On the optimality of vagueness: “around”, “between” and the Gricean maxims

Abstract

Why is ordinary language vague? We argue that in contexts in which a cooperative speaker is not perfectly informed about the world, the use of vague expressions can offer an optimal tradeoff between truthfulness (Gricean Quality) and informativeness (Gricean Quantity). Focusing on expressions of approximation such as “around”, which are semantically vague, we show that they allow the speaker to convey indirect probabilistic information, in a way that can give the listener a more accurate representation of the information available to the speaker than any more precise expression would (intervals of the form “between”). That is, vague sentences can be more informative than their precise counterparts. We give a probabilistic treatment of the interpretation of “around”, and offer a model for the interpretation and use of “around”-statements within the Rational Speech Act (RSA) framework. In our account the shape of the speaker’s distribution matters in ways not predicted by the Lexical Uncertainty model standardly used in the RSA framework for vague predicates. We use our approach to draw further lessons concerning the semantic flexibility of vague expressions and their irreducibility to more precise meanings.

Appendices

Appendix A: A limitation result about the LU model

In this appendix, we prove that in the Lexical Uncertainty model (LU model, Bergen et al., 2016), if two observations $o_1$ and $o_2$ are such that the support of the conditional distributions $P(w|o_1)$ and $P(w|o_2)$ are the same, then, at every level of the recursion, the speaker’s probability of using a message m if she observed $o_1$ is the same as if she observed $o_2$. It follows that in the LU model, only the support of the subjective probability distribution of the speaker, and not its shape, plays a role in her choice of a message, in contrast with our model.

The LU model is defined by the following equations, where $\llbracket m \rrbracket ^i(w)$ is the truth-value of the literal meaning of m, relative to the interpretation function i, in world w, and $\llbracket m \rrbracket ^i$ denotes the set of worlds where m is true relative to interpretation i (in our setting, i is what determines the interpretation of “around”, i.e. a certain value for y). The parameter $\uplambda$ is a non-null, positive real number. For any message m, c(m) is the cost of m, a null or positive real number. P is the prior distribution about the possible values of w (world state), o (observation) and i, and is such that the value taken by i is probabilistically independent from w and o.

• 1. $L^0(w,o | m,i) = \dfrac{P(w,o) \times \llbracket m \rrbracket ^i(w)}{P(\llbracket m \rrbracket ^i)}$

• 2. $U^1(m|o,i) = (\sum \limits _{w}P(w | o) \times \log (L^0(w,o | m,i))) - c(m)$

• 3. $S^1(m | o, i) = \dfrac{\exp (\uplambda U^1(m,o,i))}{ \sum \limits _{m'}\exp (\uplambda U^1(m',o,i))}$

• 4. $L^1(w,o | m) = \dfrac{P(w,o)\times \sum \limits _{i} P(i)S^1(m | o, i)}{\alpha _1(m)}$, where $\alpha _1(m) = \sum \limits _{w',o'} (P(w',o')$ $\sum \limits _{i} P(i)S^1(m | o', i))$

• 5. For $n \ge 1$, $U^{n+1}(m|o) = (\sum \limits _w P(w| o)\log (L^n(w,o | m))) - c(m)$

• 6. $S^{n+1}(m | o) = \dfrac{\exp (\uplambda U^{n+1}(m,o))}{\sum \limits _{m'} \exp (\uplambda U^{n+1}(m',o))}$

• 7. For $n \ge 2$, $L^n(w,o | m) = \dfrac{P(w,o) \times S^n(m | o)}{\alpha _n(m)}$, where $\alpha _n(m) = \sum \limits _{w',o'}P(w',o')$ $S^n(m | o')$

Results to be proven

If o is an observation, $P_o$ is the probability distribution over worlds defined by:

$P_o(w) = P(w,o|o) = P(w|o)$

• 1. If two observations $o_1$ and $o_2$ are such that $P_{o_1}$ and $P_{o_2}$ have the same support (i.e. for every w, $P(w|o_1)>0$ iff $P(w|o_2) > 0$), then for every message m and every interpretation function i, $S^1(m|o_1, i) = S^1(m|o_2, i)$

• 2. If two observations $o_1$ and $o_2$ are such that $P_{o_1}$ and $P_{o_2}$ have the same support, then for every $n \ge 2$, $S^n(m|o_1) = S^n(m|o_2)$

1.1 A.1: Proof strategy

The key ingredient of the proof is the following. Consider two observations $o_1$ and $o_2$ such that $P_{o_1}$ and $P_{o_2}$ have the same support (i.e. they assign a non-null probability to the same worlds). Consider the set of messages ${\mathcal {M}}$ that express a proposition that is entailed by this support under at least one interpretation i (a condition we call Weak Quality—as we shall see, other messages are not usable at all by a speaker who observed $o_1$ or $o_2$, as they violate Grice’s maxim of Quality, cf. (A-2) below). We prove that, at every level of recursion, the utility achieved by each message in ${\mathcal {M}}$ for a speaker who observed $o_2$ is equal to the one achieved for a speaker who observed $o_1$ plus a constant term which does not depend on the message. In other words, there is a quantity K which depends on the various parameters of the models and on $o_1$ and $o_2$ but crucially not on the message m, such that for every message m, $U^n(m|o_2) = U^n(m|o_1) + K$.

