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- Problems on conditional independence
The pair
.
exponential
, given
;
exponential
, given
; and
uniform
. Determine a
general formula for
, then evaluate for
.
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A small random sample of size
is taken to determine the proportion of
the student body which favors a proposal to expand the student Honor Council by adding twoadditional members “at large.” Prior information
indicates that this proportion is about 0.6 = 3/5. From a Bayesian point of view, the populationproportion is taken to be the value of a random variable
H . It seems reasonable to assume a
prior distribution
beta
, giving a maximum of the density at
. Seven of the twelve interviewed favor the proposition. What is
the best mean-square estimate of the proportion, given this result? What is the conditionaldistribution of
H , given this result?
Beta
,
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Let
be a random sample, given
H . Set
. Suppose
X conditionally geometric
, given
i.e., suppose
for all
. If
uniform
on [0, 1], determine the best mean square estimator for
H , given
W .
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Let
be a random sample, given
H . Set
. Suppose
X conditionally Poisson
, given
i.e., suppose
. If
gamma
,
determine the best mean square estimator for
H , given
W .
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Suppose
is independent and
. Use properties
of conditional expectation and conditional independence to show that
by
(CI6) and
by
(CE5) .
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Consider the composite demand
D introduced in the section on
Random Sums in "Random Selecton"
Suppose
is independent,
for all
i , and
, invariant with
i . Show that
.
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The transition matrix
P for a homogeneous Markov chain is as follows (in
m-file
npr16_07.m ):
- Obtain the absolute values of the eigenvalues, then consider increasing powers
of
P to observe the convergence to the long run distribution.
- Take an arbitrary initial distribution
(as a row matrix). The product
is the distribution for stage
k . Note what happens as
k becomes
large enough to give convergence to the long run transition matrix. Does the endresult change with change of initial distribution
?
ev = abs(eig(P))'
ev = 1.0000 0.0814 0.0814 0.3572 0.2429a = ev(4).^[2 4 8 16 24]
a = 0.1276 0.0163 0.0003 0.0000 0.0000% By P^16 the rows agree to four places
p0 = [0.5 0 0 0.3 0.2]; % An arbitrarily chosen p0
p4 = p0*P^4p4 = 0.2297 0.2622 0.1444 0.1644 0.1992
p8 = p0*P^8p8 = 0.2290 0.2611 0.1462 0.1638 0.2000
p16 = p0*P^16p16 = 0.2289 0.2611 0.1462 0.1638 0.2000
p0a = [0 0 0 0 1]; % A second choice of p0
p16a = p0a*P^16p16a = 0.2289 0.2611 0.1462 0.1638 0.2000
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Source:
OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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