Designing randomized response surveys to support honest answers to stigmatizing questions

Abstract

Randomized response survey methods use noise to mask respondents’ answers to stigmatizing questions in an attempt to elicit honest responses. Respondents weigh the preference for honesty against the disutility of stigmatization when deciding how to answer. Since the disutility of stigmatization depends on the degree of noise, the interviewer designs the survey to balance two goals: (i) honest reporting by respondents and (ii) maximization of the accuracy of estimates based on the survey. We fully characterize the non-linear set of design parameters that lead to truth-telling, as well as the interviewer’s equilibrium survey design.

A Appendix: Proofs

This appendix collects the proofs.

Proof of Lemma 1. Obvious and omitted. $\square$

Proof of Lemma 2. Obvious and omitted. $\square$

Proof of Proposition 3. Follows directly from Proposition 11. $\square$

Proof of Proposition 4. Towards a contradiction, suppose the first claim is false. Then, there is a $p_{0}\in (\frac{1}{2},1)$ such that $a(N,p_{0})\ge a(N,p)$ for all $p\ge p_{0}$. But, we have that (i) $a(\infty ,p)$ is strictly increasing and onto and that (ii) $a(N,p_{0})\le a(1,p_{0})=2p_{0}-1<1$ since $p_{0}<1$. Thus, there is a $\tilde{p}<1$ such that $a(N,p_{0})=a(\infty ,\tilde{p})$. But then for $p>\tilde{p}$, we have $a(N,p_{0})<a(\infty ,p)\le a(N,p)$ by Proposition 3, an impossibility. $\square$

Proof of Corollary 5. Obvious and omitted. $\square$

Proof of Corollary 6. Obvious and omitted. $\square$

Proof of Lemma 7. Obvious and omitted. $\square$

Proof of Lemma 8. Obvious and omitted. $\square$

Proof of Proposition 9. Obvious and omitted. $\square$

Proof of Lemma 10. Obvious and omitted. $\square$

Proof of Proposition 11. We only establish part (i) of the proposition since the argument for part (ii) parallels the proof of Proposition 4. The proof is by induction. It is broken into four parts:

• Part 0 describes notation as well as transformations and preliminary results on the random variables used in the proof.

• Part 1 develops the lower bound on a(1, p) and show that $a(1,p)<a(2,p)$ for $p\in (\frac{1}{2},1)$.²⁸

• Part 2 establishes that, when $a(N,p)<a(N+1,p)$ for all $p\in (\frac{1}{2},1)$, then $a(k,p)<a(k+1,p)$ for all $p\in (\frac{1}{2},1)$.

• Part 3 gives the upper bound on $a(\infty ,p)$.

Part 0: Our proof uses a transformation of a(N, p), which allows us to establish the desired result by focusing on means and variances. Key to the proof is the variable $\tau =pv+(1-p)(1-v)$. To build intuition, we rewrite the interviewer’s posterior belief under honesty. After he observes responses $\varvec{r}$, then his Bayesian posterior regarding v is $h(v|Y(\varvec{r}))$, where

$$\begin{aligned} h(v|Y)=\frac{\tau ^{Y}(1-\tau )^{N-Y}}{\int _{0}^{1}\tau ^{Y}(1-\tau )^{N-Y}f_{v}(v)dv}, \end{aligned}$$

because each honest response is an identically distributed and conditionally independent Bernoulli random variable given v. The conditional probability of success of this Bernoulli variable is $\tau$. Hence, given responses $\varvec{r}=(r_{1},\ldots ,r_{N})$, the interviewer’s posterior belief regarding $\varvec{\theta }=(\theta _{1},\ldots ,\theta _{N})$ and v may be rewritten as

$$\begin{aligned} \mu (\varvec{\theta },v|\varvec{r})=h(v|Y(\varvec{r}))\tilde{g}(\varvec{\theta }|v,\varvec{r}), \end{aligned}$$

