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Confidence Intervals II Pivotal Quantities Pivoting Using the Probability Transform Topic 16 Interval Estimation Building Confidence Intervals 1 / 16 Pivotal Quantities Pivoting Using the Probability Transform Outline Pivotal Quantities Constructing Confidence Intervals Pivoting Using the Probability Transform Optimizing the Length of the Confidence Interval 2 / 16 Pivotal Quantities Pivoting Using the Probability Transform Pivotal Quantities A random quantity Q(X; θ) is called a pivotal quality or a pivot if • is a function of X and θ where θ is the only unknown parameter. • The distribution of Q(X; θ) does not depend on θ. Examples. For X = (X1;:::; Xn), independent and identically distributed. • If the Xi arise from a location family with density f (x − θ), then n X Q(X; θ) = ai (Xi − θ) i=1 is a pivotal quantity. In particular, X¯ − θ is a pivot. 3 / 16 Pivotal Quantities Pivoting Using the Probability Transform Pivotal Quantities 1 x • If the Xi arise from a scale family with density β f β , then Pn a X Q(X; θ) = i=1 i i β Pn is a pivotal quantity. In particular, i=1 XI /β is a pivot. 1 x−θ • If the Xi arise from a location-scale family with density β f β X¯ − θ Q(X; θ) = S is a pivotal quantity. Note. Even though Q(X; θ) does not depend on θ, it is not an ancillary statistic. It is not even a statistic. 4 / 16 Pivotal Quantities Pivoting Using the Probability Transform Constructing Confidence Intervals To construct a γ-level confidence interval, • Find ` and u so that Pθf` ≤ Q(x; θ) ≤ ug = γ: • Define C(x) = fx; ` ≤ Q(X; θ) ≤ ug: Then Pθfθ 2 C(X)g = γ: If, for each x, Q(x; θ) is monotone increasing in θ, then let • θ^`(x) be the solution to Q(x; θ) = ` and • θ^u(x) be the solution to Q(x; θ) = u. Then. C(x) = fx; θ^`(x) ≤ θ ≤ θ^u(x)g: 5 / 16 Pivotal Quantities Pivoting Using the Probability Transform Constructing Confidence Intervals For Q(x; θ) is monotone decreasing in θ, then reverse the roll of θ^`(x) and θ^u(x). Example. For X = (X1;:::; Xn) from a location family with density f (x − θ), then take Q(X; θ) = X¯ − θ as the pivot. Then choose ` and u so that γ = PθfX; ` ≤ Q(X; θ) ≤ ug = PθfX; ` ≤ X¯ − θ ≤ ug = PθfX : ` − X¯ ≤ −θ ≤ u − X¯g = PθfX; X¯ − ` ≥ θ ≥ X¯ − ug 2 For X = (X1;:::; Xn) ∼ N(θ; σ0), then take z σ0 z σ0 ` = − (1−pγ)=2 and u = (1−pγ)=2 : n n 6 / 16 Pivotal Quantities Pivoting Using the Probability Transform Constructing Confidence Intervals 1 x Example. For X = (X1;:::; Xn) from a scale family with density β f β , take Pn Q(X; β) = i=1 xi /β = T (x)/β as the pivot. Then choose ` and u so that γ = PβfX; ` ≤ Q(X; β) ≤ ug = PβfX; ` ≤ T (X)/β ≤ ug = PβfX; 1=` ≥ β=T (X) ≥ 1=ug = PβfX; T (X)=` ≥ β ≥ T (X)=ug For Exp(β) random variable, T (X)/β ∼ Γ(n; 1) is a pivot. For an equal tailed 95% confidence interval, 7 / 16 Pivotal Quantities Pivoting Using the Probability Transform Constructing Confidence Intervals > gamma<-0.95 0.12 > alpha<-(1-gamma)/2 > qgamma(c(alpha,1-alpha),10) 0.08 dgamma(x, 10) 0.04 [1] 4.795389 17.084803 0.00 0 5 10 15 20 x Thus, 0:95 = PβfX; 4:795 ≤ T (X)/β ≤ 17:085g = PβfX; T (X)=4:795 ≥ β ≥ T (X)=17:085g 8 / 16 Pivotal Quantities Pivoting Using the Probability Transform Constructing Confidence Intervals Example. For X = (X1;:::; Xn) ∼ U(0; θ), take Q(x; β) = max xi /θ = T (x)/θ i n as the pivot. Then M = T (x)/θ has distribution function FM (t) = t in the interval [0; 1]. For an equal tailed γ-confidence interval, set 1 + γ 1 − γ F (u) = and F (`) = : M 2 M 2 1 + γ 1=n 1 − γ 1=n u = and ` = : 2 2 ( ) 1 − γ −1=n 1 + γ −1=n γ = P X; T (X) ≥ β ≥ T (X) : β 2 2 9 / 16 Pivotal Quantities Pivoting Using the Probability Transform Pivoting Using the Probability Transform Theorem. Let T be a continuous (test) statistic with distribution function FT (tjθ); θ 2 Θ ⊂ R. Let α1 + α2 = 1 − γ be the size of the two tails outside the confidence interval. For FT (tjθ)a decreasing function of θ, define θ^u(t) and θ^`(t) by FT (tjθ^u(t)) = α1 and FT (tjθ^`(t)) = 1 − α2; respectively. Then, the random interval [θ^`(T ); θ^u(T )] is a γ-level confidence interval. 