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Mar 20, 2020 at 14:16 comment added user9947 I think being trainable also doesn't affect much the intuition, since it is basically selecting the projections which is most useful. For example in a car moving problem a projection selecting measurements of acceleration and velocity will be much more useful in predicting the displacement for future states as compared to displacement and velocity (if we know the initial starting position, $s=ut + 0.5at^2$). This is very informal and vague but maybe useful.
Mar 20, 2020 at 14:10 comment added Alexander Soare I think I'll rewind to before you said anything. I was happy then :P
Mar 20, 2020 at 14:09 comment added user9947 That's very unlikely. But I was just asking, its a linear connection so being trainable also might not affect much. Can't really say anything.
Mar 20, 2020 at 14:04 comment added Alexander Soare What if someone got it wrong then everyone copied them... :|
Mar 20, 2020 at 14:03 comment added Alexander Soare Here's another implementation in tensorflow. Line 337 defines the projection_shortcut. I'm not too familiar with tf but it looks like a vanilla 1x1 conv to me.
Mar 20, 2020 at 14:01 comment added user9947 It kind of makes a difference, for example in Kalman Filter we use a projection/measurement to rectify a state whose dimensions are not the same. Thus all the information about the correct state is contained in the projection.
Mar 20, 2020 at 13:59 comment added Alexander Soare If it's not, that will make a big difference to the way I think about this. Maybe I'll go double check by looking at more implementations. They keep calling it a "projection" shortcut in the papers, and to me projection means collapsing one dimension of a vector down to zero, so I'm not sure it relates
Mar 20, 2020 at 13:37 comment added user9947 Are you sure the $W_s$ is trainable?
Mar 20, 2020 at 12:43 answer added Alexander Soare timeline score: 3
Mar 19, 2020 at 17:59 history asked Alexander Soare CC BY-SA 4.0