where $\tilde{g}$ is the conditional probability of types $\varvec{\theta }$ given (v, r). But, given v, the probability that respondent i is type $\theta _{i}$ only depends on his own response $r_{i}$; it is independent of other’s responses. This is because (i) types are conditionally independent given v and (ii) every other respondent j’s answer only depends on j’s type and question. It follows that $\tilde{g}(\varvec{\theta }|v,\varvec{r})=\prod _{i=1}^{N}g(\theta _{i}|v,r_{i})$, where

$$\begin{aligned} g(\theta _{i}|v,r_{i})&=\left( \frac{(pv)^{\mathbb {I}(\theta _{i}=s)}((1-p)(1-v))^{1-\mathbb {I}(\theta _{i}=s)}}{\tau }\right) ^{Y(r_{i})}\\&\quad \times \left( \frac{((1-p)v)^{\mathbb {I}(\theta _{i}=s)}(p(1-v))^{1-\mathbb {I}(\theta _{i}=s)}}{1-\tau }\right) ^{1-Y(r_{i})} \end{aligned}$$

is readily verified to be the probability that respondent i is type $\theta _{i}$ given v and his response $r_{i}$ under honesty. Hence, under honesty,

$$\begin{aligned} \mu (\varvec{\theta },v|\varvec{r})=h(v|Y(\varvec{r}))\prod _{i=1}^{N}g(\theta _{i}|v,r_{i}). \end{aligned}$$

(11)

Equation (11) implies that the interviewer’s belief regarding respondent i’s type $\mu (\theta _{i}|\varvec{r})$ depends only on (i) i’s own response and (ii) the total number of y answers (including i’s response). Thus, a(N, p) may be rewritten in terms of $r_{i}$ and Y. As is evident from Eq. (11), this transformation leverages $\tau$. Specifically, as we will see in the body of the proof, the re-expression depends on the mean and variance of $\tau$ given Y, as well as the distribution of Y. We next derive (i) the conditional mean and variance of $\tau$ and (ii) the distribution of Y. We provide details on the re-expression of a(N, p) in the body of the proof.

Since $\tau =pv+(1-p)(1-v)$ and since v has conditional distribution h(v|Y) on [0, 1], the distribution of $\tau$ given Y is

$$\begin{aligned} f_{\tau }(\tau |Y)=\frac{1}{2p-1}h\big (\frac{\tau -(1-p)}{2p-1}|Y\big ) \end{aligned}$$

and that it has support is $[1-p,p]$.²⁹ It follows that the expectation of $\tau$ given Y is $\mathbb {E}(\tau |Y)=\int _{1-p}^{p}\tau f_{\tau }(\tau |Y)d\tau$ and that the variance of $\tau$ given Y is $\mathbb {V}(\tau |Y)=\mathbb {E}(\tau ^{2}|Y)-\mathbb {E}(\tau |Y)$, where $\mathbb {E}(\tau ^{2}|Y)=\int _{1-p}^{p}\tau ^{2}f_{\tau }(\tau |Y)d\tau .$ Two observations will prove useful in the re-expression of a(N, p). First, $d\tau =(2p-1)dv$ and so $h(\nu |Y)d\nu =f_{\tau }(\tau |Y)d\tau$. Second,

$$\begin{aligned} \mathbb {E}(\tau |Y)&=p\mathbb {E}(v|Y)+(1-p)(1-\mathbb {E}(v|Y))\\ \mathbb {V}(\tau |Y)&=(2p-1)^{2}\mathbb {V}(v|Y), \end{aligned}$$

where $\mathbb {E}(v|Y)=\int _{0}^{1}vh(v|Y)dv$ and $\mathbb {V}(v|Y)=\int _{0}^{1}(v-\mathbb {E}(v|Y))^{2}h(v|Y)dv$ are respectively the expected value and variance of v given Y. Analogously,

$$\begin{aligned} \mathbb {E}(\tau )&=p\mathbb {E}(v)+(1-p)(1-\mathbb {E}(v))\\ \mathbb {V}(\tau )&=(2p-1)^{2}\mathbb {V}(v), \end{aligned}$$

are respectively the expected value variance of $\tau$, where $\mathbb {E}(v)=\int _{0}^{1}vf_{v}(v)dv$ and $\mathbb {V}(v)=\int _{0}^{1}(v-\mathbb {E}(v))^{2}f_{v}(v)dv$ are respectively the expected value and variance of v.