Now, the speaker’s strategy in the RSA model as defined in Eqs.  3 and 6 is based on the so-called SoftMax function.⁴¹ The SoftMax function turns a sequence of numerical values (in our case the utilities of each message relative to a certain observation) into a probability distribution over members of this sequence. In the RSA model, the speaker’s strategy relative to a given observation is obtained by applying the SoftMax function to the utilities of each message relative to this observation. The SoftMax function enjoys a property known as Translation Invariance, as we prove below.⁴² That is, if, starting from a sequence of numerical values, we consider the sequence obtained by adding a constant term to each value, and then apply the SoftMax function to these shifted values, the resulting probability distribution over the members of the new sequence is exactly the same as the one obtained when applying the SoftMax function to the initial sequence. Since the utilities achieved by different messages relative to $o_2$ can be obtained by adding a constant term to those achieved by the very same messages relative to $o_1$ (as explained in the previous paragraph), this means that the resulting probability distribution over messages is the same for $o_1$ and $o_2$.

We will first prove that for $S^1$, relative to a fixed interpretation i, the difference between the utility achieved by a message m in ${\mathcal {M}}$ relative to $o_1$ and the one achieved by the same message relative to $o_2$ does not depend on m, and is therefore the same across messages in ${\mathcal {M}}$. Thanks to Translation Invariance, this will ensure that for every m in ${\mathcal {M}}$ and every i, $S^1(m|o_2, i) = S^1(m|o_1, i)$. Then we will do the same with $S^2$—there has to be a separate step for $S^2$ because the variable i that appears in the definition of $S^1$ no longer appears when $n \ge 2$, so we start the inductive proof at level 2. Then we complete the proof by proving the inductive step: assuming that for every $n \ge 2$, $S^n(m|o_2) = S^n(m|o_1)$, we prove that at the level $n+1$, the difference in the utility achieved by a message m in ${{\mathcal {M}}}$ relative to $o_2$ and relative to $o_1$ does not depend on m, which, thanks to Translation Invariance, ensures that for every m in ${{\mathcal {M}}}$, $S^{n+1}(m|o_2) = S^{n+1}(m|o_1)$. At each step, the reason why the key result is observed is that the log-function which appears in the definition of the utility function turns products into sums, allowing us, together with the fact that probabilities sum up to 1, to separate terms which depend on m from those that depend on o, and all terms in which m appears cancel out when we consider the difference $U(m|o_2)-U(m|o_1)$.

We now give the detailed proof.

1.2 A.2: Quality and Weak Quality

Definitions

• 1. We say that a message m respects Quality with respect to an observation o and an interpretation i if, for every w, if $P(w | o) > 0$, then $\llbracket m \rrbracket ^i(w)=1$.

• 2. We say that a message m respects Weak Quality with respect to an observation o if there exists an interpretation i such that $P(i)>0$ and m respects Quality with respect to o and i.

Let $o_1$ and $o_2$ be two observations such that $P_{o_1}$ and $P_{o_2}$ have the same support, i.e. for every w, $P(w|o_1)>0$ iff $P(w|o_2) > 0$. From now on, $o_1$ and $o_2$ denote two such observations.

Proof of Lemma (A-1)

Obvious from the definitions of Quality and Weak Quality, and the fact that $P_{o_1}$ and $P_{o_2}$ have the same support.

Proof of the facts in (A-2)

First we prove the result for $S^1$, then for $S^2$ and then by induction for higher values of n.

Proof of (A-2-a)—If m does not respect Quality with respect to o and i, then for some w such that $P(w|o) > 0$, $\llbracket m \rrbracket ^i(w)=0$. For such a w, then, $L^0(w,o|m,i)=0$, hence $\log (L^0(w,o|m,i)) = - \infty$. So at least one term in the sum which defines $U^1(m|o,i)$ evaluates to $-\infty$, and so the sum itself does, hence $U^1(m|o,i) = - \infty$.⁴³ Since $S^1(m|o,i)$ involves exponentiating a quantity that is infinitely negative, we have $S^1(m|o,i) = 0$. Reciprocally, if m does respect Quality with respect to o and i, then no term in the sum is infinitely negative, and $U^1(m|o,i)$ is not infinitely negative either, and so $S^1(m|o,i)>0$.

Proof of (A-2-b)—The proof is by induction.

Base-case ($n=2$): Suppose now that m does not respect Weak Quality with respect to o. Then for every i such that $P(i)>0$, m does not respect Quality with respect to o, i; and so by the result just proven, every term in the sum $\sum \nolimits _{i} P(i)S^1(m | o, i)$ is equal to 0, and so is the sum as a whole. As a result, $L^1(w,o|m)=0$, for every w. From this it follows that the sum $\sum \nolimits _w P(w| o)\log (L^1(w,o | m))$ evaluates to $-\infty$, and therefore so does $U^2(m|o)$. $S^2(m|o)$ involves again exponentiating an infinitely negative value, so is equal to 0. Reciprocally, if m respects Weak Quality with respect to o, there is at least one i relative to which $P(i)\times S^1(m | o, i)>0$, and therefore $\sum \nolimits _{i} P(i)S^1(m | o, i)$ is not equal to 0. For every w such that $P(w,o)>0$, then $L^1(w,o|m)>0$, and therefore $U^2(m|o)$ is not infinitely negative, and so $S^2(m|o)>0$.

