ssrs 2012 barcode font (b) the predictive mean is 3.76 (c) the predictive mode is 4 in Software

Generation QR Code 2d barcode in Software (b) the predictive mean is 3.76 (c) the predictive mode is 4

(b) the predictive mean is 3.76 (c) the predictive mode is 4
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11.96. Prove Theorem 11-12.
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# Since Y is normal with mean u and variance s2 >m, the proof is essentially the same as for the case m 2 replaced with s2 >m. This is shown as follows. s # The predictive density f *( y) of Y is given by # f
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*( y) `
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f ( y u x) #
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# 3f ( y, u u x) du
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# 3f ( y u u) p(u u x) du ~ 3 e
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m 2s2 ( y u)2
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(u mpost)2du
After some simplification, we get
1 2(s2y2 )>(s2 my2 ) post post
*( y)
(my2post y a2mpost) s2 my2post
2(s2y2post)>(s2
my2post)
my2post y s2mpost s2 my2post
my2post y s2mpost s2 my2post
The exponent in the second factor may be further simplified to yield
f *( y) ~ 3 e
1 2(s2y2post)>(s2 uy2post)
2 (uy2post y s mpost)
my2post
2(s2 my2post)
( y mpost)2
The second factor here is free from u. The first factor is a normal density in u and integrates out to an # expression free from u and y. Therefore, we have the following normal predictive density for Y: # f *( y) ~ e #
m 2(s2 my2 ) post ( y mpost)2
CHAPTER 11 Bayesian Methods
11.97. The random variable X has a binomial distribution with n 6 and unknown success probability u which 1 , 0 u 1. An observation on X results in three successes. If has the Haldane prior p(u) u(1 u) another observation is made on X, how many successes could be expected
The predictive distribution of the number of successes in the second observation may be obtained from Theorem 11-11 (with m n 6, x 3, a b 0) as f *( y) 6 B(3 y, 9 B(3, 3) y 2 0.1948 y) y 0, 1, . . . , 6
This is shown in Table 11-12.
y f *( y) 0 0.0606 1
Table 11-12 3 0.2165 4 0.1948 5 0.1364 6 0.0606
The expectation of this distribution is 3. We could therefore expect to see three successes in the six future trials.
Miscellaneous problems 11.98. Show that the maximum likelihood estimate of a in the exponential distribution (see page 124) is 1>x. #
ane n equal to 0 gives a We have L
aaxk.
Therefore, ln L 0 or a
a xk
n ln a a a xk. Differentiating with respect to a and setting it n 1 x. # xk a
11.99. The random variable X has a gamma distribution with parameters a and b. Show that Y inverse gamma density with parameters a and b, defined by
bay g( y) From (33), 2, we have g(y) (1>y)a 1e 1>(by) 1 2 ba (a) y b a 1 y
a 1e 1 by a 1e b>y
1>X has the
0,
0 (a, b 0 0)
ba (a)
The mean, mode, and variance are: b Mean for a 1, Mode a 1
, Variance
b2 1)2(a
for a
11.100. Show that the Bayes estimate with absolute error loss function is the posterior median. Assume that the posterior distribution is continuous (see page 83).
We have to show that if m is the median of the posterior density p(u ux), then
3 uu Assume a
m u p(u ux) du
3 uu
a u p(u ux) du for all a.
m. ux
3 (u x
a u)f (x) dx
3 (m
a)p(u u x) du
2x)p(u u x) du
m)p(u ux) du
3 (m
a)p(uu x) du
a)p(uu x) du a 2x (m 0
m)p(uu x) du (x a) m x m a)
(since, in the middle integral, m
a)c 3 p(u u x) du
3p(u u x) dus
The proof when a
m is similar.
CHAPTER 11 Bayesian Methods
11.101. Generalize the results in Problem 11.91 to the sample mean of m future observations.
# # (a) Denote the mean of the future sample of size m by Y. We then have the following joint distribution of Y and the posterior density of u. y f ( # ; u) f ( y uu)p(uu x) # ume
um y(1
nbx)n aun a 1e # n a (n b a)
u Qb n xR
nbx)n aum n ae u Qb # bn a (n a)
n x m yR
Integrating out u, we have
*( y)
nbx)n aum n ae u Q b # bn a (n a) nbx)n abm (m # nbx mby)m n # nbx)n abm (m # nbx mby)m n # n (n
n x m yR
du a) a)
(1 nbx)n abm n # (1 nbx mby)m # # y 0
n abn a
n a) (n a)
(1 (1
(b) The mean of this distribution is (1 # 3y (1
n (n
a) dy # a)
(m n m2 (n
m a 2) b 1 nbx # a)
SUPPLEMENTARY PROBLEMS
Subjective probability
11.102. Identify the type of probability used: (a) I have no idea whether I will or will not pass this exam, so I would say I am 50% sure of passing. (b) The chances are two in five that I will come up with a dime because I know the box has two dimes and three nickels. (c) Based on her record, there is an 80% chance that she will score over 40 baskets in tomorrow s game. (d) There is a 50-50 chance that you would run into an economist who thinks we are headed for a recession this year. (e) My investment banker believes the odds are five to three that this stock will double in price in the next two months.
Prior and posterior probabilities
11.103. A box contains a biased coin with P(H) 0.2 and a fair coin. A coin is chosen at random from the box and tossed once. If it comes up heads, what is the probability of the event B that the chosen coin is biased 11.104. The random variable X has a Poisson distribution with an unknown parameter l. As shown in Table 11-13, the parameter l has the subjective prior probability function, indicating prior ignorance. A random sample of size 2 yields the X-values 2 and 0. Find the posterior distribution of l. Table 11-13 l p(l) 0.5 1>3 1.0 1>3 1.5 1>3
11.105. X is a binomial random variable with known n and unknown success probability u. Find the posterior density of u assuming a prior density p(u) 4u3, 0 u 1.
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