The random variable Y is binomially distributed, i.e., the probability of Y honest answers of y from N respondents is

$$\begin{aligned} \pi (Y)={N \atopwithdelims ()Y}\int _{0}^{1}\tau ^{Y}(1-\tau )^{1-Y}f_{v}(v)dv. \end{aligned}$$

(12)

Simply, given a response vector $\varvec{r}$, there are ${N \atopwithdelims ()Y(\varvec{r})}-1$ other response vectors that have $Y(\varvec{r})$ answers of y. Since the probability of each of these vectors is $\int _{0}^{1}\tau ^{Y(\varvec{r})}(1-\tau )^{N-Y(\varvec{r})}f_{v}(v)dv$, Eq. (12) obtains.

Part 1: To show that $a(1,p)<a(2,p)$ for all $p\in (\frac{1}{2},1)$, with equality at one-half and one, we re-write a(1, p) and a(2, p). For $a(1,p)=\mu (s|y)-\mu (s|n)$, it is readily verified that³⁰:

$$\begin{aligned} \mu (s|y)&=\frac{\mu (s,y)}{\mu (y)}=\frac{\int _{0}^{1}\mu (s,y,\nu )d\nu }{\int _{0}^{1}\mu (y,\nu )d\nu }=\frac{\int _{0}^{1}p\nu f_{\nu }(\nu )d\nu }{\int _{0}^{1}[p\nu +(1-p)(1-\nu )]f_{\nu }(\nu )d\nu }\\&=\frac{p\mathbb {E}(\nu )}{p\mathbb {E}(\nu )+(1-p)(1-\mathbb {E}(\nu ))}\text { and }\\ \mu (s|n)&=\frac{\mu (s,n)}{\mu (n)}=\frac{\int _{0}^{1}\mu (s,n,\nu )d\nu }{\int _{0}^{1}\mu (n,\nu )d\nu }=\frac{\int _{0}^{1}(1-p)\nu f_{\nu }(\nu )d\nu }{\int _{0}^{1}[p(1-\nu )+(1-p)\nu ]f_{\nu }(\nu )d\nu }\\&=\frac{(1-p)\mathbb {E}(\nu )}{p\mathbb {E}(\nu )+(1-p)(1-\mathbb {E}(\nu ))}. \end{aligned}$$

Thus,

$$\begin{aligned} a(1,p)=\frac{p\mathbb {E}(\nu )}{p\mathbb {E}(\nu )+(1-p)(1-\mathbb {E}(\nu ))}-\frac{(1-p)\mathbb {E}(\nu )}{p\mathbb {E}(\nu )+(1-p)(1-\mathbb {E}(\nu ))}, \end{aligned}$$

which simplifies to $a(1,p)=2p-1$ when $\mathbb {E}(v)=\frac{1}{2}$, as in Proposition 3. Changing variables to $\tau$ and simplifying gives

$$\begin{aligned} a(1,p)=\frac{1}{2p-1}\Big [1-p(1-p)\Big (\frac{1}{\mathbb {E}(\tau )}+\frac{1}{1-\mathbb {E}(\tau )}\Big )\Big ]. \end{aligned}$$

The display equation is readily verified to be increasing in p.

Regarding a(2, p), let $\bar{r}$ denote the response of the other respondent. Then,

$$\begin{aligned} a(2,p)=\pi (\bar{y})(\mu (s|y,\bar{y})-\mu (s|n,\bar{y}))+\pi (\bar{n})(\mu (s|y,\bar{n})-\mu (s|n,\bar{n})), \end{aligned}$$

where $\mu (s|r,\bar{r})$ denotes interviewer’s posterior belief when respondent i states r and the other respondent states $\bar{r}$. It is readily verified that:

