Perform posterior predictive checks by choosing your own statistics to compare,

Question

Perform posterior predictive checks by choosing your own statistics to compare, and assess the fit of the Normal model for these data.

Example: Midge wing length Grogan and Wirth (1981) provide data on the wing length in millimeters of nine members of a species of midge (small, two-winged flies). From these nine measurements we wish to make inference on the population mean . Studies from other populations suggest that wing lengths are typically around 1.9 mm, and so we set = 1.9. We also know that lengths must be positive. implying that > 0" Therefore, ideally we would use a prior distribution for that has mass only on > 0, We can approximate this restriction with a normal prior distribution for as follows: Since for any normal distribution 5.3 Joint inference for the mean and variance 73 most of the probability is within two standard deviations of the mean, we choose T so that 0-2 × To 0, or equivalently To 1.9/2 = 0.95. For now, we take 70-0.95, which somewhat overstates our prior uncertainty about . The observations in order of increasing magnitude are (1.64, 1.70, 1.72 1.74, 1.82, 1.82, 1.82, 1.90, 2.08), giving y= 1.804. Using the formulae above for Hm and , we have31, , 40, 2} ~ normal (Hm, T ), where 0 ,y 1.11 1.94 1.804 1.11 0 rt 0 If 2 82-0.017, then yi, , 40, 2-0.017) ~ normal (1.805, 0.002) A 95% quantile-based confidence interval for based on this distribution is (1.72, 1.89). However, this interval assumes that we are certain that 2-82 when in fact s2 is only a rough estimate of 2 based on only nine observations To get a more accurate representation of our information we need to account for the fact that 2 is not known

Explanation / Answer

We assume that 2 is not known

For the given 9 readings, = 1.804

For n c 9, 2 = 0.017

Using the formula 2 = 1/ (1.11 + n/2) and using first estimate 2 = 0.017

at n = 10, 2 = 0.0016968 or = 0.0412

1 =  [0/ 02 + (n/02)y']/ [1/02 + n/02] = 1.8525

1 21 > 0

1 < 1.8525/2 or 0.92625

12 = 1/ (1.11 + 11/12)

Using 1 as 0.92625, we can estinmate the next 2

We have to keep iterating till, the value of 2 starts converging

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