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Consider again the basic statistical model, in which we have a random experiment with an observable random variable taking values in a set . Once again, the experiment is typically to sample objects from a population and record one or more measurements for each item. In this case, the outcome variable has the form
where is the vector of measurements for the item. In general, we suppose that the distribution of depends on a parameter taking values in a parameter space . The parameter may also be vector-valued. We will use subscripts in probability density functions, expected values, etc. to denote the dependence on
Let be a statistic taking values in a set . Intuitively, is sufficient for if contains all of the information about that is available in the entire data variable . Formally, is sufficient for if the conditional distribution of given does not depend on .
Sufficiency is related to the concept of data reduction. Suppose that takes values in . If we can find a sufficient statistic that takes values in , then we can reduce the original data vector (whose dimension is usually large) to the vector of statistics (whose dimension is usually much smaller) with no loss of information about the parameter .
The following result gives a condition for sufficiency that is equivalent to this definition.
Let be a statistic taking values in , and let and denote the probability density functions of and respectively. Show that is sufficient for if and only if the function
is independent of . Hint: The joint distribution of is concentrated on the set .
The definition precisely captures the intuitive notion of sufficiency given above, but can be difficult to apply. We must know in advance a candidate statistic , and then we must be able to compute the conditional distribution of given . The factorization theorem given in the next exercise frequently allows the identification of a sufficient statistic from the form of the probability density function of .
Let denote the probability density function of and suppose that is a statistic taking values in . Show that is sufficient for if and only if there exists and such that
Note that depends only on the data but not on the parameter .
Show that if and are equivalent statistics and is sufficient for then is sufficient for .
We will determine sufficient statistics for several parametric families of distributions.
Suppose that is a random sample of size from the Bernoulli distribution with success parameter . Thus, if trial is a success, and if trial is a failure. Let denote the number of successes, and recall that has the binomial distribution with parameters and . Show directly from the definition that is sufficient for . Specifically, show that the conditional distribution of given is the uniform distribution on the set of points
.The result in the previous exercise is intuitively appealing: in a sequence of Bernoulli trials, all of the information about the probability of success is contained in the number of successes . The particular order of the successes and failures provides no additional information. Of course, the sufficiency of follows more easily from the factorization theorem, but the conditional distribution provides additional insight.
Suppose that the distribution of is a -parameter exponential familiy with the natural statistic . Show that is sufficient for . Because of this result, is referred to as the natural sufficient statistic for the exponential family.
Suppose that is a random sample of size from the normal distribution with mean and variance .
Suppose that is a random sample of size from the Poisson distribution with mean . Show that is sufficient for .
Suppose that is a random sample from the gamma distribution with shape parameter and scale parameter
Suppose that is a random sample from the beta distribution with left parameter and right parameter . Show that is sufficient for where and .
Suppose that is a random sample from the Pareto distribution with shape parameter . Show that is sufficient for .
Suppose that is a random sample from the uniform distribution on the interval where is the unknown parameter. Show that (the order statistic) is sufficient for .
The entire data variable is trivially sufficient for . However, as noted above, there usually exists a statistic that is sufficient for and has smaller dimension, so that we can achieve real data reduction. Naturally, we would like to find the statistic that has the smallest dimension possible. In many cases, this smallest dimension will be the same as the dimension of the parameter vector . However, as we will see, this is not necessarily the case; can be smaller or larger than .
Formally, suppose that a statistic is sufficient for . Then is minimally sufficient if is a function of any other statistic that is sufficient for . Once again, the definition precisely captures the notion of minimal sufficiency, but is hard to apply. The following exercise gives an equivalent condition.
Let denote the probability density function of corresponding to the parameter value and suppose that is a statistic taking values in . Show that is minimally sufficient for if the following condition holds: for and
Hint: If is another sufficient statistic, use the factorization theorem and the condition above to show that implies for and . Then conclude that is a function of .
Show that if and are equivalent statistics and is minimally sufficient for then is minimally sufficient for .
Suppose that the distribution of is a -parameter exponential family with natural sufficient statistic . Show that is a minimally sufficient for .
Show that the sufficient statistics given above for the Bernoulli, Poisson, normal, gamma, and beta families are minimally sufficient for the given parameters.
