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and 1; will be closer to X.
Knowing the mean and standard deviation, we can use the normal distribu-
tion to find an interval with 95% probability for u. This 95% credibility interval is
[110.502, 123.038]. For comparison the 95% confidence interval using ¥ = 118.28
and o = 15 is ¥+ 1.960/,/n = [111.35, 125.21]. Notice that this interval must
be wider. Because the precisions add to give the posterior precision, the posterior
precision is greater than the prior precision and it is greater than the data precision.
Therefore, it is guaranteed that the posterior standard deviation ¢, will be less than
@ and less than the data standard deviation a/ \/n.
Both the credibility interval and the confidence interval exclude 110, so we
can be pretty sure that jt exceeds 110. Another way of looking at this is to calculate
the posterior probability of 4 being less than or equal to 110. Using 4, = 116.77
and o; = 3.198, we obtain the probability .0171, so this too supports the idea that jo
exceeds 110.
How should we go about choosing Ho and oo for the prior distribution?
Suppose we have four prior observations for which the mean is 110. The standard
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782 = cuarrer 14 Alternative Approaches to Inference
deviation of the mean is 15/V/4. We therefore choose fg = 110 and og = 7.5,
the same values used for this example. If the four values are combined with the 18
values from the data set, then the mean of all 22 is 116.77 = y, and the standard
deviation is 15/22 = 3.198 =. The 95% confidence interval for the mean,
based on the average of all 22 observations, is the same as the Bayesian 95%
credibility interval. This says that if you have some preliminary data values that are
just as good as the regular data values that will be obtained, then base the prior
distribution on the preliminary data. The posterior mean and its standard deviation
will be the same as if the preliminary data were combined with the regular data, and
the 95% credibility interval will be the same as the 95% confidence interval.
It should be emphasized that, even if the confidence interval is the same as
the credibility interval, they have different interpretations. To interpret the Bayes-
ian credibility interval, we can say that the probability is 95% that y is in the
interval [110.502, 123.038]. However, for the frequentist confidence interval such a
probability statement does not make sense because j and the endpoints of the
interval are all constants after the interval has been calculated. Instead we have the
more complicated interpretation that, in repeated realizations of the confidence
interval, 95% of the intervals will include the true y in the long run.
What should be done if there are no prior observations and there are no strong
opinions about the prior mean jig? In this case the prior standard deviation og can be
taken as some large number much bigger than , such as gy = 1000 in our example.
The result is that the prior will have essentially no effect, and the posterior distribu-
tion will be based on the data, “4, = = 118.28 and o, =o = 15. The 95%
credibility interval will be the same as the 95% confidence interval based on the 18
observations, [111.35, 125.21], but of course the interpretation is different. i |
In both examples it turned out that the posterior distribution has the same form as
the prior distribution. When this happens we say that the prior distribution is
conjugate to the data distribution. Exercises 31 and 32 offer additional examples
of conjugate distributions.
Exercises | Section 14.4 (23-32)
23. For the data of Example 14.7 assume a beta prior using the 19 observations. Compare with the
distribution and assume that the 1574 observa- result of (b).
tions are a random sample from the Bernoulli d. Change the prior so the prior precision is very
distribution. Use Bayes’ theorem to derive the small but positive, and then recompute (a) and (b).
posterior distribution, and compare your answer e. Find a 95% confidence interval for using the
with the result of Example 14.7. 15 observations and compare with the credibil-
24. Here are the IQ scores for the 15 first-grade girls ifjintecvalor (dy:
from the study mentioned in Example 14.8. 25. Laplace’s rule of succession says that if there
102 96 106 118 108 122 115 113 have been n Bernoulli trials and they have all
109 113. 82 110 121 110. 99 been successes, then the probability of a success
on the next trial is (m + 1)/(n + 2). For the deri-
Assume the same prior distribution used in vation Laplace used a beta prior with a = I and
Example 14.8, and assume that the data is a ran- b = 1 for binomial data, as in Example 14.7.
dom sample from a normal distribution with mean a. Show that, if @ = 1 and b = | and there are n
wand o = 15. successes in n trials, then the posterior mean of
a, Find the posterior distribution of j.. pis(n+1)/(n +2).
b. Find a 95% credibility interval for p. b. Explain (a) in terms of total successes and
¢. Add four observations with average 110 to the failures; that is, explain the result in terms of
data and find a 95% confidence interval for jv two prior trials plus 7 later trials.
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Supplementary Exercises 783
¢. Laplace applied his rule of succession to f. Calculate a 95% confidence interval for p using
compute the probability that the sun will rise Equation (8.11) of Section 8.2, and compare
tomorrow using 5000 years, or n = 1,826,214 with the results of (d) and (e).
days of history in which the sun rose every day. g. Compare the interpretations of the credibility
Is Laplace’s method equivalent to including interval and the confidence intervals.
two prior days when the sun rose once and h. Based on the prior in (c), test the hypothesis
failed to rise once? Criticize the answer in p < 5 using the posterior distribution to find
terms of total successes and failures. P(p <5).