When are concepts comparable across minds?

Abstract

In communication, people cannot resort to direct reference (e.g., pointing) when using diffuse concepts like democracy. Given that concepts reside in individuals’ minds, how can people share those concepts? We argue that concepts are comparable across a social group if they afford agreement for those who use it; and that agreement occurs whenever individuals receive evidence that others conceptualize a given situation similarly to them. Based on Conceptual Agreement Theory, we show how to compute an agreement probability based on the sets of properties belonging to concepts. If that probability is sufficiently high, this shows that concepts afford an adequate level of agreement, and one may say that concepts are comparable across individuals’ minds. In contrast to other approaches, our method considers that inter-individual variability in naturally occurring conceptual content exists and is a fact that must be taken into account, whereas other theories treat variability as error that should be cancelled out. Given that conceptual variability will exist, our approach may establish whether concepts are comparable across individuals’ minds more soundly than previous methods.

Appendices

Appendix 1: Calculation of p(a1) and p(a2) when properties are non-equiprobable

Note that in order to use (1) and (2) we must previously calculate the probabilities of obtaining each of the different possible samples drawn from C or Cn (i.e. the p(sCO _j ) and p(sCnA _l )). Although this is simple to do, that calculation involves many summations and multiplications, making it cumbersome to do by hand. To show it here for the example corresponding to the concepts “right and left political views” depicted in Fig. 1 for the non-equiprobable case, we will only calculate the probabilities for the C properties. Recall that for C the possible samples are (ab), (ac) and (bc). Thus:

$$ p\left( {ab} \right) = p\left( {ab} \right) + p\left( {ba} \right) = p\left( a \right) p\left( {b/a} \right) + p\left( b \right) p\left( {a/b} \right) $$

(7)

$$ p\left( {ab} \right) = p\left( a \right)\frac{p\left( b \right)}{p\left( b \right) + p\left( c \right)} + p\left( b \right)\frac{p(a)}{p\left( a \right) + p\left( c \right)} $$

(8)

In (7) and (8), remember that the order in which properties a and b are sampled is irrelevant, thus the two permutations of properties (ab) and (ba) are equivalent. Thus, to calculate the probability of combination (ab), we calculate the probability of permutation (ab) (p(ab)) and of permutation (ba) (p(ba)), and given that they are mutually exclusive, we can add them. In the case of combination p(ab), putting the corresponding values of p(a), p(b) and p(c) in (8), we obtain:

$$ p\left( {ab} \right) = \frac{1}{2} \frac{\frac{3}{10}}{\frac{3}{10} + \frac{1}{5}} + \frac{3}{10}\frac{\frac{1}{2}}{\frac{1}{2} + \frac{1}{5}} = \frac{18}{35}$$

(9)

Using expressions similar to (8), the value for p(ac) is 13/40 and for p(bc) equals 9/56. Recalling that the number of common properties for all independent samples drawn from C is {2,1,1,1,2,1,1,1,2} and applying (1), we obtain:

$$ p\left( {a1} \right) = \frac{1}{2}\left( {\left( \frac{18}{35} \right)^{2} { \times }2 + \frac{18}{35} \frac{13}{40 }{ \times }1 + \frac{18}{35} \frac{9}{56 }{ \times }1 + \cdots + \left( \frac{9}{56} \right)^{2} { \times }2} \right) = 0.69797 $$

(10)

Using the same reasoning involved in expression (8), we can obtain the probabilities of obtaining the samples from Cn, which are: p(bcd) = 0.0337, p(bce) = 0.0917, p(bde) = 0.4373 and p(cde) = 0.4373. Then, we can input those probabilities and the number of common elements between all samples drawn from C and those obtained from Cn, which we already calculated to be {1,1,1,0,1,1,0,1,2,2,1,1}, into (2) and compute an exact value for p(a2):

$$ p\left( {a2} \right) = \frac{1}{3}\left( {\frac{18}{35} 0.0337{ \times }1 + \frac{18}{35} 0.0917{ \times }1 + \cdots + \frac{9}{56 } 0.4373{ \times }1} \right) = 0.21771 $$

(11)

Appendix 2: Expressions for calculating p(a1) and p(a2) when properties are equiprobable

The demonstration that, for equiprobable properties, expression (1) reduces to (5) (i.e., $p\left( {a1} \right) = s_{1} /k_{1}$) and that Eq. (2) reduces to (6) (i.e., $p\left( {a2} \right) = p\left( {a1} \right) \, u/k_{2}$) is very lengthy, and thus it is available upon request. However, another way of arriving at Eq. (5) is to calculate it through a combinatorial approach that assumes that all properties in C are equiprobable. Given that (1) is stated in terms of samples, we must note that when the properties in C are equiprobable, then it is intuitive to see that all independent possible samples drawn from C are also equiprobable. Consequently, we will base our demonstration on the simple example illustrated in Fig. 1, which shows hypothetical concepts C and Cn. From that figure and the definition of p(a1), it is clear that p(a1) is the probability that a property of a sample of size s ₁ taken from C is also included in a second independent sample of size s ₁ taken from the same set C.

