A Help with the statistics of Upper Limits?

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ChrisVer
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This could as well go to the statistics, but I am looking at it from particle physics point of view...
Why adding systematic uncertainties worsen the expected upper limits to the signal strength?
I am trying to find where the flaw enters in the following logic:

0. The model most analyses use is the following likelihood:
L( N_{obs} | b(\theta ) + \mu s(\theta ) ) = P(N_{obs} |b(\theta ) + \mu s(\theta ) ) U(\mu) \Pi_i Gaus(\theta_i | 0,1)
Where N_{obs} is the observed events, b/s are the background/signal expected events, \theta_i are the different nuisance parameters and \mu is called the signal strength. In a Bayesian approach, one has to also to feed in a prior distribution for the signal strength parameter, which is the U(\mu)- let's consider it Uniform. P(x|n) is the poisson probability to get x observed given the expectation of n, and Gaus is a way to represent the variation of the nuisance parameters (given you have symmetric errors).

1. In order for one to get the expected limits, they would set N_{obs}=N_{exp}=b.

2. Once they do it, they can start varying the background+signal uncertainties, \theta_{stat} (+\theta_{sys}) [these uncertainties don't affect the signal and background in the same way]

3. On the varied result, they would try to figure out what is the \mu so that the b'+\mu s' = N_{obs}

4. Doing that several times, you get a distribution for \mu (after you marginalize over the uncertainties) which is called the posterior pdf...

5. From μ-distribution get the 95-quantile point.

Now for some reason, adding nuisance parameters (such as \theta_{sys} on top of the statistical), moves the \mu_{.95} higher.
Is that because the uncertainties are not the same for bkg/signal?
Intuitively I can see how eg by subtracting the background from the observed result with higher uncertainties is going to give a more unclear picture of how much signal you allow in the game... but I don't see where this fits in the above logic.
 
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Nuisance parameters make your µ distribution broader. Ignoring asymmetries in the Poisson distribution, they should not shift the expected µ (which should be zero if your method is sound), you just get another uncertainty that gets added in quadrature.
 
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