Inductive step: Finally, let assume that the result holds for $S^n$ (Induction Hypothesis). Suppose again that m does not respect Weak Quality with respect to o. By the Induction Hypothesis, $S^n(m|o)=0$, and therefore for every w, $L^n(w,o|m)=0$ (given the definition of $L^n$). Then $U^{n+1}(m|o) = - \infty$, as in each term of the sum that defines $U^{n+1}(m|o)$, the $\log$-function takes 0 as its argument. It follows that $S^{n+1}(m|o)=0$. Reciprocally, if m does respect Weak Quality with respect to o, then $S^n(m|o)>0$, and therefore for every w such that $P(w|o)>0$, $L^n(w,o|m)>0$, from which it follows that $U^{n+1}(m|o)$ is not infinitely negative and therefore that $S^{n+1}(m|o) >0$.

1.3 A.3: Reformulating utility functions in terms of SoftMax, SoftMax invariance

The SoftMax function

The SoftMax function takes three arguments: a sequence of real numbers $\textbf{x} = (x_1, \ldots , x_i)$, a member of this sequence, and the paratemer $\uplambda$. It is defined as follows:

$$\begin{aligned} \text {SoftMax}(x_k, \textbf{x}, \uplambda ) = \dfrac{\exp (\uplambda x_k)}{\sum \limits _{x_i \in \textbf{x}} \exp (\uplambda x_i)} \end{aligned}$$

Reformulating the utility function in terms of SoftMax

Equation 3 is repeated here:

$$\begin{aligned} S^1(m | o, i) = \dfrac{\exp (\uplambda U^1(m,o,i))}{ \sum \limits _{m'}\exp (\uplambda U^1(m',o,i))} \end{aligned}$$

Note that in the denominator, every term of the sum corresponding to a message which does not respect Quality with respect to o and i is equal to 0 [because its utility is infinitely negative, and so after exponentiation we get 0, cf. (A-2-a)]. This means that we can restrict the denominator to the messages that respect Quality with respect to o and i.

Let ${\mathcal {M}}_o = <m_1, \ldots , m_j, \ldots>$ be an ordered sequence of messages which contains all and only the messages that respect Quality with respect to o and i.⁴⁴ The above equation can then be rewritten as:

$$\begin{aligned} S^1(m | o, i) = \dfrac{\exp (\uplambda U^1(m,o,i))}{ \sum \nolimits _{m_j \in {\mathcal {M}}_o}\exp (\uplambda U^1(m_j,o,i))} \end{aligned}$$

Let $\overrightarrow{U^1_{o,i}}$ be the sequence of numbers that one gets by applying the function $U^1(\ldots |o,i)$ pointwise to the sequence ${\mathcal {M}}_o$ (i.e. $\overrightarrow{U^1_{o,i}}= <U^1(m_1|o, i), \ldots , U^1(m_j|o,i), \ldots>$)

Then Eq. 3 can be rewritten as follows (when m respects Weak Quality with respect to o and i):

Consider now Eq. 6, which we repeat here:

$$\begin{aligned} S^{n+1}(m | o) = \dfrac{\exp (\uplambda U^{n+1}(m|o))}{\sum \limits _{m'} \exp (\uplambda U^{n+1}(m'|o))} \end{aligned}$$

For any n, if $m'$ does not respect Weak Quality with respect to o, $U^{n+1}(m'|o) = -\infty$ (cf. (A-2-b)), and every term in the sum in the denominator that corresponds to such a message $m'$ is equal to 0 (exponentiation of $-\infty$). With $\mathcal {M'}_o$ a sequence $\{m_1,\ldots ,m_j, \ldots \}$ containing all and only the messages that respect Weak Quality with respect to o, we can restrict the sum to the members of $\mathcal {M'}_o$. Let $\overrightarrow{U^{n+1}_o}$ be the sequence that one gets by applying the function $U^{n+1}(\ldots |o)$ pointwise to the sequence $\mathcal {M'}_o$ (i.e. $\overrightarrow{U^{n+1}_o}= <U^{n+1}(m_1|o), \ldots , U^{n+1}(m_j|o), \ldots>$). We have (for m in $\mathcal {M'}_o$):

Translation Invariance of SoftMax

Notation:  If $\textbf{x}$ is a sequence of real numbers $<x_1, \ldots , x_i, \ldots>$ and a is a real number, we notate $\textbf{x} \oplus a$ the sequence $<x_1 +a,\ldots , x_i + a, \ldots>$.