Now, if we take any property in C, for example “a”, which is part of a first sample, then in order for “a” to appear in a second sample, that second sample must contain “a” and the rest of the properties of that sample must come from the k ₁  − 1 other properties contained in C (the rest of the properties that are not “a”). Because in the second sample we already have property “a”, the number of the other properties that must be sampled is s ₁  − 1, i.e., the size of the sample minus one property, which is “a”. Then, it is straightforward to see that the number of such second samples that can be obtained from C is simply:

$$ m = \left( {\begin{array}{*{20}c} {k_{1} - 1} \\ {s_{1} - 1} \\ \end{array} } \right) $$

(12)

Thus, the probability p(a1) is the number in (12) divided by the total number of samples of size s ₁ that can be obtained from C, which we know is $\left( {\begin{array}{*{20}c} {k_{1} } \\ {s_{1} } \\ \end{array} } \right)$.

Thus, we can write:

$$ p\left( {a1} \right) = \frac{{\left( {\begin{array}{*{20}c} {k_{1} - 1} \\ {s_{1} - 1} \\ \end{array} } \right)}}{{\left( {\begin{array}{*{20}c} {k_{1} } \\ {s_{1} } \\ \end{array} } \right)}} = \frac{{s_{1} }}{{k_{1} }} $$

(13)

where we expanded the expressions and used the property of factorials $m!/\left( {m - 1} \right)! = m$.

In the case of p(a2), we will also show that (6) is correct by using a combinatorial approach similar to that used in arriving to Eq. (13). From the definition of p(a2), one can see that p(a2) is the probability that one property of a sample of size s ₂ drawn from Cn is also present in one sample of size s ₁ obtained from C. For that to happen, that property must belong to the intersection of C and Cn. For example, for the situation depicted in Fig. 1, the property may be “b” or “c”.

Thus, to start computing p(a2), we first need to calculate the probability of equiprobably obtaining a specific property of the sample of size s ₂ equiprobably drawn from Cn. To this end, we can easily see that we can draw $\left( {\begin{array}{*{20}c} {k_{2} } \\ {s_{2} } \\ \end{array} } \right)$ samples of size s ₂ from the k ₂ properties that belong to Cn, and that since we assume equiprobability, each of the samples has a probability equal to $1/\left( {\begin{array}{*{20}c} {k_{2} } \\ {s_{2} } \\ \end{array} } \right)$ of being obtained.

Now, a property contained in a given sample drawn from Cn, has a probability equal to 1/s ₂ of being selected. Thus, the probability of obtaining a property obtained from a sample of size s ₂ drawn from a set Cn that contains k ₂ properties is

$$ p1 = \frac{1}{{\left( {\begin{array}{*{20}c} {k_{2} } \\ {s_{2} } \\ \end{array} } \right)s_{2} }} $$

(14)

However, we want to compute the probability of obtaining a specific property (i.e., not only any property). Thus, if we select one property, then the number of all the samples that can contain that specific property is $\left( {\begin{array}{*{20}c} {k_{2} - 1} \\ {s_{2} - 1} \\ \end{array} } \right)$, i.e., we need to draw samples of size s ₂  − 1 to complete the other properties of it from the other k ₂  − 1 properties, without taking into account the specific property. Hence, the probability of obtaining a specific property from a sample of size s ₂ drawn from k ₂ properties that belong to Cn is

$$ p2 = \frac{{\left( {\begin{array}{*{20}c} {k_{2} - 1} \\ {s_{2} - 1} \\ \end{array} } \right)}}{{\left( {\begin{array}{*{20}c} {k_{2} } \\ {s_{2} } \\ \end{array} } \right)s_{2} }} = \frac{1}{{k_{2} }} $$

(15)

An interesting feature of (15) is that it means that equiprobably drawing a property from a sample equiprobably obtained from k ₂ properties is equivalent to directly drawing it from the k ₂ properties, i.e., from Cn. Using (15) we can now state that the probability of obtaining a property that belongs to the intersection between C and Cn in a sample equiprobably drawn from Cn is equal to the number of common properties between C and Cn, i.e., $u = \# \left( {C \cap Cn} \right)$, divided by the total number of properties of Cn: u/k ₂ . Now, event a2 will occur when that same property is contained in the sample of size s ₁ drawn from C. Expression (13) gives that probability, i.e., the probability of obtaining a given property in a sample equiprobably drawn from C, which is s ₁ /k ₁ . Therefore, p(a2) is the multiplication of u/k ₂ times s ₁ /k ₁ , which corresponds to:

$$ p\left( {a2} \right) = \frac{{s_{1 } u}}{{k_{1 } k_{2} }} = p\left( {a1} \right)\frac{u}{{k_{2} }} $$

(16)