I saw this exercise and its solution, but I'm not sure about it. (I'm gonna quote the solution and after I'll say why I have doubts)

According http://mathfaculty.fullerton.edu/sbehseta/Math538_Elena_Quiz%203.pdf, I have the following problem.

Problem 10.1: Number of simulation draws. Suppose that scalar variable $\theta$ is approximately normally distributed in a posterior distribution that is summarized by n independent simulation draws. How large does n have to be so that the 2.5% and 97.5% quantiles of $\theta$ are specified to an accuracy of 0.1sd($\theta|y$). Figure this out mathematically.

And the following solution (also given by the same author of the paper)

(Without mention why) The text says that the posterior probabilities are estimated to a standard deviation of $\sqrt{p(1-p)/S}$, where S is the number of draws. Suppose we have priors $\mu=\tau=\sigma=1$ and ${y}_{1},...,{y}_{n}|\theta\sim N(\theta,\sigma=1)$ and $\theta\sim N(\mu=1,\tau=1)$. Then the posterior is $\theta|y\sim N(\frac{1+n\overline{y}}{1+n},\frac{1}{2})$.

Suppose we want to know the amount of samples needed to find the probability of $\theta\ge 2$. We know, based on our assumptions above that $\mathbb{P}(\theta\ge 2|$y)=0.91. Then, $S=\frac{p(1-p)}{0.01^{2}}=\frac{.91(1-.91)}{0.01^{2}}\approx 853.3$ and we need 854 draws.

Now, these are my doubts:

  1. Why can we estimate the sd with $\sqrt{p(1-p)/S}$?

  2. Why $\mathbb{P}(\theta\ge 2|$y)$=0.91$? To me, this value don't match, don't care the parameters of the distribution (At least at this example). Also, I think p should be 0.975 because we are looking for 97.5% quantile.

  3. Clearly the calculus of the number of draws is incorrect, but whatever, why are they using 0.01 (instead of 0.1)? According to the proposed posterior, $sd(\theta|y)=\sqrt{1/2}$

Summarizing, I thought the solution of the problem assuming that que estimation of s.d. proposed by the paper is correct (Even when I'm not sure). And i'd only make changes that I wrote at 2. and 3.

  • 1
    $\begingroup$ Since this is clearly an exercise, please add the self-study tag and try to expand on what you attempted and where exactly you get stick. Reference for the exercise would also be needed. $\endgroup$ – Xi'an Mar 18 '17 at 12:36
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    $\begingroup$ Thanks @Xi'an , I edited my question. I attempted almost the same thing but changing (mainly) the probability given. (I can't figure out why they are using 0.91). I understand the priors election, but the development of the given solution is not clear for me $\endgroup$ – student Mar 18 '17 at 16:51

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