Corrections to: Interval Estimation of Population Means under Unknown but Bounded Probabilities of Sample Selection

The problem

The outcomes are in increasing order. At $k=0$ the left side of the printed inequality is the sample mean in Proposition 1, and the $\Pi$-weighted mean of the distinct outcomes in Proposition 2. Since neither is smaller than the smallest outcome, $k=0$ satisfies the printed inequality in every case, and both printed rules return the cut point $0$. The vector $w^{\mathrm L}$, or $W_-$, then has every entry equal to $1/\beta$. Evaluating these vectors in (3) and (7) gives the unweighted sample mean and the $\Pi$-weighted mean, respectively. When $\gamma>1$ and the outcomes are not all equal, this vector does not minimize (3) or (7), since a larger weight on the smallest outcome lowers the ratio. The left endpoint of (5), computed with the printed $k^{\mathrm L}$, then leaves out values that $\hat\mu$ can take. The paper does not display the correct rule elsewhere. The Appendix says that the solution to (3) “can be derived in a similar manner”, and the supplementary proof of Proposition 2 treats $\mu_+$ only. When $\gamma=1$ or all outcomes are equal, the ratio has the same value at every feasible weight vector, and the printed rules return a minimizer.

Evidence an exact counterexample

The evidence is an exact counterexample for each proposition and a derivation of the corrected rule.

Counterexample. Let $y=(0,1)$, $\alpha=1/2$ and $\beta=1$, so that $\gamma=2$. I write $(u,v)=(\beta w_1,\beta w_2)\in[1,2]^2$ for the scaled weights. The ratio in (3) is $v/(u+v)$. The printed rule accepts $k=0$, since $1/2\geqslant 0$, and returns $w^{\mathrm L}=(1,1)$ and $\hat\mu^{\mathrm L}=1/2$. For all $(u,v)\in[1,2]^2$, I have

with equality at $(u,v)=(2,1)$. The minimum of (3) is $1/3$, and the printed lower bound exceeds it by $1/6$. For Proposition 2, I consider a population of two units with outcomes $(0,1)$ and selection probabilities $(1/2,1)$. Then $(\Pi_1,\Pi_2)=(1/2,1)$, the ratio in (7) is $v/(u/2+v)$, and

The minimum of (7) is $1/2$, which is the population mean, and the ratio at the printed $W_-=(1,1)$ is $2/3$. The table lists the values at the four vertices.

Derivation of the corrected rule. Both propositions are instances of one problem. Let $x_1\leqslant\cdots\leqslant x_m$, let $a_j>0$, and let $\gamma\geqslant 1$. Proposition 1 has $m=n$, $x_j=y_j$ and $a_j=1$, and Proposition 2 has $m=|Y|$, $x_j=y_{(j)}$ and $a_j=\Pi_j$. After the weights are multiplied by $\beta$, the lower problem is to minimize

For $0\leqslant k\leqslant m$, I define

so that $F_k$ is the left side of the inequality in each rule. By these definitions, I have

For $\gamma>1$, the value $F_k$ decreases at the step $k$ exactly when $F_k>x_{k+1}$. Let $k_*$ be the smallest $k\in\{0,\ldots,m-1\}$ such that $F_k\leqslant x_{k+1}$. Such a $k$ exists, since $F_{m-1}$ is a weighted average of outcomes that are at most $x_m$. I set $\lambda=F_{k_*}$, and $q^*_j=\gamma$ for $j\leqslant k_*$ and $q^*_j=1$ for $j>k_*$. If $k_*>0$, then $F_{k_*-1}>x_{k_*}$ because $k_*$ is the smallest such index, and

It follows that $x_j<\lambda$ for all $j\leqslant k_*$, and $x_j\geqslant x_{k_*+1}\geqslant\lambda$ for all $j>k_*$. Let $q$ be any feasible vector. For $j\leqslant k_*$, I have $q_j-q^*_j\leqslant 0$ and $x_j-\lambda<0$, and for $j>k_*$, I have $q_j-q^*_j\geqslant 0$ and $x_j-\lambda\geqslant 0$. Since $\sum_ja_jq^*_j(x_j-\lambda)=0$, I obtain

The vector $q^*$ minimizes $R$ over the whole box, including when outcomes are tied and when $\gamma=1$.

