Precision Threshold and Noise: An Alternative Framework of Sensitivity Measures

Abstract

At many national statistical organizations, linear sensitivity measures such as the prior-posterior and dominance rules provide the basis for assessing statistical disclosure risk in tabular magnitude data. However, these measures are not always well-suited for issues present in survey data such as negative values, respondent waivers and sampling weights. In order to address this gap, this paper introduces the Precision Threshold and Noise framework, defining a new class of sensitivity measures. These measures expand upon existing theory by relaxing certain restrictions, providing a powerful, flexible and functional tool for national statistical organizations in the assessment of disclosure risk.

Appendix

1.1 Proof of Theorem 1

Proof

We start with the first statement, assuming $\tau _1 \ne \sigma _1$. As $f_t(\tau _1)\ge f_t (t)$ for any t and $f_s (\sigma _1 ) \ge f_s (s)$ for any s, it should be clear from (6) that

$$\begin{aligned} S(\tau _1,\sigma _1) \ge S(t,s) \end{aligned}$$

for any pair (t, s), proving the first part of the theorem.

For the second part, we begin with the condition that $\tau _1 = \sigma _1$. Now, suppose $(\tau _1,\sigma _2 )$ is not maximal. Then there exists maximal $(\tau _i,\sigma _j )$ where $(i,j) \ne (1,2)$ such that $f_t (\tau _i )+f_s (\sigma _j )>f_t (\tau _1 )+f_s (\sigma _2 )$. As $f_t (\tau _1 )\ge f_t (\tau _i )$ by definition, it follows that $f_s (\sigma _j )>f_s (\sigma _2 )$ and we can conclude that $j=1$. Then $(\tau _i,\sigma _j )=(\tau _i,\sigma _1 )$ for some $i \ne 1$. But we know that $f_t (\tau _2 )\ge f_t (\tau _i )$ and so $S(\tau _2,\sigma _1 )\ge S(\tau _i,\sigma _1 )$ for any $i\ne 1$. This shows that if $(\tau _1,\sigma _2 )$ is not maximal, $(\tau _2,\sigma _1 )$ must be, completing the proof.    $\square$

1.2 Proof of Theorem 2

Proof

When all respondents are self-aware, $f_t=PT+N$ and $f_s=N$, and consequently any ordering that results in non-ascending PT, N also results in non-ascending $f_t$, $f_s$. Setting $\tau =\sigma =\eta$ and applying Theorem 1, we conclude that one of $(\eta _1,\eta _2)$ or $(\eta _2,\eta _1)$ is maximal. From (6) we can see that

$$S(\eta _1,\eta _2)-S(\eta _2,\eta _1) =PT(\eta _1)-PT(\eta _2) \ge 0$$

showing $S(\eta _1,\eta _2) \ge S(\eta _2,\eta _1)$ and $(\eta _1,\eta _2)$ is maximal.   $\square$

1.3 Proof of Theorem 3

Proof

The proof is self-evident for cells with two or fewer respondents, so we will assume there are at least three. Applying Theorem 1 and noting $f_s=N$ we can conclude that there exists a maximal pair of the form $(\eta _i,\eta _j)$ for $j \le 2$. As this pair is maximal it can be used to calculated cell sensitivity:

$$ S^1_1=S(\eta _i,\eta _j)=PT(\eta _i)-\sum _{r \ne i,j} N(\eta _r) $$

As $j \le 2$, if $i \ge 3$ then exactly one of $N(\eta _1)$ or $N(\eta _2)$ is included in the summation above. Both of these are $\ge N(\eta _i)$ by ordering $\eta$, which is $\ge PT(\eta _i)$ by assumption. This means $S^1_1 < 0$ and the cell is safe. Conversely, if the cell is sensitive, there must exist a maximal pair of the form $(\eta _i,\eta _j)$ with both $i,j \le 2$, completing the proof.    $\square$

1.4 Interpreting Arbitrary Linear Sensitivity Measures in $S^{n_t}_{n_s}$ Form

All linear sensitivity measures of the form $\sum _r \alpha _r x_r$ can be expressed in PTN form, provided they satisfy the following conditions:

• Finite number of non-negative coefficients

• All positive coefficients have the same value, say $\alpha _+$

• All negative coefficients have the same value, say $\alpha _-$.

Assuming these conditions are met, an equivalent PTN sensitivity measure can be defined as follows:

• Set $n_t$ equal to the number of positive coefficients

• Set $n_s$ equal to the number of coefficients equal to zero

• Set $PT(r)=\alpha _+ x_r$ for all r

• Set $N(r)=|\alpha _-| x_r$ and $SN(r)=0$ for all r

We show that the resulting PTN cell sensitivity measure is equivalent to $\sum _r \alpha _r x_r$ by first writing (7) as follows:

$$\begin{aligned} S(T,S) = \sum _{t \in T}\left( PT(t)+N(t) \right) +\sum _{s \in S} \left( N(s) - SN(s) \right) - \sum _{r } N(r) \end{aligned}$$

(9)

Substituting in the appropriate PTN values gives

(10)

It is easy to see that T and S should be selected from the largest $n_t + n_s$ respondents to maximize S(T, S). If they are already indexed in non-ascending order, then sensitivity is maximized when $T=\left\{ 1,\ldots ,n_t \right\}$ and $S= \left\{ n_t+1,\ldots ,n_t+n_s \right\}$. Then cell sensitivity is given by

$$\begin{aligned} S^{n_t}_{n_s} = \sum _{r =1}^{n_t} \alpha _+ x_r -\sum _{r > n_t + n_s} |\alpha _-| x_r \end{aligned}$$

(11)

which is exactly $\sum _r \alpha _r x_r$.

1.5 Proof of Theorem 4

We begin with a simple lemma:

Lemma 1

Let T and S be non-intersecting sets of respondents. Let k be a respondent in neither, and assume $SN(k) \le N(k)$. Then

(12)

Proof

We write (7) in maximal form, substituting in the target and suspect functions:

$$\begin{aligned} S(T,S) = \sum _{t \in T} f_t(t) + \sum _{s \in S} f_s(s)- \sum _{r} N(r) \end{aligned}$$

(13)

Then $S(T, S \cup k) - S(T,S) = f_s(k)$. As $SN(k) \le N(k)$ by assumption (we expect this to be true anyway, as a respondent should never know less about their own contribution than the general public), $f_s \ge 0$ proves the first inequality. The second inequality holds because $f_t \ge f_s$ for all respondents, including k.    $\square$

With this lemma, the proof of Theorem 4 is almost trivial:

Proof

Let (T, S) be maximal with respect to $S^{n_t}_{n_s}$. We know there exists at least one respondent $k \notin T \cup S$, and by Lemma 1, $S(T, S) \le S(T, S \cup k)$, proving that$S^{n_t}_{n_s} \le S^{n_t}_{n_s+1}$.

For the second inequality, we note that any set pair that is maximal with respect to $S^{n_t}_{n_s+1}$ can be written in the form $(T, S \cup k)$ for some T of size $n_t$, S of size $n_s$ and single respondent k. Once again applying Lemma 1 we see that $S(T, S \cup k) \le S(T \cup k , S)$ and consequently $S^{n_t}_{n_s+1} \le S^{n_t+1}_{n_s}$.    $\square$