In this section we study the Bayesian strategy of choosing an optimal test among all the admissible tests and a practical method which enables us to find a best test of a given size. The latter is due to Neyman and Pearson



figure 9.4 A set of admissible characteristics

and is stated in the lemma that bears their names. A Bayesian interpreta­tion of the Neyman-Pearson lemma will be pedagogically useful here.

We first consider how the Bayesian would solve the problem of hypothe­sis testing. For her it is a matter of choosing between HQ and Hx given the posterior probabilities P(H0 | x) and P{H | x) where x is the observed value of X. Suppose the loss of making a wrong decision is as given in Table 9.2. For example, if we choose H0 when Hx is in fact true, we incur a loss y2.


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The nonlinear regression model is defined by (13.4.1) yt = /Др) + ut, t = 1, 2, . . . , T,

where /,(•) is a known function, P is a ^-vector of unknown parameters, and {ut} are i. i.d. with Eut = 0 and Vut = a2. In practice we often specify /*(P) = /(x(, P), where x, is a vector of exogenous variables which, unlike the linear regression model, may not necessarily be of the same dimension as p.

An example of the nonlinear regression model is the Cobb-Douglas pro­duction function with an additive error term,

(13.4.2) d = + Щ,

where Q, K, and L denote output, capital input, and labor input, respec­tively. Another example is the CES production function (see Arrow et al., 1961):

(13.4.3) Qt = + (1 – р2)АГРзГр4/Рз + щ.

We can write (13.4.1) in vector notation as


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Maximum Likelihood Estimators

In this section we show that if we assume the normality of {ut} in the model

(10.1.1) , the least squares estimators a, (3, and a are also the maximum likelihood estimators.

The likelihood function of the parameters (that is, the joint density of Уъ Уъ ■ ■ ■ , Ут) is given by

1 ,

(10.2.78) / = 0 ■ ■ — exp t=l V2tt ct

– (yt~ 0L – fix,)2

L 2ct2 j

= (2ttct2) r/2exp

– – Ц – Z(yt – a – |3x()2


Taking the natural logarithm of both sides of (10.2.78), we have

(10.2.79) log L = log 2tt – ^ log ct2 – Z(yt – cl – fix,)2.

2 2 2a2

Since log L depends on a and (3 only via the last term of the right-hand side of (10.2.79), the maximum likelihood estimators of a and (3 are identical to the least squares estimators.

Inserting a and (3 into the right-hand side of (10.2...

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Best Linear Unbiased Estimator

Neither of the two strategies discussed in Section 7.2.4 is the primary strategy of classical statisticians, although the second is less objectionable to them. Their primary strategy is that of defining a certain class of estimators within which we can find the best estimator in the sense of Definition 7.2.1. For example, in Example 7.2.1, if we eliminate W and Z from our consideration, T is the best estimator within the class consisting of only T and S. A certain degree of arbitrariness is unavoidable in this strategy. One of the classes most commonly considered is that of linear unbiased estimators. We first define

DEFINITION 7.2.4 0 is said to be an unbiasedestimator of 0 if £0 = 0 for

all 0 Є 0. We call £0-0 bias.

Among the three estimators in Example 7.2...

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Student’s t Test

The t test is ideal when we have a single constraint, that is, q = 1. The F test, discussed in the next section, must be used if q > 1.

Since (J is normal as shown above, we have

(12.4.7) Q’0 ~ N[c, a2Q’ (X’X^Q] under the null hypothesis (that is, if Q’3 = c). Note that here Q’ is a row vector and c is a scalar. Therefore,


(12.4.8) т z ~ N(0, 1).


The random variables defined in (12.4.2) and (12.4.8) are independent because of Theorem 12.4.2. Hence, by Definition 2 of the Appendix, we have

Подпись:QP ~ c


Student’s t with T — К degrees of freedom, where a is the square root of


the unbiased estimator of a defined in equation (12.2.29). Note that the denominator in (12.4.9) is an estimate of the standard deviation of the numerator...

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We have so far considered only situations in which both the null and the alternative hypotheses are simple in the sense of Definition 9.1.1. Now we shall turn to the case where the null hypothesis is simple and the alterna­tive hypothesis is composite.

We can mathematically express the present case as testing H0: 0 = 0O against Hi: 0 Є 01; where 0X is a subset of the parameter space. If 0j consists of a single element, it is reduced to the simple hypothesis consid­ered in the previous sections. Definition 9.2.4 defined the concept of the most powerful test of size a in the case of testing a simple against a simple

(i +e,)2

Подпись: FIGURE 9.6 The Neyman-Pearson critical region in a counterintuitive case


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