What survives

The correction needs no new assumption. The definitions (2), (3), (6) and (7), the rules for $k^{\mathrm H}$ and $k_+$, the upper bounds, and the closed form for dichotomous outcomes are unaffected by this sign correction. All 42 entries of Table 1 are reproduced from the reported percentages by the closed form, and this sign correction does not change the table. The estimated bounds still converge to the limits defined by (6) and (7). The rules for $k^{\mathrm L}$ and $k_-$ change, and with them the left endpoint of (5) and computations of the lower bound that use either printed rule. The summation limits in the displayed formula for $\mu_-$ are corrected separately in P025-E002.

The problem

The sentence makes two claims about the selection of the samples. It says that infected shooting gallery needles were more than three times as likely to be tested as uninfected ones, and that uninfected exchange needles were more than three times as likely to be tested as infected ones. Table 1 does not show this. The parameter $\gamma$ is a common bound on how far the selection probabilities may differ within a population. On p. 236 the paper says that $\gamma=5$ “indicates that some units may be up to five times more likely to be sampled than others”. The table shows that the two intervals overlap only when this bound is larger than three. It says nothing about the ratio that each sample attains. Equal reweighted estimates fix the product of the two effective ratios, which must equal the odds ratio of the two sample proportions, about $10.9$. The product can be split unevenly, and one ratio can be below three when the other is larger. The paper does not assume that the two ratios are equal.

Evidence an exact counterexample

The evidence is a derivation and two exact designs.

Derivation. Let a sample have an infected fraction $p$, and write $\pi_1$ and $\pi_0$ for the effective probabilities obtained by averaging inverse selection probabilities within each sampled infection group and taking reciprocals. By (1), the Hajek estimate $q$ of the prevalence satisfies $q/(1-q)=\{p/(1-p)\}(\pi_0/\pi_1)$. I write $r_G=\pi_{G,1}/\pi_{G,0}$ for the shooting gallery and $r_E=\pi_{E,0}/\pi_{E,1}$ for the needle exchange. The two estimates are equal exactly when

With the printed proportions $0.917$ and $0.503$, the right side is $455749/41749\approx 10.92$, and with the count-derived proportions $44/48$ and $291/579$, it is $1056/97\approx 10.89$. By the closed form on p. 237, the two intervals overlap exactly when $\gamma^2$ is at least this number. The condition is $\gamma\geqslant\sqrt{455749/41749}\approx 3.304$ with the printed proportions and $\gamma\geqslant\sqrt{1056/97}\approx 3.299$ with the count-derived proportions. At the smallest such $\gamma$ both ratios equal $\gamma$, and at a larger bound, such as the row $\gamma=4$ of Table 1, they can differ.

Two designs. In each design below, a fixed number of needles is drawn uniformly from the infected and from the uninfected needles of each population. The samples have the sizes $48$ and $579$ reported in Section 4 and positive counts $44$ and $291$ implied by rounding its percentages to the nearest tenth of a percent. The two populations have the same prevalence, the two Hajek estimates equal this prevalence on every draw, and all selection probabilities lie in $[1/4,1]$, so that $\gamma=4$. In design A the ratio for the needle exchange is below three, and in design B the ratio for the shooting gallery is below three. The conclusion is the same for two further choices of the selection probabilities, with the proportions $0.917$ and $0.503$ taken as exact. There the ratios $(r_G,r_E)$ are $(917/249,\ 1491/503)$ at prevalence $3/4$, with $1491/503<3$, and $(917/332,\ 1988/503)$ at prevalence $4/5$, with $917/332<3$.

What survives

Table 1 is unchanged, and so is the statement that the shooting gallery and needle exchange intervals overlap only when $\gamma>3$, with the first overlap in the table at $\gamma=4$. The comparison of the street and needle exchange samples and the method of sensitivity analysis are unaffected. I replace the inference in the last sentence about the two selection processes by a weaker statement about the common bound, without adding an assumption.