$$\begin{aligned} \mu (s|y,\bar{y})&=\frac{\mu (s,y,\bar{y})}{\mu (y,\bar{y})},=\frac{\int _{0}^{1}\mu (s,y,\bar{y},\nu )d\nu }{\int _{0}^{1}\mu (y,\bar{y},\nu )d\nu }\\&=\frac{\int _{0}^{1}\mu (s,y|\nu )\mu (\bar{y}|\nu )f_{\nu }(\nu )d\nu }{\int _{0}^{1}\mu (y|\nu )\mu (\bar{y}|\nu )f_{\nu }(\nu )d\nu }\\&=\frac{\int _{0}^{1}p\nu (p\nu +(1-p)(1-\nu ))f_{\nu }(\nu )d\nu }{\int _{0}^{1}[p\nu +(1-p)(1-\nu )]^{2}f_{\nu }(\nu )d\nu } \end{aligned}$$

because a respondent’s type is conditionally independent of other respondents’ answers given v. Analogous logic gives

$$\begin{aligned} \mu (s|n,\bar{y})&=\frac{\int _{0}^{1}(1-p)\nu (p\nu +(1-p)(1-\nu )f_{\nu }(\nu )d\nu }{\int _{0}^{1}[p(1-\nu )+(1-p)\nu ][(p\nu +(1-p)(1-\nu )]f_{\nu }(\nu )d\nu },\\ \mu (s|y,\bar{n})&=\frac{\int _{0}^{1}p\nu (p(1-\nu )+(1-p)\nu )f_{\nu }(\nu )d\nu }{\int _{0}^{1}[p(1-\nu )+(1-p)\nu ][p\nu +(1-p)(1-\nu )]f_{\nu }(\nu )d\nu },\text { and }\\ \mu (s|n,\bar{n})&=\frac{\int _{0}^{1}(1-p)\nu (p(1-\nu )+(1-p)\nu )f_{\nu }(\nu )d\nu }{\int _{0}^{1}[p(1-\nu )+(1-p)\nu ]^{2}f_{\nu }(\nu )d\nu }. \end{aligned}$$

Changing variables to $\tau$ and simplifying gives

$$\begin{aligned} \mu (s|y,\bar{y})&=\frac{\int _{0}^{1}p\nu (p\nu +(1-p)(1-\nu ))f_{\nu }(\nu )d\nu }{\int _{0}^{1}[p\nu +(1-p)(1-\nu )]^{2}f_{\nu }(\nu )d\nu }\\&=\frac{\int _{1-p}^{p}p(\tau -(1-p))\tau f_{\tau }(\tau )d\tau }{\int _{1-p}^{p}(2p-1)\tau ^{2}f_{\tau }(\tau )d\tau }\\&=\frac{1}{2p-1}\Big [p-\frac{p(1-p)\mathbb {E}(\tau )}{\mathbb {E}(\tau ^{2})}\Big ]. \end{aligned}$$

Analogous reasoning gives

$$\begin{aligned} \mu (s|n,\bar{y})&=\frac{1}{2p-1}\Big [-(1-p)+\frac{p(1-p)\mathbb {E}(\tau )}{\mathbb {E}(\tau )-\mathbb {E}(\tau ^{2})}\Big ],\\ \mu (s|y,\bar{n})&=\frac{1}{2p-1}\Big [p-\frac{p(1-p)(1-\mathbb {E}(\tau ))}{\mathbb {E}(\tau )-\mathbb {E}(\tau ^{2})}\Big ],\text { and }\\ \mu (s|n,\bar{n})&=\frac{1}{2p-1}\Big [-(1-p)+\frac{p(1-p)(1-\mathbb {E}(\tau ))}{1-2\mathbb {E}(\tau )+\mathbb {E}(\tau ^{2})}\Big ]. \end{aligned}$$

Since $\pi (y)=\mathbb {E}(\tau )$ and $\pi (n)=1-\mathbb {E}(\tau )$, the above display equations imply that

$$\begin{aligned} a(2,p)= & {} \frac{1}{2p-1}\bigg [1-p(1-p)\bigg (\frac{\mathbb {E}(\tau )^{2}}{\mathbb {V}(\tau )+\mathbb {E}(\tau )^{2}}+\frac{\mathbb {E}(\tau )^{2}+(1-\mathbb {E}(\tau ))^{2}}{\mathbb {E}(\tau )(1-\mathbb {E}(\tau ))-\mathbb {V}(\tau )}\\{} & {} \qquad +\frac{(1-\mathbb {E}(\tau ))^{2}}{(1-\mathbb {E}(\tau ))^{2}+\mathbb {V}(\tau )}\bigg )\bigg ]. \end{aligned}$$