Proof

$$\begin{aligned} \text {SoftMax}(x_k + a, \textbf{x} \oplus a, \uplambda )= & {} \dfrac{\exp (\uplambda x_k + a)}{\sum \limits _{y_j \in \textbf{x} \oplus a} \exp (\uplambda y_j)}\\= & {} \dfrac{\exp (\uplambda x_k + a)}{\sum \limits _{x_j \in \textbf{x}} \exp (\uplambda (x_j + a))}\\= & {} \dfrac{\exp (\uplambda x_k) \times \exp (\uplambda a)}{\sum \limits _{{x_j} \in \textbf{x}} \exp (\uplambda x_j) \times \exp (\uplambda a)}\\= & {} \dfrac{\exp (\uplambda x_k)}{\sum \limits _{{x_j} \in \textbf{x}} \exp (\uplambda x_j)}\\= & {} \text {SoftMax}(x_k, \textbf{x}, \uplambda ) \end{aligned}$$

1.4 A.4: Proving the result for $S^1$

Recall that $o_1$ and $o_2$ are two observations such that $P_{o_1}$ and $P_{o_2}$ have the same support.

Proof of the Lemma in (A-6)

Let us assume that m, i and $o_1$ and $o_2$ meet the condition stated in (A-6), i.e. that m respects Quality with respect to i and $o_1$, and with respect to i and $o_2$.

$$\begin{aligned} U^1(m | o_1, i)= & {} \sum \limits _{w}P(w | o_1)\log (L^0(w,o_1 | m,i)) - c(m)\\= & {} \sum \limits _{w}P(w | o_1)\log \left( \dfrac{P(w,o_1)\times \llbracket m \rrbracket ^i(w)}{P(\llbracket m \rrbracket ^i)}\right) - c(m)\\ \end{aligned}$$

Let us notate S the support of $P_{o_1}$ and $P_{o_2}$. Note that $P(w|o_1) = 0$ if $w \notin S$. Furthemore, since m respects Quality with respect to $o_1$, $o_2$, and i, then if $w \in S$, $\llbracket m \rrbracket ^i(w) = 1$. It follows that we can restrict the sum to the worlds in S (because all the terms in the sum are equal to 0 when w is not in S), and that, having done this, we can remove $\llbracket m \rrbracket ^i(w)$ from the equation (since $\llbracket m \rrbracket ^i(w)$ is always equal to 1 when $w \in S$). We can therefore continue as follows:

$$\begin{aligned} U^1(m | o_1, i)= & {} \sum \limits _{w \in S}P(w | o_1)\log \left( \dfrac{P(w,o_1)}{P(\llbracket m \rrbracket ^i)}\right) - c(m) \\= & {} \sum \limits _{w \in S}P(w | o_1)[\log (P(w,o_1)) - \log (P(\llbracket m \rrbracket ^i))] - c(m)\\= & {} \sum \limits _{w \in S}P(w | o_1)\log (P(w,o_1)) - \sum \limits _{w \in S}P(w | o_1)\log (P(\llbracket m \rrbracket ^i)) - c(m)\\= & {} \sum \limits _{w \in S}P(w | o_1)\log (P(w,o_1)) - \log (P(\llbracket m \rrbracket ^i))\times \underbrace{\sum \limits _{w \in S}P(w | o_1)}_{=1} - c(m)\\= & {} \sum \limits _{w \in S}P(w | o_1)\log (P(w,o_1)) - \log (P(\llbracket m \rrbracket ^i)) - c(m). \end{aligned}$$

The same formula of course holds for $o_2$, replacing every occurrence of $o_1$ with $o_2$. Given this, when we subtract $U^1(m | o_1, i)$ from $U^1(m | o_2, i)$, the terms that depend on m ($- \log (P(\llbracket m \rrbracket ^i)) - c(m)$) cancel out, and we get:

$$\begin{aligned} U^1(m | o_2, i) - U^1(m | o_1, i)= & {} \sum \limits _{w \in S}P(w | o_2)\log (P(w,o_2)) \\{} & {} - \sum \limits _{w \in S}P(w | o_1)\log (P(w,o_1)) \end{aligned}$$

As promised, then, this difference does not depend on m or i.

Proof of (A-7)

First consider the case where m does not respect Quality with respect to $o_1$, $o_2$, i (again, relative to a fixed i, either it respects Quality for both $o_1$ and $o_2$, or for neither, cf. (A-1)). In this case, given the first fact in (A-2), $S^1(m | o_1, i) = S^1(m | o_2, i) = 0$.

