[Solved] Why is the output of `conv2()` is divided by `sz^2`?

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Note that

conv2(double(im(:,:,q)),mask,'same')./sz^2

is the same as

conv2(double(im(:,:,q)),mask./sz^2,'same')

This is because the convolution and the multiplication commute.

Thus, the convolution operation computes a local mean. Without the division, it would be a local sum.

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solved Why is the output of `conv2()` is divided by `sz^2`?