Thus, by algebraic simplification, $a(1,p)\ge a(2,p)$ if and only if

$$\begin{aligned} \frac{1}{\mathbb {E}(\tau )}+\frac{1}{1-\mathbb {E}(\tau )}\le & {} \frac{\mathbb {E}(\tau )^{2}}{\mathbb {V}(\tau )+\mathbb {E}(\tau )^{2}}+\frac{\mathbb {E}(\tau )^{2}+(1-\mathbb {E}(\tau ))^{2}}{\mathbb {E}(\tau )(1-\mathbb {E}(\tau ))-\mathbb {V}(\tau )}\nonumber \\{} & {} \qquad +\frac{(1-\mathbb {E}(\tau ))^{2}}{(1-\mathbb {E}(\tau ))^{2}+\mathbb {V}(\tau )} \end{aligned}$$

(13)

when $p<1$. (It is clear that $a(1,p)=a(2,p)$ when $p=1$ because $p(1-p)=0$.) Yet, it is readily verified that

$$\begin{aligned} \frac{1}{E[\tau ]}+\frac{1}{1-E[\tau ]}=\frac{E[\tau ]^{2}}{E[\tau ]^{2}}+\frac{E[\tau ]^{2}+(1-E[\tau ])^{2}}{E[\tau ](1-E[\tau ])}+\frac{(1-E[\tau ])^{2}}{(1-E[\tau ])^{2}}. \end{aligned}$$

Thus, Eq. (13) holds with equality when $\mathbb {V}(\tau )=0$. When $\mathbb {V}(\tau )>0$, then algebra shows that Eq. (13) holds with strict inequality. Since $\mathbb {V}(\tau )=(2p-1)^{2}\mathbb {V}(v)$ and since $\mathbb {V}(v)>0$ by the full support assumption on $f_{v}(v)$, we have that $\mathbb {V}(\tau )=0$ if and only if $p=\frac{1}{2}$. Consequently, $a(1,p)>a(2,p)$ on $(\frac{1}{2},1)$, with equality at one-half and one.

Part 2: We show that $a(k,p)<a(k+1,p)$ for all $p\in (\frac{1}{2},1)$, with equality at one-half and one, provided the induction hypothesis holds. To these ends, write

$$\begin{aligned} a(k,p)=\sum _{Z=0}^{k-1}\pi (Z)\big (\mu (s|y,Z)-\mu (s|n,Z)\big ), \end{aligned}$$

where $\mu (s|r_{i},Z)$ is the interviewer’s posterior belief after observing respondent i’s answer $r_{i}$ and Z answers of y among the other $k-1$ respondents. Further, we have that

$$\begin{aligned} \mu (s|y,Z)-\mu (s|n,Z)&=\frac{p\mathbb {E}(\nu |Z)}{p\mathbb {E}(\nu |Z)+(1-p)(1-\mathbb {E}(\nu |Z))}\\&\quad -\frac{(1-p)\mathbb {E}(\nu |Z)}{p(1-\mathbb {E}(\nu |Z))+(1-p)\mathbb {E}(\nu |Z)}\\&=\frac{1}{2p-1}\Big [1-p(1-p)\Big (\frac{1}{\mathbb {E}(\tau |Z)}+\frac{1}{1-\mathbb {E}(\tau |Z)}\Big )\Big ], \end{aligned}$$

where the first equality follows from reasoning akin to that used for a(1, p) and the second equality follows from changing variables to $\tau$. Thus,

$$\begin{aligned} a(k,p)=\sum _{Z=0}^{k-1}\pi (Z)\Bigg (\frac{1}{2p-1}\Big [1-p(1-p)\Big (\frac{1}{\mathbb {E}(\tau |Z)}+\frac{1}{1-\mathbb {E}(\tau |Z)}\Bigg )\Big ]\bigg ). \end{aligned}$$