10 / 16 Pivotal Quantities Pivoting Using the Probability Transform Pivoting Using the Probability Transform Proof. Because, for any given value of t, FT (tjθ)a decreasing function of θ. Thus, 1 − α2 > α1 implies θ^`(t) < θ^u(t): Moreover, FT (tjθ) < α1 , θ > θ^u(t) FT (tjθ) > 1 − α2 , θ < θ^`(t): Consequently, fθ; α1 ≤ FT (T jθ) ≤ 1 − α2g = fθ; θ`(T ) ≤ θ ≤ θu(T )g Remark. The proof can be modified to handle the case of FT (tjθ) an increasing function of θand for one-sided confidence intervals. 11 / 16 Pivotal Quantities Pivoting Using the Probability Transform Pivoting Using the Probability Transform Example. For independent X = (X1;:::; Xn) ∼ U(−θ; θ), we will use T (X) = X(n) − X(1) to determine the confidence interval. Set X + θ U = i : i 2θ Then the independent random variables U = (U1;:::; Un) ∼ U(0; 1). The T (X) = X(n) − X(1) = 2θ((U(n) − θ) − (U(1) − θ)) = 2θ(U(n) − U(1)) = 2θD: The distribution FT (tjθ) = PθfT ≤ tg = Pθf2θD ≤ tg = PfD ≤ t=(2θ)g = FD (t=(2θ)); which is a decreasing function of θ. 12 / 16 Pivotal Quantities Pivoting Using the Probability Transform Pivoting Using the Probability Transform We have seen that the distribution of D = U(n) − U(1) ∼ Beta(n − 1; 2). Let QD be the quantile function for D, T (X) 1 2θ 1 QD (α1) ≤ ≤ QD (1 − α2) = ≥ ≥ 2θ QD (α1) T (X) QD (1 − α2) T (X) T (X) = ≥ θ ≥ 2QD (α1) 2QD (1 − α2) For n = 16 and γ = 0:95, then the confidence interval with equal probability tails, > gamma<-0.95; alpha<-(1-gamma)/2 > qbeta(c(alpha,1-alpha),15,2) [1] 0.6976793 0.9844864 13 / 16 Pivotal Quantities Pivoting Using the Probability Transform Optimizing the Length of the Confidence Interval As you see the choice of endpoints has the only constraint that the sum of the tail probabilities equals1 − γ. One goal is to have the shortest expected length.This leads to the following constrained optimization problem. minfE[θ^u(T ) − θ^`(T )]; Pfθ^u(T ) − θ^`(T )g ≥ γgg For the case where the density of T is unimodal, to maximize the probability over an interval of length c, we differentiate. d (F (a + c) − F (a)) = f (a + c) − f (a) = 0: da T T T T If the mode tm of the density, then a < tm < a + c. Also, 0 0 fT (a) > 0 and fT (a + c) < 0 showing that the probability is maximized. Notice that if the distribution ofT is symmetric about its mode, then the equal tailed probabilities are also the shortest. 14 / 16 Pivotal Quantities Pivoting Using the Probability Transform Optimizing the Length of the Confidence Interval Returning to the previous example, we want to find the shortest interval for D that has probability γ. > gamma<-0.95 > diff<-function(a) qbeta(gamma+a,15,2)-qbeta(a,15,2) >(astar<-optimize(diff,interval=c(0.001,0.05),maximum=FALSE)) $minimum [1] 0.04840239 $objective [1] 0.2621124 > (u<-qbeta(gamma+astar$minimum,15,2)) [1] 0.9962875 > (l<-qbeta(astar$minimum,15,2)) [1] 0.7341751 > dbeta(c(l,u),15,2) [1] 0.8433389 0.8457852 15 / 16 Pivotal Quantities Pivoting Using the Probability Transform Optimizing the Length of the Confidence Interval • Our first confidence interval had equal tails α1 = α2 = 0:025 and width 0.2868071. • The density at the left endpoint is lower than on the right. • This suggests that we can shorten the confidence interval by moving the interval to the right. 16 / 16.
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