Hi,
No this is not correct. Each kernel has same amount of channels as the input.



 Wesley_Neill:

my filter is a 3D stack, where each layer is an identical copy of that 2D kernel.


In our earlier discussion, Why add an extra dimension to convolution layer weights? 4, we agreed that if input has [3, h, w], and if we want to have 10 output channels, then we run 10 different filters which each has size of [3, k, k] which creates [1, h, w] responses then we stack all of them which in our case we wanted 10 output channels so we have 10 [1, h, w] responses then by stacking them along channel dim we get [10, h ,w]. About the values, assume a 2D gaussian for [1, k, k] filter, now assume a 3D gaussian for [3, k, k] filter.
So,

All values distributed from a function which leads to different values in each channel (can be identical but no reason for that)
Based on my example, no, we have [4, 3, f, f] filters but output is [4, h ,w] and there is no averaging as responses have only one and channel we stack them.
based on 2
About this I am not really sure about the idea, do you mean skipping some windows in convolution? or skipping some filters?

Bests