The problem

Proposition 2 states that the first $k_-$ entries of $W_-$ equal $1/\alpha$ and the remaining $|Y|-k_-$ entries equal $1/\beta$. The value of (7) at $W_-$ therefore has the classes $j\leqslant k_-$ with the factor $\gamma$ and the classes $j\geqslant k_-+1$ without it. As printed, the second sums start at $k_--1$. For $k_-\geqslant 2$, the classes $k_--1$ and $k_-$ are counted in both sums, and for $k_-=0$ or $k_-=1$, the lower limit is an index with no class. The displays for $\mu_+$, for the cut points and in Assumption 2 start their second sums at $k+1$.

Evidence an exact counterexample

The evidence is a substitution and an exact example. For $0\leqslant k\leqslant|Y|$, let $W^{(k)}$ have its first $k$ entries equal to $1/\alpha$ and the others equal to $1/\beta$. Substituting $W^{(k)}$ into the ratio in (7) and multiplying the numerator and the denominator by $\beta$, I obtain

which is the corrected display at $k=k_-$. For the example, let $\alpha=1/2$, $\beta=1$, and let the population have outcomes $(0,0,1,1,2,3)$ and selection probabilities $(1/2,1/2,1/2,1/2,1,1)$. The distinct outcomes are $(0,1,2,3)$ and $\Pi=(1,1,1,1)$. With the weights scaled by $\beta$, the ratio in (7) is $F(W)=(W_2+2W_3+3W_4)/(W_1+W_2+W_3+W_4)$ on $[1,2]^4$, and

with equality only at $W=(2,2,1,1)$. The minimum of (7) is $7/6$, and it is attained at the cut point $k_-=2$. At $k_-=2$ the printed display equals $\{2(0+1)+(0+1+2+3)\}/\{2(1+1)+(1+1+1+1)\}=1$, and the corrected display equals $\{2(0+1)+(2+3)\}/\{2(1+1)+(1+1)\}=7/6$.

What survives

The definitions (6) and (7), the description of $W_-$, the bounds of Section 3.1 and Table 1 are unaffected. The supplementary material works with the full weighted sums and does not use the printed limits. The rule that selects $k_-$ is corrected in P025-E001.

The problem

Proposition 1 allows $\gamma=1$, and Table 1 includes it. Let $F_k=(w^{k\prime}y)/(w^{k\prime}1_n)$ and $D_k=k+\gamma(n-k)$. Then

For $\gamma>1$, this identity implies that $F_{k+1}-F_k\leqslant 0$ if and only if $F_k\leqslant y_{k+1}$, and the two minima on p. 240 agree. For $\gamma=1$, the vectors $w^k$ coincide and the difference is zero for every $k$. The printed implication and the equality of the two minima can then fail. The steps (B1) and (B2) have the same gap. Each difference of objective values equals $1/\beta-1/\alpha$ times a positive factor times the left side of (B1) or (B2), and for $\alpha=\beta$ the difference is zero whatever the sign of that left side. The results are not affected. For $\gamma=1$ the only feasible weight vector has every entry equal to $1/\alpha$, both bounds in (2) and (3) equal the sample mean, and both (6) and (7) equal $\mu$.

Evidence an exact counterexample

The evidence is an exact counterexample to the four printed steps and the identities that show where $\gamma>1$ is used. Let $n=N=2$, $y=(0,1)$ and $\alpha=\beta=1$, so that $\gamma=1$ and $w^0=w^1=w^2=(1,1)$. Then $F_0=F_1=F_2=1/2$. At $k=0$, I have $F_1-F_0=0\leqslant 0$, and the conclusion $F_0=1/2\leqslant y_1=0$ is false. For $k\in\{0,1\}$, the first minimum on p. 240 is $0$ and the second is $1$. In the supplementary material, the two classes have $y_{(1)}=0$, $y_{(2)}=1$ and $\Pi_1=\Pi_2=1$. With $k=1$, I have $c_1=c_2=0$, the left sides of (B1) and (B2) both equal $-1$, and $\mu(1/\beta,1/\beta)=\mu(1/\alpha,1/\beta)=\mu(1/\alpha,1/\alpha)=1/2$. For the general case, let $a=1/\beta$, $b=1/\alpha$, $p=\Pi_k$, $q=\Pi_{k+1}$, and let $E_1$ and $E_2$ be the left sides of (B1) and (B2). A direct computation gives

Both denominators are positive. Since $a-b<0$ for $\alpha<\beta$, the two equivalences hold in this case. For $\alpha=\beta$, both differences vanish.