Consider now the case were m respects Quality with respect to $o_1$, $o_2$, i. Let us define $K(o_1, o_2) = \sum \nolimits _{w \in S}P(w | o_2)\log (P(w,o_2)) - \sum \nolimits _{w \in S}P(w | o_1)\log (P(w,o_1))$. From the lemma in (A-6), we have: for every $m'$ which respects Quality with respect to $o_1$, $o_2$, i,

$$\begin{aligned} U^1(m' | o_2, i) =U^1(m' | o_1, i) + K(o_1, o_2). \end{aligned}$$

Given that this result holds for all the messages that respect Quality with respect to $o_1$, $o_2$ and i (recall that the messages that respect Quality w.r.t. $o_1$, i are the same as those that respect it w.r.t. $o_2$, i), it can be restated as follows, using the notations introduced in Sect. 1:⁴⁵

$$\begin{aligned} \overrightarrow{U^1_{o_2,i}} = \overrightarrow{U^1_{o_1,i}} \oplus K(o_1,o_2) \end{aligned}$$

Given (A-3) and Translation Invariance (A-5), we have:

$$\begin{aligned} S^1(m | o_2, i)= & {} \text {SoftMax}(U^1(m|o_2, i), \overrightarrow{U^1_{o_2,i}},\uplambda )\\= & {} \text {SoftMax}(U^1(m|o_1, i) + K(o_1, o_2), \overrightarrow{U^1_{o_1,i}} \oplus K(o_1,o_2),\uplambda )\\= & {} {}_{(\text {{Translation Invariance}})}~ \text {SoftMax}(U^1(m|o_1, i), \overrightarrow{U^1_{o_1,i}},\uplambda )\\= & {} S^1(m|o_1, i) \end{aligned}$$

1.5 A.5: Proving the result for $n \ge 2$

This will be a proof by induction.

Recall again that $o_1$ and $o_2$ are two observations such that $P_{o_1}$ and $P_{o_2}$ have the same support.

First, note that if a certain message m does not satisfy Weak Quality with respect to $o_1$, $o_2$, then given the second fact in (A-2), for any $n \ge 2$, $S^n(m|o_1)= S^n(m|o_2)=0$, so we can now ignore this case and assume for the rest of the proof that m does satisfy Weak Quality with respect to $o_1$ and $o_2$ (recall that it either respects it for both or neither, cf. Lemma (A-1)).

1.5.1 A.5.1: Base-Case: $n =2$

We start with a counterpart to the Lemma in (A-6).

Proof of (A-9)

Assume that m, $o_1$ and $o_2$ meet the condition of the above Lemma. Note that, since $P_{o_1}$ and $P_{o_2}$ have the same support, the interpretations i such that m respects Weak Quality with respect to $o_1$ and i are the same as the interpretations i such that m respects Weak Quality with respect to $o_2$ and i (cf. Lemma (A-1)). As before, we call S the support of $P_{o_1}$ and $P_{o_2}$. We have, given the definitions:⁴⁶

$$\begin{aligned} U^2(m | o_1)= & {} \sum \limits _{w \in S} P(w | o_1) \times \log \left( \frac{P(w,o_1)\sum \limits _{i}P(i)S^1(m | o_1, i)}{\alpha _1(m)} \right) \\= & {} \sum \limits _{w \in S} P(w | o_1) \times [\log (P(w,o_1)) + \log \left( \sum \limits _{i}P(i)S^1(m | o_1, i)\right) - \log (\alpha _1(m))] \end{aligned}$$

Likewise, we have:

$$\begin{aligned} U^2(m | o_2)= & {} \sum \limits _{w \in S} P(w | o_2) \times [\log (P(w,o_2)) \!+\! \log \left( \sum \limits _{i}P(i)S^1(m | o_2, i)\right) \\ {}{} & {} - \log (\alpha _1(m))] \end{aligned}$$

Recall that for every i, $S^1(m | o_1,i) = S^1(m | o_2, i)$ (Theorem in (A-7)). It follows that the quantities $\log (\sum \nolimits _{i}P(i)S^1(m | o_1, i))$ and $\log (\sum \nolimits _{i}P(i)S^1(m | o_2, i))$ are equal. Let us call this quantity X. We can rewrite the above formulae as:

$$\begin{aligned} U^2(m | o_1) = \sum \limits _{w \in S} P(w | o_1) \times [\log (P(w,o_1)) + X - \log (\alpha _1(m))]\\ U^2(m | o_2) = \sum \limits _{w \in S} P(w | o_2) \times [\log (P(w,o_2)) + X - \log (\alpha _1(m))] \end{aligned}$$

A few lines of computation yield the lemma stated in (A-9):

$$\begin{aligned} U^2(m | o_2) - U^2(m | o_1)= & {} (X - \log (\alpha _1(m))\times \underbrace{\left( \underbrace{\sum \limits _{w \in S} P(w | o_2)}_{=1} - \underbrace{\sum \limits _{w \in S}P(w | o_1)}_{=1}\right) }_{=0}\\+ & {} \sum \limits _{w \in S} P(w | o_2)\log (P(w,o_2)) - \sum \limits _{w \in S} P(w | o_1)\log (P(w,o_1))\\= & {} \sum \limits _{w \in S} P(w | o_2)\log (P(w,o_2)) - \sum \limits _{w \in S} P(w | o_1)\log (P(w,o_1)) \end{aligned}$$

As promised, this difference does not depend on m.

Using Translation Invariance to prove the base-case ($n=2$)

To complete the proof for the base case ($S^2(m|o_1) = S^2(m|o_2)$), we just need to exploit again the Translation-Invariance property of SoftMax. The computation proceeds in exactly the same way as in the proof of the Theorem in (A-7):

With $K(o_1, o_2) = \sum \nolimits _{w \in S} P(w | o_2)\log (P(w,o_2)) - \sum \nolimits _{w \in S} P(w | o_1)\log (P(w,o_1))$, we have, for any message m that respects Weak Quality with respect to $o_1$, $o_2$, $U^2(m|o_2) = U^2(m|o_1) + K(o_1, o_2)$.

