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Papers like Briggs et al. 2002 say that logical constraints on inputs such as probability parameters exclude the the Normal distribution from consideration due to its unboundedness. In this example, values below 0 and above 1. They briefly describe the Bayesian derivation from the proportion data (binomial) to the beta, for use as the sampling distribution for probabilities in the probabilistic sensitivity analysis via Monte Carlo simulation. The beta can take on a normal-like shape, but it can also have very non-normal shapes.

  1. Can this Bayesian derivation be reconciled with the frequentist central limit theorem that says with a sufficiently large sample size the sampling distribution of the sampling mean will be approximately normal? For resource use (ie. number of visits to the doctor) Briggs et al 2002 use a Gamma distribution, as shaped in Table 4, which is certainly not Normal. Normality is what would be expected for sufficiently large samples in the frequentist interpretation.

  2. In a Monte Carlo simulation, if a Normal distribution was used to draw realized values for probabilities, there would be some iterations where values <0 or >1 are observed. However, for symmetric distributions this should cancel out asymptotically. The distribution is still centered around the mean which is within its logical bounds, and Monte Carlo considers aggregate results not results of a single iteration. So I'm wondering why we care about the logical constraints of parameters such as probabilities, costs, etc. within the context of a Monte Carlo simulation?

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For your first point: Not sure what you mean by reconciled, as I think Bayesians do not challenge the CLT. The use of the Gamma appears fine...note that the Gamma $\rightarrow$ Normal as the mean approaches $\infty$. For a Bayesian analysis, your prior should exclude parameter values that cannot happen, hence they are correct than an unbounded normal would not be applicable. What the CLT says is that the normal distribution becomes a better and better approximation to the sample mean distribution, hence also to the sum. The Beta, Gamma and Normal form an asymptotic triad, with the Gamma being the limit of the Beta as you increase the right tail to infinity, holding the mean and variance constant. The Normal is the limit of the Gamma as you allow the mean to go to infinity holding the variance constant (hence the centered Gamma gets a longer and longer left tail too). So I don't see an issue here.

Item 2: Depending on your problem, the tail values can make a large impact on your analysis. Also, if you used the normal to generate probabilities, how would you actaully use a value of -4 or 50? Those are not proabiliites and I don't know how you would use them to make the simulatoin "balance out". Also, having these illicit tails in your simulation will jack up your variance without increasing yoru accuracy, which is not helpful. If you think that the normal distribution, as a shape, holds well, I would use a truncated Normal. If it is a good approximation, the tail probability you are truncating should be very small.

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  • $\begingroup$ With regard to your first paragraph: what do you mean by "as the mean approaches infinity?" $\endgroup$ – kirk Nov 18 '13 at 15:07
  • $\begingroup$ And again, a value of -4 is definitely not logical, but my question is why does that matter in the context of a Monte Carlo simulation? It is the aggregate of the iterations not the individual draws that matter. That illogical draw will not be used in isolation $\endgroup$ – kirk Nov 18 '13 at 15:13
  • $\begingroup$ If you select $k$ and $\alpha$ for the Gamma such that the mean approaches infinity with constant standard deviation, it will approach a normal distribution. $\endgroup$ – user31668 Nov 18 '13 at 20:07
  • $\begingroup$ As for monte carlo, it doesn't consider anything. you need to tell it what to calculate. What statistics are you calculating from the Monte Carlo simulation? $\endgroup$ – user31668 Nov 18 '13 at 20:08
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    $\begingroup$ (cont'd): Also note that they are using the beta in a sensitivity analysis. The sensitivity depends not only on the mean of the input, but the model's response to changes in that input. In general, the function of the mean of a random quantity does not equal to mean of a function of a random quanity: $f(E[X])\neq E[f(X)]$, so while simply drawing proportions and averaging them will be insensitive to outliers, the model's response to such outliers may not be linear and symmetric. Also, its good form to not allow your model to take on unrealistic values, inviting more criticism $\endgroup$ – user31668 Nov 19 '13 at 14:24

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