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CHAPTER 11 Bayesian Methods
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11.82. In Problem 11.74, perform the test in each case using the Bayes factor rule to reject the null hypothesis if BF 4.
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5P(H0 u x)>[1 P(H0 u x)]6 (a) BF reject the null hypothesis. (b) BF 5P(H0 u x)>[1 P(H0 u x)]6 cannot reject the null hypothesis. (c) BF 5P(H0 u x)>[1 P(H0 u x)]6 cannot reject the null hypothesis. 5P(H0)>[1 5P(H0)>[1 5P(H0)>[1 P(H0)]6 P(H0)]6 P(H0)]6 (0.12>0.88) (0.07>0.93) (0.04>0.96) (0.04>0.96) < 3.27. We (0.008>0.992) < 9.33. We (0.002>0.998) < 20.79. We
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we know that the prior distribution of u is normal with mean 0.4 and variance 0.13; therefore, 0.3 0.4 < 0.39. We cannot reject the null hypothesis. P(H0) P Z 0.361
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11.83. In Problem 11.21, determine what can be concluded using the Bayes factor criterion: Reject H0 if BF
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Since the prior distribution in this problem is improper, the prior odds ratio is not defined. Therefore, the Bayes factor criterion cannot be employed here.
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11.84. Suppose that in Example 11.18 a second sample of 100 observations yielded a mean reaction time of 0.25 sec. Test the null hypothesis H0 : u 0.3 against the alternativeH1 : u 0.3 using the Bayes factor criterion to reject the null hypothesis if BF 0.05.
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We take the prior distribution of u to be the posterior distribution obtained in Example 11.18: Normal with mean 0.352 and variance 0.004. Applying Theorem 11-3 with this prior and the new data, we get a posterior mean 0.269 and variance 0.0007. Using this, we obtain the posterior probability of the null hypothesis as 0.12. Note that the prior probability of the null hypothesis that is needed for calculating the Bayes factor in this problem should be based on the prior distribution given in Example 11.18: Normal with mean 0.4 and variance 0.13. Using this, we get the prior probability of the null hypothesis as 0.61. The Bayes factor is 0.087. Since this is larger than 0.05, we cannot reject the null hypothesis.
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11.85. In Problem 11.48, test the null hypothesis H0 : u to reject the null hypothesis if BF 1.
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0.7 against H1 : u
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0.7 using the Bayes factor rule
The prior distribution of u is gamma with parameters 0.16 and 2.5. The posterior distribution of u is gamma with parameters 10.16 and 0.04. The prior and posterior probabilities of the null hypothesis are respectively 0.154 and 0.022. Since P(H0 u x) P(H0), we reject the null hypothesis (see Problem 11.79).
Bayesian predictive distributions 11.86. The random variable X has a Bernoulli distribution with success probability u, which has a prior beta distribution with parameters a b 2. A sample of n trials on X yielded n successes. A future sample of two trials is being contemplated. Find (a) the predictive distribution of the number of future successes and (b) the predictive mean.
(a) P(Y a f
*(y)
y u u) n
as follows:
2 y 3 y u (1
2 uy (1 y 2 and b 2.
u)2 y, y
0, 1, 2 and by Theorem11-1, p(u u x) is beta with parameters 2 B( y n B(n y
un 1(1 u) du B(n 2, 2)
2, 4 2, 2)
0, 1, 2. This is tabulated
Table 11-9
f *(y)
6>[(n
4)(n
2)>[(n
20)>[(n
4)(n
4)(n
5)].
2)(n
3)]>[(n
4)(n
(b) The predictive mean is (2n2
CHAPTER 11 Bayesian Methods
11.87. X is a Poisson random variable with parameter l. An initial sample of size n gives l a gamma posterior distribution with parameters nx a and b>(1 nb). It is planned to make one further observation on # the original population. (a) Find the predictive distribution of this observation. (b) Show that the predictive mean is the same as the posterior mean.
(a) Let Y denote the future observation. The predictive density of Y is given by
f *( y) ~
e lly (1 y! (1
nb)nx al(nx bnx
a) 1e l(nb 1)>b
b 1)>b
nb)nx al(nx y a) 1e l(nb y!bnx a (nx a) # (1 nb)nx (nx # y nx # a a 1
dl (nx # b y
y!bnx a f *( y) With u u (nx # nx # 1 a) 1 a nx #
a) 1
a) ,y
which is a negative binomial probability function with parameters r r (b) The mean of this distribution is p (nx # (nx # a)(nb b (nb 1) 1) (nx # a) 1)
nx nb 1 nb b 1
y the right hand side above may be written as
nx nb 1 nb b 1
1)nx y a
0, 1, c
u (n x a)
nx # nx #
a, (nx #
1, c nb 1 . nb b 1
a and p
. The predictive mean of Y is therefore b(nx # (nb a) 1)
a)(nb b (nb 1)
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