Since the messages that respect Weak Quality with respect to $o_1$ and with respect to $o_2$ are the same, we have:

$$\begin{aligned} \overrightarrow{U^2_{o_2}} = \overrightarrow{U^2_{o_1}} + K(o_1, o_2) \end{aligned}$$

We therefore have, thanks to Translation Invariance and (A-4):

$$\begin{aligned} S^2(m |o_2)= & {} \text {SoftMax}(U^2(m|o_2), \overrightarrow{U^2_{o_2}},\uplambda )\\= & {} \text {SoftMax}(U^2(m|o_1) + K(o_1, o_2), \overrightarrow{U^2_{o_1}} \oplus K(o_1,o_2),\uplambda )\\= & {} {} _{(\text {{Translation Invariance}})}~ \text {SoftMax}(U^2(m|o_1), \overrightarrow{U^2_{o_1}},\uplambda )\\= & {} S^2(m|o_1) \end{aligned}$$

1.5.2 A.5.2: Inductive step for the Theorem in (A-8)

Recall again that $P_{o_1}$ and $P_{o_2}$ have the same support.

Induction Hypothesis: We assume that $S^n(m | o_2) = S^n(m | o_1)$.

We want to prove that $S^{n+1}(m| o_2) = S^{n+1}(m | o_1)$.

As before, the key intermediate result is the following:

Proof of (A-10)

We have:

$$\begin{aligned} U^{n+1}(m | o_1)= & {} \sum \limits _{w \in S} P(w | o_1) \times \log \left( \dfrac{P(w,o_1)S^n(m | o_1)}{\alpha _n(m)}\right) \\= & {} \sum \limits _{w \in S} P(w | o_1) \times [\log (P(w,o_1)) + \log (S^n(m | o_1)) - \log (\alpha _n(m))] \end{aligned}$$

Likewise, we have:

$$\begin{aligned} U^{n+1}(m | o_2)= & {} \sum \limits _{w \in S} P(w | o_2) \times [\log (P(w,o_2) + \log (S^n(m | o_2)) - \log (\alpha _n(m))] \end{aligned}$$

By the induction hypothesis, $\log (S^n(m | o_2)) = \log (S^n(m | o_1))$. Calling this quantity X, we then have:

$$\begin{aligned} U^{n+1}(m | o_1)= & {} \sum \limits _{w \in S} P(w | o_1) \times [\log (P(w,o_1)) + X - \log (\alpha _n(m))]\\ U^{n+1}(m | o_2)= & {} \sum \limits _{w \in S} P(w | o_2) \times [\log (P(w,o_2)) + X - \log (\alpha _n(m))] \end{aligned}$$

The computation then proceeds exactly as in the proof for the Lemma in (A-9)—terms that depend on m cancel out when we take the difference between the two lines, and we so we can conclude the proof of (A-10):

$$\begin{aligned} U^{n+1}(m | o_2) \!-\! U^{n+1}(m | o_1) \!=\! \sum \limits _{w \in S}P(w | o_2)\log (P(w,o_2)) \!-\! \sum \limits _{w \in S}P(w | o_1)\log (P(w,o_1)) \end{aligned}$$

Completing the proof using Translation Invariance

On the basis of (A-10), the proof that $S^{n+1}(m|o_2) = S^{n+1}(m|o_1)$ proceeds in exactly the same way as in the case of $n=2$, and boils down to Translation Invariance again.

With $K(o_1, o_2) = \sum \nolimits _{w \in S} P(w | o_2)\log (P(w,o_2)) - \sum \nolimits _{w \in S} P(w | o_1)\log (P(w,o_1))$, we have, for any message m that respects Weak Quality with respect to $o_1$, $o_2$, $U^{n+1}(m|o_2) = U^{n+1}(m|o_1) + K(o_1, o_2)$.

Since the messages that respect Weak Quality with respect to $o_1$ and with respect to $o_2$ are the same, we have:

$$\begin{aligned} \overrightarrow{U^{n+1}_{o_2}} = \overrightarrow{U^{n+1}_{o_1}} \oplus K(o_1, o_2) \end{aligned}$$

Given Translation Invariance and (A-4), we have:

$$\begin{aligned} S^{n+1}(m |o_2)= & {} \text {SoftMax}(U^{n+1}(m|o_2), \overrightarrow{U^{n+1}_{o_2}},\uplambda )\\= & {} \text {SoftMax}(U^{n+1}(m|o_1) + K(o_1, o_2), \overrightarrow{U^{n+1}_{o_1}} \oplus K(o_1,o_2),\uplambda )\\= & {} {}_{(\text {{Translation Invariance}})}~ \text {SoftMax}(U^{n+1}(m|o_1), \overrightarrow{U^{n+1}_{o_1}},\uplambda )\\= & {} S^{n+1}(m|o_1) \end{aligned}$$

This completes the proof of (A-8).

