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In a standard convolution there is no averaging of values. Help me understand this. What am I missing?
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the average part is related to the filter size of (3,3) and not related to convolution layer averaging.
Element-wise multiplication, summation and average of filter size (3,3)
But I agree the answer should have better presentation of filter calculation instead of using multiplication, summation and average
@Mubsi can the answer be summarised in more representative of filter operative calculation in relation to convolution layer.
Maybe I’m also missing something, but the only division shown on those slides is in computing the output image size:
n_{out} = \displaystyle \lfloor \frac {n_{in} + 2p - f}{s} \rfloor + 1
My expectation is that the correct answer would have said “add the bias term” in place of “average” there. I agree with @tanishk007 that I didn’t think there were any averages done in the convolution layer itself. Of course you may well follow that up with a separate average pooling layer, but that’s a different operation and is optional, right?
yes, infact I remember this answer and I actually didn’t find any of the given answer to.be completely correct.
Sounds like we need to file a bug. I’ll DM you about that @Deepti_Prasad.
Also note that this thread is tagged as about week-module-5, but I think it is meant to be week-module-4. There is no module 5 in that course. 
(Update: I just edited the title to fix that.)
Thank you for reporting this. I have raised issue in the GitHub repo and assigned @Mubsi to update the issue.
Thank you @paulinpaloalto for contributing to this discussion thread topic.
Regards
DP
