proportionality to area of intersection of areas under functions
I love your work and am so grateful. I've been staring at this (particularly the correlation ones) and trying to think through why the nice color-coded area of intersection (i.e. overlap) is proportionate to (not equal to, is it?) the correlation. Here's how far I think I've gotten:

Correlation is covariance divided by product of standard deviations. You can get covariance by going over each point and averaging the product of the distances from the mean for each function: (1/n) * Σ(f_i - μ_f)*(g_i - μ_g). In your example functions, if we imagine they're zero forever to the left and right, the mean is zero, so this simplifies to just averaging the product of the values at each point: (1/n) * Σ f_i * g_i. That's not exactly the area of intersection, but it's the area of the product. I'm trying to think when that would be the same as the area of the intersection. It's zero when one of them is zero, at least. If one is 1 and the other's 0.5, then the product is also the intersection. But if they're both 0.5, the intersection is 0.5 but the product is 0.25. If one is negative and one is positive at that point, then the overlap is zero but the contribution to covariance is the product, which is nonzero negative.

And then ultimately you're dividing by the product of the standard deviations, but that just sorta rescales it the whole thing; I don't *think* it would make product and overlap equal, but maybe....?

That's as far as I've gotten and I don't feel very confident in even that hahah. Does that sorta make sense? How would you explain it? Thank you so much.

Tophtucker (talk) 00:23, 27 August 2018 (UTC)