Sorry for the delay - I have been away.
Yes, it starts out as ordinary photon noise, but the question is why it is obvious in the final image. That is because (1) the red channel is not silent, as ideally it would be when blue light falls on the sensor, because the spectral responses of the channels are not perfectly separated, and (2) noise is imported from the red channel during raw processing. The effect of the second point is that channel noise is different for different raw converters. http://www.libraw.org/articles/channel-noise-and-raw-converters.html has examples.
Thanks for replying.
Have you looked at the RGB composition of the "blue colour" of a "blue-sky" ? I just looked at one of my pictures and found, for the "blue sky", Blue=200, Green=100, Red=50. The numbers are rounded a bit and obviously one expects some variation depending upon "conditions". However the approximate 4:2:1 ratio is seen in a number of reports when searching for "blue sky spectrum" (
https://www.google.co.uk/search?q=blue+sky+spectrum&client=firefox-b&source=lnms&tbm=isch&sa=X&ved=0ahUKEwjm8Jqqgt3PAhVfFMAKHarcDvYQ_AUICCgB&biw=1920&bih=1091#imgrc=uwXQaJSamV9nqM%3A ).
So the "blue sky" contains 4 times as much blue light as red light. Looking at photon (shot) noise alone, this means that the signal-to noise (SNR) ratio of the blue channel is twice that of the red channel. We describe the red channel as being more "noisy" - because what is noticeable is the SNR and not the absolute amount of noise.
I do not say that demosaicing does not contribute to noise, but I'm not sure that it's the most important factor here.
Are you suggesting that perfect spectral separation between channels is desirable ? How would a sensor which achieved this recognize yellow ?