Steins two-stage procedure. (i) If m S2/2 has a 2-distribution with m degrees of freedom, and if

Stein’s two-stage procedure.

(i) If m S2/σ2 has a χ2-distribution with m degrees of freedom, and if the conditional distribution of Y given S = s is N(0, σ2/S2), then Y has Student’s t-distribution with m degrees of freedom.

(ii) Let X1, X2,... be independently distributed as N(ξ, σ2). Let X¯

0 = n0 i=1 Xi /n0, S2 = n0 i=1(Xi − X¯ 0)2/(n0 − 1), and let a1 =···= an0 = a, an0+1 =···= an = b and n ≥ n0 be measurable functions of S. Then Y =

n i=1 ai(Xi − ξ)

S2 n i=1 a2 i

has Student’s distribution with n0 − 1 degrees of freedom.

(iii) Consider a two-stage sampling scheme /

1, in which S2 is computed from an initial sample of size n0, and then n − n0 additional observations are taken. The size of the second sample is such that n = max 

n0 + 1, S2 c

+ 1



, where c is any given constant and where [y] denotes the largest integer ≥ y.

There then exist numbers a1,..., an such that a1 =···= an0 , an0+1 =··· an, n i=1 ai = 1, n i=1 a2 i = c/S2. It follows from (ii) that n i=1 ai(Xi − ξ)/√c has Student’s t-distribution with n0 − 1 degrees of freedom.

(iv) The following sampling scheme /

2, which does not require that the second sample contain at least one observation, is slightly more efficient than /

1, for the applications to be made in Problems 5.24 and 5.25. Let n0, S2, and c be defined as before; let n = max 

n0, S2 c

+ 1



ai = 1/n (i = 1,..., n), and X¯ = n i=1 ai Xi . Then √n(X¯ − ξ)/S has again the t-distribution with n0 − 1 degrees of freedom.

[(ii): Given S = s, the quantities

a, b, and n are constants, n i=1 ai(Xi − ξ) =

n0a(X¯ 0 − ξ) is distributed as N(0, n0a2σ2), and the numerator of Y is therefore normally distributed with zero mean and variance σ2 n i=1 a2 i . The result now follows from (i).]