Suppose that is a random sample from the uniform distribution on the interval where is the unknown parameter. Show that , the vector consisting of the first and last order statistics, is minimally sufficient for . Note that we have a single parameter, but the minimally sufficient statistic is a vector of dimension 2.
Sufficiency is related to several of the methods of constructing estimators that we have studied.
Suppose that is sufficient for and that there exists a maximum likelihood estimator of . Show that there exists a MLE that is a function of . Hint: Use the factorization theorem.
In particular, suppose that is the unique MLE of and that is sufficient for . If is sufficient for then is a function of by the previous exercise. Hence it follows that is minimally sufficient for .
Suppose that the statistic is sufficient for the parameter and that is a Bayes' estimator of . Show that is a function of . Hint: Use the factorization theorem.
The following exercise gives the Rao-Blackwell theorem, named for CR Rao and David Blackwell. The theorem shows how a sufficient statistic can be used to improve an unbiased estimator.
Suppose that is sufficient for and that is an unbiased estimator of a real parameter . Use sufficiency and properties of conditional expectation and conditional variance to show that
Suppose that is a statistic taking values in a set . Then is a complete statistic for if for any real-valued function on
To understand this rather strange looking condition, suppose that is a statistic constructued from that is being used as an estimator of 0 (thought of as a function of ). The completeness condition means that the only such unbiased estimator is the statistic that is 0 with probability 1.
Show that if and are equivalent statistics and is complete for then is complete for .
Suppose that is a random sample of size from the Bernoulli distribution with success parameter . Show that the number or successes is complete for . Hint: Note that can be written as a polynomial in If this polynomial is 0 for all in an open interval, then the coefficients must be 0.
Suppose that is a random sample of size from the Poisson distribution with parameter . Show that the sum of the sample values is complete for . Hint: Note that can be written as a power series in . If this series is 0 for all in an open interval, then the coefficients must be 0.
Suppose that is a random sample of size from the exponential distribution with scale parameter . Show that the sum of the sample values is complete for . Hint: Note that is the Laplace transform of a certain function. If this transform is 0 for all in an open interval, then the function must be 0.
The results in the previous exercises generalize to exponential families, although the proof is complicated. Specifically, if the distribution of is a -parameter exponential family with the natural sufficient statistic then is complete for (as well as minimally sufficient for ). This applies to random samples from the Bernoulli, Poisson, normal, gamma, and beta distributions discussed above.
The notion of completeness depends very much on the parameter space.
Suppose that is a random sample of size 3 from the Bernoulli distribution with success parameter . Show that is not complete for .
The next exercise shows the importance of complete sufficient statistics; it is known as the Lehmann-Scheffé theorem, named for Erich Lehmann and Henry Scheffé.
Suppose that is sufficient and complete for and that is an unbiased estimator of a real parameter . Show that is a uniformly minimum variance unbiased estimator of . The proof is based on the following steps:
Suppose that is a random sample of size from the Bernoulli distribution with success parameter . As usual, let denote the number of successes. Show that an UMVUE for , the variance of the sample distribution, is
Suppose that is a random sample of size from the Poisson distribution with parameter . Let . Show that an UMVUE for is
Hint: Use the probability generating function of .
Suppose that is a statistic taking values in a set . If the distribution of does not depend on , then is called an ancillary statistic for . Thus, the notion of an ancillary statistic is complementary to the notion of a sufficient statistics (which contains all information about the parameter that is contained in the sample). Thus, the result in the following exercise, known as Basu's Theorem and named for Debabrata Basu, makes this point more precisely.
Suppose that is complete and sufficient for a parameter and that is an ancillary statistic. Show that and are independent. The following steps sketch the proof:
Show that if and are equivalent statistics and is ancillary for then is ancillary for .
Suppose that is a random sample from a scale family with scale parameter . Show that is an ancillary statistic for if is a function of
Suppose that is a random sample of size from the gamma distribution with shape parameter and scale parameter . Let denote the sample arithmetic mean of and let denote the sample geometric mean of . Show that is ancillary for , and thus conclude that and are independent. Hint: Use the previous exercise.