Regarding $a(k+1,p)$, let $\bar{r}$ denote the response of the $k+1$-st respondent. We have $a(k+1,p)$ equal to the following:

$$\begin{aligned}{} & {} \sum _{Z=0}^{k-1}\big (\pi (Z,\bar{y})(\mu (s|y,\bar{y},Z)-\mu (s|n,\bar{y},Z))+\pi (Z,\bar{n})(\mu (s|y,\bar{n},Z)-\mu (s|n,\bar{n},Z))\big )\\{} & {} \quad =\sum _{Z=0}^{k-1}\bigg [\pi (Z,\bar{y})\bigg (\frac{1}{2p-1}\bigg [1-p(1-p)\bigg (\frac{\mathbb {E}(\tau |Z)}{\mathbb {V}(\tau |Z)+\mathbb {E}(\tau |Z)^{2}}+\frac{\mathbb {E}(\tau |Z)}{\mathbb {E}(\tau |Z)(1-\mathbb {E}(\tau |Z))-\mathbb {V}(\tau |Z)}\bigg )\bigg ]\bigg )\\{} & {} \quad +\pi (Z,\bar{n})\bigg (\frac{1}{2p-1}\bigg [1-p(1-p)\bigg (\frac{(1-\mathbb {E}(\tau |Z))}{\mathbb {E}(\tau |Z)(1-\mathbb {E}(\tau |Z))-\mathbb {V}(\tau |Z)}+\frac{(1-\mathbb {E}(\tau |Z))}{(1-\mathbb {E}(\tau |Z))^{2}+\mathbb {V}(\tau |Z)}\bigg )\bigg ]\bigg )\bigg ] \end{aligned}$$

by reasoning akin to that used for a(2, p), where $\pi (Z,\bar{r})$ is the joint probability of Z answers of y among the first $k-1$ other respondents and a response of $\bar{r}$ by $k+1$. Yet, $\pi (Z,\bar{y})=\pi (Z)\pi (\bar{y}|Z)=\pi (Z)\mathbb {E}(\tau |Z)$ and $\pi (Z,\bar{n})=\pi (Z)(1-\mathbb {E}(\tau |Z))$ so we have

$$\begin{aligned} a(k+1,p)&=\sum _{Y=0}^{k-1}\pi (Z)\bigg (\frac{1}{2p-1}\bigg [1-p(1-p)\bigg (\frac{\mathbb {E}(\tau |Z)^{2}}{\mathbb {V}(\tau |Z)+\mathbb {E}(\tau |Z)^{2}}\\&\quad +\frac{\mathbb {E}(\tau |Z)^{2}+(1-\mathbb {E}(\tau |Z))^{2}}{\mathbb {E}(\tau |Z)(1-\mathbb {E}(\tau |Y))-\mathbb {V}(\tau |Z)}+\frac{(1-\mathbb {E}(\tau |Z))^{2}}{(1-\mathbb {E}(\tau |Z))^{2}+\mathbb {V}(\tau |Z)}\bigg )\bigg ]\bigg ). \end{aligned}$$

Algebraic simplification gives that $a(k,p)\ge a(k+1,p)$ when

$$\begin{aligned}{} & {} \frac{1}{\mathbb {E}(\tau |Z)}+\frac{1}{1-\mathbb {E}(\tau |Z)}\le \frac{\mathbb {E}(\tau |Z)^{2}}{\mathbb {V}(\tau |Z)+\mathbb {E}(\tau |Z)^{2}}+\frac{\mathbb {E}(\tau |Z)^{2}+(1-\mathbb {E}(\tau |Z))^{2}}{\mathbb {E}(\tau |Z)(1-\mathbb {E}(\tau |Y))-\mathbb {V}(\tau |Z)}\nonumber \\{} & {} \qquad \qquad +\frac{(1-\mathbb {E}(\tau |Z))^{2}}{(1-\mathbb {E}(\tau |Z))^{2}+\mathbb {V}(\tau |Z)} \end{aligned}$$

(14)

for each $Z\in \{0,1,\ldots ,k-1\}$ when $p<1$. (If $p=1$, then $a(k,p)=a(k+1,p)$ since $p(1-p)=0$.) Yet, Eq. (14) is of the same form as Eq (14) and thus Eq. (14) holds with strict inequality whenever $\mathbb {V}(\tau |Z)>0$. Since $\mathbb {V}(\tau |Z)=(2p-1)^{2}\mathbb {V}(v|Z)$ and since $\mathbb {V}(v|Z)>0$ by the full support assumption on $f_{v}(v)$, we have that $\mathbb {V}(\tau |Z)=0$ if and only if $p=\frac{1}{2}$. Consequently, $a(k,p)>a(k+1,p)$ on $(\frac{1}{2},1)$, with equality at one-half and one.