Appendix B: An alternative model for the literal listener

The model presented in Sect. 3.2 was originally derived from a distinct model of the listener that we first came up with, which we present in this appendix for comparison. Although that model makes basically the same qualitative predictions, it is not a Bayesian model, and it makes different quantitative predictions.

Like the Bayesian model, this model assumes that the listener has a probability distribution P over the intervals selected by “around”, and over the possible values some variable x can take. However, the alternative model (written WIR, for Weighted Interpretation Rule) says that the listener’s posterior value of x upon hearing “x is around n”, notated $L(x = \ldots |\ x\ \text {is around}\ n)$, is the sum of the conditional probabilities that x takes that value given that x belongs to a given interval, weighted by the probability of that interval:⁴⁷

$$\begin{aligned} {L}(x=k\ |\ x\ \text {is around}\ n)=_{df}\sum _{i} P(x=k\ |\ x\in [n-i, n+i])P(y=i) \end{aligned}$$

(WIR)

To see the difference with the Bayesian model, recall Eq. BIR, which is reproduced here (with an explicit formula for the proportionality factor):

$$\begin{aligned} P(x=k\mid x ~\hbox {is}\, \hbox {around}\, n) = \dfrac{P(x=k) \times \sum \limits _{i \ge |n-k|} P(y=i)}{\sum \limits _k P(x=k)\times \sum \limits _{i \ge |n-k|} P(y=i)} \end{aligned}$$

The two models are distinct. For instance, when P is uniform over values of x as well as over candidate meanings for “around”, the distribution $L(x = \ldots |\ x\ \text {is around}\ n)$ is distinct from the one depicted in Fig. 2, and it is no longer linear, as represented in Fig. 5.

Fig. 5

$L(x=k|\ x\ \text {is around}\ n)$, with maximum interval [0, 40]

To see more precisely how the two models differ conceptually, the following observation will be useful. Let $P_{post}$ be the posterior probability distribution resulting from updating P with the “around n” message in the Bayesian model, i.e.:

$$\begin{aligned} P_{post}(x=k, y=i) =_{df} P(x=k, y=i\ |\ x \in [n-y, n+y]). \end{aligned}$$

It can be proved that:

$$\begin{aligned} P_{post}(x=k) = \sum \limits _{i} P(x=k\ |\ x \in [n-i, n+i])P_{post}(y=i) \end{aligned}$$

(BIR')

Equation WIR looks almost like BIR’, except that the first term in WIR is weighted by the prior distribution on the values of y (the candidate meanings for “around”) instead of the posterior. This is the sense in which the model proposed in WIR is not Bayesian. Instead of the listener updating also her probability of intervals after hearing “x is around n”, the listener does not make her interval probability depend on that information. The model is not illegitimate for that matter. Regarding our explananda, it makes the same central prediction: this model can be used to derive the Ratio Inequality. If we used this model in order to characterize the literal listener, we could still build an RSA model, in the same way as we did in Sect. 6, and we would derive qualitatively similar predictions.

The proof of BIR’ goes as follows. $P_{post}$ being a probability distribution, it satisfies, for any k:

$$\begin{aligned} P_{post}(x=k) = \sum \limits _{i}[P_{post}(x=k\ |\ y=i)\times P_{post}(y=i)] \end{aligned}$$

Let us develop the first factor in the sum, $P_{post}(x=k\ |\ y=i)$. Since, in general, $P_C(A|B) = P(A|B\wedge C)$ [where $P_C$ is P conditionalized on event C] we have:

$$\begin{aligned} P_{post}(x=k\ |\ y=i)&= P(x=k\ |\ y=i \wedge x \in [n-y, n+y])\\&= \dfrac{P(x=k \wedge y=i \wedge x \in [n-y, n+y])}{P(y=i \wedge x \in [n-y, n+y])}\\&= \dfrac{P(x=k \wedge y=i \wedge x \in [n-i, n+i])}{P(y=i \wedge x \in [n-i, n+i])} \end{aligned}$$

Since the random variables x and y are independent, the events $\ulcorner y=i \urcorner$ and $\ulcorner x=k \wedge x \in [n-i, n+i] \urcorner$ are independent, thanks to which we can simplify the formula above:

$$\begin{aligned} P_{post}(x=k\ |\ y=i)&= \dfrac{P(y=i)\times P(x=k \wedge x \in [n-i, n+i])}{P(y=i) \times P(x \in [n-i, n+i])}\\&= \dfrac{P(x=k \wedge x \in [n-i, n+i])}{P(x \in [n-i, n+i])}\\&= P(x=k\ |\ x \in [n-i, n+i]) \end{aligned}$$

Plugging this equality into the sum above, we get exactly BIR’.

