www.downloadslide.com 9.3 Consistency 453 THEOREM 9.3 Suppose that U n has a distribution function that converges to a standard normal distribution function as n →∞.IfW n converges in probability to 1, then the distribution function of U n /W n converges to a standard normal distribution function. This result follows from a general result known as Slutsky’s theorem (Serfling, 2002). The proof of this result is beyond the scope of this text. However, the usefulness of the result is illustrated in the following example. EXAMPLE 9.4 Suppose that Y 1 , Y 2 ,..., Y n is a random sample of size n from a distribution with E(Y i ) = µ and V (Y i ) = σ 2 . Define Sn 2 as Sn 2 = 1 n∑ (Y i − Y n ) 2 . n − 1 i=1 Show that the distribution function of ) √ (Y n − µ n S n converges to a standard normal distribution function. Solution In Example 9.3, we showed that Sn 2 converges in probability to σ 2 . Notice that g(x) = + √ x/c is a continuous function of x if both x and c are positive. Hence, it follows from Theorem 9.2(d) that S n /σ =+ √ Sn 2/σ 2 converges in probability to 1. We also know from the central limit theorem (Theorem 7.4) that the distribution function of U n = √ ( n Y n − µ σ converges to a standard normal distribution function. Therefore, Theorem 9.3 implies that the distribution function of )/ √ (Y n − µ n (S n /σ ) = √ ( ) Y n − µ n σ S n converges to a standard normal distribution function. ) The result of Example 9.4 tells us that, when n is large, √ n(Y n − µ)/S n has approximately a standard normal distribution whatever is the form of the distribution from which the sample is taken. If the sample is taken from a normal distribution, the results of Chapter 7 imply that t = √ n(Y n − µ)/S n has a t distribution with n − 1 degrees of freedom (df). Combining this information, we see that, if a large sample is taken from a normal distribution, the distribution function of t = √ n(Y n −µ)/S n can be approximated by a standard normal distribution function. That is, as n gets large and hence as the number of degrees of freedom gets large, the t-distribution function converges to the standard normal distribution function.
www.downloadslide.com 454 Chapter 9 Properties of Point Estimators and Methods of Estimation If we obtain a large sample from any distribution, we know from Example 9.4 that √ n(Y n − µ)/S n has approximately a standard normal distribution. Therefore, it follows that [ P −z α/2 ≤ √ ( ) ] Y n − µ n ≤ z α/2 ≈ 1 − α. S n If we manipulate the inequalities in the probability statement to isolate µ in the middle, we obtain [ ( ) ( )] Sn Sn P Y n − z α/2 √ ≤ µ ≤ Y n + z α/2 √ ≈ 1 − α. n n Thus, Y n ± z α/2 (S n / √ n) forms a valid large-sample confidence interval for µ, with confidence coefficient approximately equal to 1 − α. Similarly, Theorem 9.3 can be applied to show that √ ˆpn ˆq n ˆp n ± z α/2 n is a valid large-sample confidence interval for p with confidence coefficient approximately equal to 1 − α. In this section, we have seen that the property of consistency tells us something about the distance between an estimator and the quantity being estimated. We have seen that, when the sample size is large, Y n is close to µ, and Sn 2 is close to σ 2 , with high probability. We will see other examples of consistent estimators in the exercises and later in the chapter. In this section, we have used the notation Y n , Sn 2, ˆp n, and, in general, ˆθ n to explicitly convey the dependence of the estimators on the sample size n. We needed to do so because we were interested in computing lim P(|ˆθ n − θ| ≤ε). n→∞ If this limit is 1, then ˆθ n is a “consistent” estimator for θ (more precisely, ˆθ n a consistent sequence of estimators for θ). Unfortunately, this notation makes our estimators look overly complicated. Henceforth, we will revert to the notation ˆθ as our estimator for θ and not explicitly display the dependence of the estimator on n. The dependence of ˆθ on the sample size n is always implicit and should be used whenever the consistency of the estimator is considered. Exercises 9.9 Applet Exercise How was Figure 9.1 obtained? Access the applet PointSingle at www. thomsonedu.com/statistics/wackerly. The top applet will generate a sequence of Bernoulli trials [X i = 1, 0 with p(1) = p, p(0) = 1 − p] with p = .5, a scenario equivalent to successively tossing a balanced coin. Let Y n = ∑ n i=1 X i = the number of 1s in the first n trials and ˆp n = Y n /n. For each n, the applet computes ˆp n and plots it versus the value of n. a If ˆp 5 = 2/5, what value of X 6 will result in ˆp 6 > ˆp 5 ? b Click the button “One Trial” a single time. Your first observation is either 0 or 1. Which value did you obtain? What was the value of ˆp 1 ? Click the button “One Trial” several more