What survives

The upper bound (4) and its optimal weights hold for all $\gamma\geqslant 1$. Lemma 1, Propositions 2 and 3, the closed form for dichotomous outcomes and Table 1 are unaffected by this boundary case; the rows $\gamma=1$ are the sample percentages. The identity for the sum of the left sides of (B1) and (B2) in the supplementary material is correct. One case is added to each proof, and no assumption or result changes.

The problem

There are three points. First, the estimated bounds $\hat\mu_h^{\mathrm H}$ and $\hat\mu_h^{\mathrm L}$ converge almost surely to $\mu_+$ and $\mu_-$ under Assumption 1 alone. The supplementary material uses Assumption 2 to make the estimated cut point equal to $k_+$ for all large $h$, which the convergence of the bounds does not require. Proposition 3 is correct as a statement of sufficient conditions, with a hypothesis that it does not need. Second, Proposition 3 prints $\mu_h^{\mathrm H}$ and $\mu_h^{\mathrm L}$ without hats. Assumption 1 defines the estimators $\hat\mu_h^{\mathrm H}$ and $\hat\mu_h^{\mathrm L}$, and no quantity without a hat is defined. Third, $y_{(k+1)}$ is defined only for $0\leqslant k<|Y|$, and “for any $k\geqslant 0$” includes indices at which Assumption 2 has no meaning.

Evidence an exact counterexample

The evidence is an exact example in which Assumption 2 fails and the bounds converge, a derivation that Assumption 1 suffices, and a check of the page for the hats and the range of $k$.

Example. Let the population have outcomes $(0,0,2,2,3,3)$. Each replication selects one unit from each pair of units with equal outcomes, each with probability $1/2$, and the three choices are independent. Then $\pi_i=1/2$ for all $i$, and I take $\alpha=1/4$ and $\beta=1/2$, so that $\gamma=2$. The distinct outcomes are $(0,2,3)$ and $\Pi=(1,1,1)$. After $h$ replications the sample contains exactly $h$ units at each outcome. With class weights $(w_0,w_2,w_3)\in[2,4]^3$, the ratio is $f=(2w_2+3w_3)/(w_0+w_2+w_3)$ for the sample and in the limit, and

By these two inequalities, I have $5/4\leqslant f\leqslant 2$, with equality at $(4,2,2)$ and at $(2,4,4)$. I conclude that $[\hat\mu_h^{\mathrm L},\hat\mu_h^{\mathrm H}]=[\mu_-,\mu_+]=[5/4,2]$ for every $h$ and every sample path. Assumption 2 fails at $k=1$, since the first ratio there equals $\{1\cdot 0+2(2+3)\}/\{1+2(1+1)\}=2=y_{(2)}$.

Derivation. Let $P_{h,j}$ be the number of sampled units of outcome $y_{(j)}$ after $h$ replications, divided by $h$. By Assumption 1, the counts of class $j$ in the successive replications are independent and identically distributed, they are bounded, and their mean is $\Pi_j$. The strong law of large numbers implies that $P_{h,j}\to\Pi_j$ almost surely for every $j$. For a vector $p$ with nonnegative entries and positive sum, I define

In (2) and (3) the weights of the units in one class can be replaced by their average, which lies in $[1/\beta,1/\alpha]$. Whenever the sample is nonempty, $\hat\mu_h^{\mathrm H}=\max_{W\in B}f(W,P_h)$ and $\hat\mu_h^{\mathrm L}=\min_{W\in B}f(W,P_h)$, and by (6) and (7), $\mu_+=\max_{W\in B}f(W,\Pi)$ and $\mu_-=\min_{W\in B}f(W,\Pi)$. Let $R=y_{(|Y|)}-y_{(1)}$ and $S=\sum_j\Pi_j$. With $m=f(W,p)$, I have

and therefore

In the last inequality, I used that $|y_{(j)}-m|\leqslant R$ and $1/\beta\leqslant W_j\leqslant 1/\alpha$. The maximum and the minimum over $B$ change by at most the same amount. With $p=P_h$, I obtain $\hat\mu_h^{\mathrm H}\to\mu_+$ and $\hat\mu_h^{\mathrm L}\to\mu_-$ almost surely, including when the ratios of Assumption 2 tie.