SEQUENTIAL RISK-EFFICIENT ESTIMATION OF THE PARAMETER IN THE UNIFORM DENSITY

We develop a risk-efficient sequential procedure for estimating the parameter θ of the uniform density on (0,θ). We give explicit expressions for the distribution of the stopping time and derive its expectation and variance. We also tabulate the values of the expected stopping time and its standard deviation for some selected values of the parameter. Asymptotic properties such as efficiency and risk-efficiency are established.


Introduction.
The problem of obtaining a confidence interval having a specified width for the parameter in the density that is uniform on (0, θ)(θ > 0) or on (ξ, 1)(ξ < 1) has been considered by Govindarajulu and others.For earlier references on this problem (see [2]).In this paper, we provide a risk-efficient sequential procedure for estimating θ or ξ.
2. Notation and the sequential procedure.Let X be distributed uniformly on (0,θ), where θ > 0 and let X 1 ,X 2 ,...,X n be a random sample from that distribution.It is well known that X nn = max(X 1 ,...,X n ) is sufficient for θ.Consider θ = bX nn (2.1) and minimize E(bx nn − θ) 2 with respect to b.We find the optimal value of b to be (n + 2)/(n + 1).Then, the mean squared error of (n + 2)Y nn /(n + 1) is (2.2) Next consider where c is proportional to the cost of a single observation.We can write The value of n which minimizes R can be obtained by solving ∂R/∂n = 0.This gives However, since θ is unknown, we cannot compute the optimal n given by (2.4).Hence, we resort to the following adaptive sequential rule.Stop at N, where (2.7)

Properties of the stopping time.
Consider where which tends to zero as n → ∞ since X nn converges to θ almost surely (a.s.).Thus, the sequential procedure terminates finitely with probability 1.Now, let a = (c/2) 1/2 .N = n implies that X nn ≤ a(n + 1) 3/2 and X n−1,n−1 > an 3/2 .Hence, Next, we explicitly evaluate the first two moments of N. Recall that where The sample (U 1 ,...,U k ) is a random sample of size k from the standard uniform distribution.Also note that P (N > n) = 0 whenever n > (θ/a) 2/3 − 1 (because, then U nn > 1).In the following lemma, we obtain a recurrence relation for evaluating the values of be an increasing sequence of real numbers, i.e., where P 0 = 0.
Proof.From the addition law, we have (3.8) We also note that (3.10)Then, we can express P n − P n−1 as Further expressing the right-hand side of (3.11) as a polynomial in α n , we obtain (3.7).This completes the proof of Lemma 3.1.Now, using (3.7) recurrently, we can evaluate the values of the P i 's in terms of the α i 's.In particular, we have (3.12) Thus where In Table 3.1, we provide numerical values of EN and σ N for certain selected values of η.From Table 3.1, we can see that the expected stopping time is increasing in η, whereas its standard deviation is tending to 1 as η becomes large (or as c tends to 0).

Asymptotic consideration.
In this section, we derive some of the asymptotic properties of the sequential procedure which are given in the following theorem.
Let the stopping rule be Then there exists a p > 2, B > 1 and k 0 ≥ 1 such that E|S k 0 | p < ∞ and for some r > 1 with r > p(1 − 2α)/(2α(p − 2)), where α and p are given by (4.4) and (4.7), respectively, and ζn is an estimate of ζ.In the following, we establish (4.6) and (4.7).Proof.We have (4.9) Since U nn is equivalent in distribution to e −δ/n , where δ has the standard exponential distribution Proof.We have where C = max(2 p−1 , 1) by the c r inequality (cf.[4, page 155]).
Now, we can write where Consequently, Thus, it readily follows that as n becomes large.
The proof for (iii) of Theorem 4.1 is now complete by using Lemmas 4.2 and 4.3.
Next, we present a lemma giving the exact distribution of S kn which might be of interest elsewhere and from which we can also assert (4.6).Lemma 4.4.We have  Next we consider the cases y > 0 and y < 0. Case 1.Let y > 0. Then This completes the proof of Lemma 4.4.
Corollary 4.5.The probability density function of S k,n is given by if y > 0. Proof.We can write   Concluding remark.In order to solve the dual problem of estimating ξ when X is distributed uniformly on (ξ, 1), change θ to 1 − ξ and X nn to 1 − X 1n , where X 1n = min(X 1 ,...,X n ).

Table 3 .
1. Exact values of EN and σ N for selected values of η.