Comments (7)
Provided the integral is integrated between [0, 1] (which it always should be), then where does this become an issue? The log-signature never needs to be normalised by the interval length.
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The integral isn't always integrated over [0, 1]. (Why should it always be?)
I think all that's necessary is removing the multiplication on this line:
Line 81 in 1cc566c
(Or possibly make it a divide? Would need to check more carefully.)
from torchcde.
The log-ODE method is defined as:
Z_b = Z_a + int_0^1 f(Z)logsig_[a, b](X) ds
There is never any need to evaluate an integral with limits that are not [0, 1].
Perhaps I am misunderstanding, but if it is done in this way then you do not need to worry about dividing by b-a.
from torchcde.
What you've written is correct if a=0, b=1.
We can certainly reparameterise for that to be true each time... by dividing by b - a.
from torchcde.
Check definition A.8. in https://arxiv.org/pdf/2009.08295.pdf
The integral is over [0, 1] for any interval [a, b]. We divide by (b-a) if we integrate from a to b.
from torchcde.
Okay, flipping this around: integrating over [0, 1] (or rather [n, n+1] later on) is indeed what torchcde will be doing by default later on.
Over that region it should be just the logsignature, as you say. Therefore the multiplication I highlighted above should be removed?
To emphasise this: the tests for this check that the derivative on each piece is equal to the logsignature -- whilst using the optional t
argument. (They predate the t
argument being optional.) So the tests are verifying that over an interval determined by t (i.e. not of unit length), the vector field is the logsignature (i.e. without a b-a normalisation). Which is a mismatch.
Do you agree?
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Closed in version 0.2.0 with the introduction of logsig_windows
.
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Related Issues (20)
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