If only the mean if the likelihood distribution is involved, not the variance, then truly the sample size used when creating the likelihood distribution has no influence on the Bayesian update.
Then the next question is: is it a problem? If I understand you correctly then your answer is: "not really, because ".
Then it's only the part I don't get.
You ask me if it's clear to me why only the mean if the likelihood distribution is involved in the Bayesian update. Well, it isn't currently, but I'll read the article "Continuous Bayes" and see if it then becomes more clear to me:
http://www.sidhantgodiwala.com/blog/2015/03/14/continuous-bayes/
In the introductory example in the Wikipedia article on the Bayesian theorem, they start out with a prior distribution for P(machine_ID | faulty_product)* and then updates this using a likelihood distribution P(faulty_product | machine_ID) to acquire a posterior distribution for P(machine_ID | faulty_product).
How did they come up with the likelihood distribution? Maybe they sampled 100 products from each machine and for each sample counted the number of faulty products. Maybe they sampled 1.000.000 products from each machine...
We don't know which sample size is used: the likelihood distribution doesn't reveal this. Thus this matter doesn't influence the weight of the Bayesian update. But shouldn't it do so? Uncertain likelihood distributions should have a small influence and vice versa. How do I make the bayesian update reflect this?
I read the links provided by somervta in the 'Error margins' discussion from yesterday, but I'm not skillful enough to adapt them to this example.
* technically they just make the prior distribution a clone of the distribution P(machine_ID) but I like to keep the identity across the Bayesian update so I gave the prior and the posterior distribution the same form: P(machine_ID | faulty_product).