Interval Estimation of Bounded Variable Means via Inverse Sampling

When the experiment of inverse sampling is completed, it is desirable to construct a confidence interval for the mean value µ in terms of the random sampling number n. For this purpose, we have Theorem 1 Let H (z, µ) = ln µ z + 1 z -1 ln 1-µ 1-z for 0 < z < 1 and 0 < µ < 1. Let δ ∈ (0, 1). Let µ be a random variable such that µ = 1 for n ≤ γ + 1 and that

for n > γ + 1. Let µ be a random variable such that

The proof is given in Appendix A. It should be noted that, due to the monotone property of the function H (z, µ) with respect to µ, the confidence limits can be readily determined by a bisection search method.

In the special case that X is a Bernoulli random variable and γ is an integer, the confidence interval can be constructed in a slightly different way as follows.

Theorem 2 Let δ ∈ (0, 1). Let µ be a random variable such that µ = 1 for n = γ and that

for n > γ. Let µ be a random variable such that

The proof is given in Appendix B. As mentioned earlier, the confidence limits can be readily determined by a bisection search method.

Theorems 1 and 2 are established by employing Hoeffding’s inequality [5]. If we replace the Hoeffding’s inequality by Massart’s inequality (i.e., Theorem 2 at page 1271 of [6]), which is slightly more conservative, we can obtain via analogy arguments explicit formulas for interval estimation. In this regard, we have Theorem 3 for the general inverse sampling scheme.

for n > γ + 1,

Then,

For the inverse binomial sampling scheme (with integer γ), we have

We have established rigorous and simple interval estimation methods for means of bounded random variables. The construction of confidence intervals is based on inverse sampling and requires little computation. The nominal coverage probability of confidence intervals is always guaranteed.

We need some preliminary results.

The following lemma is a classical result, known as Hoeffding’s inequality [5].

The following lemma has been established by Chen [1].

) is monotonically decreasing with respect to ε ∈ (0, 1). Lemma 3 For any γ > 0 and ε ∈ (0, 1),

Proof. Since n is an integer, we have

where ε * is a number depends on µ and ε such that

where

Now we shall consider three cases.

(i):

(ii): In the case of z = 1, we have µ

(iii): In the case of µ < z < 1, by Lemma 1, we have

Therefore, we have shown

for all cases.

✷ Lemma 4 For any γ > 0 and ε ∈ (0, 1),

Proof. Since n is an integer, we have

where ⌈.⌉ denotes the ceiling function. Let ζ be a number such that γ

for any µ ∈ (0, 1). Hence,

. Then, m is a positive integer and

Note that

Then,

Let z = γµ γ(1+̺)-µ and m = γ z . For ̺ > µ γ , we have 0 < z < µ, ∂m ∂̺ > 0 and

which implies that H (z, µ) is monotonically decreasing with respect to ̺ > µ γ . By the relation between ̺ and ε, we have that ̺ increases as ε increases and that ̺ > µ γ if and only if ε > µ γ+µ . This proves the lemma.

The following lemma can be shown by direct computation.

which is negative for 1 > µ > z > 0, and positive for 0 < µ < z < 1. Moreover,

Lemma 7 For any γ > 0 and δ ∈ (0, 1), there exists ε * ∈ µ γ+µ , 1 such that

Proof. Note that lim

For any γ > 0 and δ ∈ (2µ γ , 1), there exists ε * ∈ 0, 1 µ -1 such that

For any γ > 0 and δ ∈ (0, 1),

Proof. By Lemma 7, there exists

Therefore, to show Lemma 9, it suffices to show

Note that

which can be written as 1 -

By the definition of ε, we have

As a result, H

, µ , and by Lemma 5 ε ≥ ε * .

By this inequality and the definition of ε,

.

This shows the inclusion relationship of the sets. The lemma is thus proved. ✷ Lemma 10 For any γ > 0 and δ ∈ (0, 1),

Proof. There are two cases: Case (i) 0 < δ < 2µ γ ; Case (ii) δ ≥ 2µ γ . We first consider Case (i). Note that H (z, µ) is monotonically decreasing with respect to z ∈ (µ, 1) and that lim z→1 H (z, µ) = ln µ.

Since 0 < δ < 2µ γ , we have ln µ >

To show Lemma 10, it suffices to show

as a result of n ≥ γ > nµ and 0 < µ < 1. By the definition of ε, we have γ n = (1 + ε)µ and thus

Hence,

By this inequality and the definition of ε,

Then, µ ≤ µ <

Finally, Theorem 1 is justified by invoking the Bonferroni’s inequality.

To show Theorem 2, we need a modified version of Lemma 4 as follows.

Lemma 11 For any γ > 0 and ε ∈ (0, 1),

Proof. Since n is an integer, we have

The remainder of the proof is similar to that of Theorem 1 and is thus omitted.

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