Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

4.3.∗Confidence Intervals for Parameters of Discrete Distributions 249

random sample on a discrete random variableX with pmfp(x;θ),θ∈Ω, where
Ω is an interval of real numbers. LetT=T(X 1 ,X 2 ,...,Xn) be an estimator ofθ
with cdfFT(t;θ). Assume thatFT(t;θ) is a nonincreasing and continuous function
ofθfor everytin the support ofT. For a given realization of the sample, lettbe
the realized value of the statisticT.Letα 1 >0andα 2 >0begivensuchthat
α=α 1 +α 2 < 0 .50. Letθandθbe the solutions of the equations

FT(t−;θ)=1−α 2 andFT(t;θ)=α 1 , (4.3.1)

whereT−is the statistic whose support lags by one value ofT’s support. For

instance, ifti<ti+1are consecutive support values ofT,thenT=ti+1if and only

ifT−=ti. Under these conditions, the interval (θ,θ) is a confidence interval forθ

with confidence coefficient of at least 1−α. We sketch a proof of this at the end of

this section.

Before proceeding with discrete examples, we provide an example in the con-

tinuous case where the solution of equations (4.3.1) produces a familiar confidence

interval.

Example 4.3.1.AssumeX 1 ,...,Xnis a random sample from aN(θ, σ^2 ) distri-
bution, whereσ^2 is known. LetXbe the sample mean and letxbe its value for a
given realization of the sample. Recall, from expression (4.2.6), thatx±zα/ 2 (σ/

√
n)
is a (1−α)100% confidence interval forθ. Assumingθis the true mean, the cdf
ofXisFX;θ(t)=Φ[(t−θ)/(σ/

√

n)], where Φ(z) is the cdf of a standard normal

distribution. Note for the continuous case thatX−has the same distribution asX.
Then the first equation of (4.3.1) yields

Φ[(x−θ)/(σ/

√

n)] = 1−(α/2);

i.e.,
(x−θ)/(σ/

√

n)=Φ−^1 [1−(α/2)] =zα/ 2.

Solving forθ, we obtain the lower bound of the confidence intervalx−zα/ 2 (σ/

√

n).

Similarly, the solution of the second equation is the upper bound of the confidence

interval.

For the discrete case, generally iterative algorithms are used to solve equations
(4.3.1). In practice, the functionFT(T;θ) is often strictly decreasing and continuous
inθ, so a simple algorithm often suffices. We illustrate the examples below by using
the simplebisection algorithm, which we now briefly discuss.

Remark 4.3.1(Bisection Algorithm). Suppose we want to solve the equation
g(x)=d,whereg(x) is strictly decreasing. Assume on a given step of the algorithm
thata<bbracket the solution; i.e.,g(a)>d>g(b). Letc=(a+b)/2. Then on
the next step of the algorithm, the new bracket valuesaandbare determined by

if(g(c)>d)then{a←candb←b}

if(g(c)<d)then{a←aandb←c}.

The algorithm continues until|a−b|<,where>0 is a specified tolerance.