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554 NP-hard inferenceHere let us assume that a probit regression model is adequate to capturethe responding behavior, which depends on only the individual’s x value.That is, we can express R i = 1(Z i ≤ α + βx i ), where the Z i ’s form an i.i.dsample from N (0, 1). We could imagine Z i being, e.g., the ith individual’slatent “refusal tendency,” and when it is lower than a threshold that is linearin x i ,theindividualresponds.Theinterceptα allows us to model the overallpercentage of respondents, with larger α implying more respondents. The slopeβ models the strength of the self-selecting mechanism. In other words, as longas β ≠ 0, we have a non-ignorable missing-data mechanism (Rubin, 1976).Given that ¯x s is unbiased, its MSE is the same as its variance (Cochran,2007), viz.var(¯x s )= 1 − f sn sS 2 N (x), where S 2 N (x) = 1N − 1N∑(x i − ¯x N ) 2 . (45.9)The MSE of ¯x a is more complicated, mostly because R i depends on x i .Butunder our assumption that N is very large and f a = n a /N stays (far) awayfrom zero, the MSE is completely dominated by the squared bias term of ¯x a ,which itself is well approximated by, again because N (and hence n a )isverylarge,{ ∑N } 2Bias 2 i=1(¯x a )=(x i − ¯x N )p(x i )∑ Ni=1 p(x , (45.10)i)where p(x i )=E(R i |x i )=Φ(α + βx i ), and Φ is the CDF for N (0, 1).To get a sense of how this bias depends on f a ,letusassumethatthefinitepopulation {x 1 ,...,x N } itself can be viewed as an SRS of size N from a superpopulation X ∼N(µ, σ 2 ). By the Law of Large Numbers, the bias term in(45.10) is essentially the same as (again because N is very large)⎛ ⎞φ ⎝˜α√ ⎠cov{X, p(X)} σE{ZΦ(˜α + ˜βZ)} σ=E{p(X)} E{Φ(˜α + ˜βZ)}= √˜β 1+ ˜β 2⎛ ⎞1+ ˜β, (45.11)2Φ ⎝˜α√ ⎠1+ ˜β 2where ˜α = α + βµ, ˜β = σβ, Z ∼ N(0, 1), and φ is its density function.Integration by parts and properties of Normals are used for arriving at (45.11).An insight is provided by (45.11) when we note Φ{˜α/(1 + ˜β 2 ) 1/2 } is wellestimated by f a because N is large, and hence ˜α/(1+ ˜β 2 ) 1/2 ≈ Φ −1 (f a )=z fa ,where z q is the qth quantile of N (0, 1). Consequently, we have from (45.11),MSE(¯x a )σ 2 ≈ Bias2 (¯x a )σ 2 = ˜β 2 φ 2 (z fa )1+ ˜β 2 fa2 = ˜β 2 e −z2 fa1+ ˜β 2 2πfa2 , (45.12)i=1