Part 3: We close by deriving $a(\infty ,p)$. Since $\pi (Y)=\int _{1-p}^{p}\pi (Y|\tau )f_{\tau }(\tau )d\tau$, where $\pi (Y|\tau )$ is the conditional distribution of Y given $\tau$, write

$$\begin{aligned} a(\infty ,p)=\frac{1}{2p-1}\bigg [1-p(1-p)\int _{1-p}^{p}f_{\tau }(\tau )\lim _{N\rightarrow \infty }\sum _{Z=0}^{N-1}\Big (\frac{\pi (Z|\tau )}{\mathbb {E}(\tau |Z)}+\frac{\pi (Z|\tau )}{1-\mathbb {E}(\tau |Z)}\Big )d\tau . \end{aligned}$$

We will establish that

$$\begin{aligned} \lim _{N\rightarrow \infty }\sum _{Z=0}^{N-1}\Big (\frac{1}{\mathbb {E}(\tau |Z)}\pi (Z|\tau )+\frac{1}{1-\mathbb {E}(\tau |Z)}\pi (Z|\tau )\Big )=\frac{1}{\tau }+\frac{1}{1-\tau }. \end{aligned}$$

Consequently,

$$\begin{aligned} a(\infty ,p)=\frac{1}{2p-1}\Big [1-p(1-p)\int _{1-p}^{p}\Big (\frac{1}{\tau }+\frac{1}{1-\tau }\Big )f_{\tau }(\tau )d\tau \Big ], \end{aligned}$$

which is (i) readily verified to be increasing in p (via Leibniz’s rule) and (ii) reduces to the specific form given in Proposition 4 when $f_{v}=1$.

Consider

$$\begin{aligned} \lim _{N\rightarrow \infty }\sum _{Z=0}^{N-1}\frac{1}{\mathbb {E}(\tau |Z)}\pi (Z|\tau ), \end{aligned}$$

which is a conditional expectation of the random variable Z, which has binomial distribution and probability of success $\tau$. The Normal Approximation to the Binomial thus gives $\frac{Z}{N}\sim N(\tau ,\frac{\tau (1-\tau )}{N})$ as N grows large; see Mood et al. (1974) for details. Let $R=\frac{Z}{N}$ and let r denote a realization of R. For each r, $\mathbb {E}(\tau |Nr)$ is a Bayesian estimator and so converges to r as $N\rightarrow \infty$ by the full support of $f_{v}$.³¹ Hence, we write

$$\begin{aligned} \lim _{N\rightarrow \infty }\sum _{Z=0}^{N-1}\frac{1}{\mathbb {E}(\tau |Z)}\pi (Z|\tau )= & {} \lim _{N\rightarrow \infty }\int _{-\infty }^{\infty }\frac{1}{\mathbb {E}(\tau |Nr)}\phi _{R}(r,N)dr\\= & {} \lim _{N\rightarrow \infty }\int _{-\infty }^{\infty }\frac{1}{r}\phi _{R}(r,N)dr=\frac{1}{\tau }, \end{aligned}$$

where $\phi _{R}$ is the density of R at N and where the last equality follows from the fact that the distribution of R collapses to $\tau$ as N grows large. Analogous reasoning gives that

$$\begin{aligned} \lim _{N\rightarrow \infty }\sum _{Z=0}^{N-1}\frac{1}{1-\mathbb {E}(\tau |Z)}\pi (Z|\tau )=\frac{1}{1-\tau }. \end{aligned}$$

This concludes the proof. $\square$

Proof of Proposition 12. Obvious and omitted. $\square$