Appendix C: Two alternative RSA models (discussed in Sect. 8)

1.1 C.1. A variant of our model which uses the standard utility function

In our official model, the utility function for the speaker is defined by the following equations, where $P_o$ is understood to be the posterior distribution over the variable of interest (here notated w, for world) induced by observation o, and $L^n_m$ is the posterior distribution over w of the level-n listener who has processed a message m. These distributions, importantly, are not joint distributions over w and o.

$$\begin{aligned} U^{n+1}(m,o_j) =-D_{KL}(P_{o}||L^n_m) \end{aligned}$$

Develo** the formula for KL-divergence, this is more explicitly cashed out as:

$$\begin{aligned} U^n(m,o)&= \sum \limits _w P(w|o) \times [\log (L^n(w|m)) - \log (P(w|o))]\\&= \sum \limits _w P(w|o) \times [\log (\sum \limits _{o'}L^n(w,o'|m)) - log(P(w|o))]\\&= \sum \limits _w P(w|o) \times \log (\sum \limits _{o'}L^n(w,o'|m)) - \sum \limits _w P(w|o) \times \log (P(w|o)) \end{aligned}$$

Note that that the second term, $- \sum \nolimits _w P(w|o) \times \log (P(w|o))$, does not depend on the message m. For this reason it can be dropped: drop** this term amounts to adding a constant term to the utility of each message (relative to a fixed observation o), which has no effect when we apply the SoftMax function in order to derive the speaker’s behavior (Translation Invariance of SoftMax). So we can as well use the following utility function, with no change whatsoever in the behavior of the model:

$$\begin{aligned} U^n(m,o) = \sum \limits _w P(w|o) \times \log (\sum \limits _{o'}L^n(w,o'|m)) \end{aligned}$$

Now, we can also consider a model whose general architecture is like ours, where the literal listener $L^0$, in particular, is exactly the same as the one we defined, but where we use the utility function of Bergen et al. (2016), which is based on the KL-divergence between the joint distribution over (w, o) of the level-n listener, and the joint distribution of the speaker which results from an observation o (such a joint distribution assigns probability 0 to all pairs $(w,o')$ where $o' \ne o$, i.e. $P(w,o'|o) = P(w|o)$ if $o'=o$, otherwise $P(w,o'|o)=0$).

This amounts to moving to the following utility function, as in Bergen et al. (2016) (ignoring the cost term):

$$\begin{aligned} U^n(m,o) = \sum \limits _w P(w|o) \times \log (L^n(w,o|m)) \end{aligned}$$

Kee** all the other ingredients of the model presented in Sects. 6 and 7, we obtain, with such a model, numerically different results from those of our main model, but qualitatively similar ones, in the following sense: the pragmatic speaker (at different recursive depths) can have a preference for an “around”-statement over any “between”-statement in some situations where she is able to exclude the peripheral values 0 and 8 (and so could have said, e.g., between 1 and 7) but has a private distribution that is strongly biased towards values closer to 4. Crucially, for this model, the limitation result proved in Appendix A for the standard LU model does not hold.

1.2 C.2. A variant of the LU model where the utility function is as in our own model

We present here a modified LU model. The crucial difference with Bergen et al.’s (2016) LU model shows up in the utility functions (lines 2 and 5), where $L^0(w,o|m,i)$ and $L^n(w,o|m)$ have been replaced, respectively, with $L^0(w|m,i)$ and $L^n(w|m,i)$, which are themselves equal, respectively, to $\sum \nolimits _{o'}L^0(w,o'|m,i)$ and $\sum \nolimits _{o'}L^n(w,o'|m)$.

Importantly, the limitation result reported in Appendix A for the standard LU model no longer holds for this model.

• 1. $L^0(w,o | m,i) = \dfrac{P(w,o) \times \llbracket m \rrbracket ^i(w)}{P(\llbracket m \rrbracket ^i)}$

• 2. $U^1(m|o,i) = (\sum \limits _{w}P(w | o) \times \log (L^0(w | m,i))) - c(m)$                   $= (\sum \limits _{w}[P(w | o) \times \log (\sum \nolimits _{o'}L^0(w,o' | m,i))]) - c(m)$

• 3. $S^1(m | o, i) = \dfrac{\exp (\uplambda U^1(m,o,i))}{ \sum \limits _{m'}\exp (\uplambda U^1(m',o,i))}$

• 4. $L^1(w,o | m) = \dfrac{P(w,o)\times \sum \nolimits _{i} P(i)S^1(m | o, i)}{\alpha _1(m)}$, where $\alpha _1(m) = \sum \nolimits _{w',o'} (P(w',o')\times$ $\sum \nolimits _{i} P(i)S^1(m | o', i))$

• 5. For $n \ge 1$, $U^{n+1}(m|o) = \sum \nolimits _w P(w| o)\log (L^n(w | m)) - c(m)$                                     $=\sum \nolimits _{w}[P(w | o) \times \log (\sum \nolimits _{o'}L^n(w,o' | m))] - c(m)$

• 6. $S^{n+1}(m | o) = \dfrac{\exp (\uplambda U^{n+1}(m,o))}{\sum \nolimits _{m'} \exp (\uplambda U^{n+1}(m',o))}$

• 7. For $n \ge 2$, $L^n(w,o | m) = \dfrac{P(w,o) \times S^n(m | o)}{\alpha _n(m)}$, where $\alpha _n(m) = \sum \nolimits _{w',o'}P(w',o')\times$ $S^n(m | o')$