Supplementary Information for - PNAS

Supplementary Information for Cognitive resource allocation determines the organisation of personal networks Ignacio Tamarit, José A. Cuesta, Robin I.M. Dunbar & Angel Sánchez Ignacio Tamarit. E-mail: [email protected]

This PDF file includes:

Supplementary text Figs. S1 to S6 References for SI reference citations

Ignacio Tamarit, José A. Cuesta, Robin I.M. Dunbar & Angel Sánchez

www.pnas.org/cgi/doi/10.1073/pnas.1719233115

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Supporting Information Text 1. Supplementary Text A. Prior Distribution. There are r social layers and individuals distribute their links among them at will. Without further assumptions, and

given that there are N (assumed large) individuals in the population, the prior distribution of links is a random choice. Thus the probability that an individual has `k links in layer k ∈ {1, 2, . . . , r} will be P0 (``|N) = (r + 1)−N+1

N −1 (N − 1)!(r + 1)−N+1 = , ` `1 !`2 ! · · · `r !(N − 1 − `1 − `2 − · · · − `r )!

[1]

where ` = (`1 , `2 , . . . , `r ). B. Information: Cost of ties and Dunbar’s number. Individuals have limited time capacity or limited cognitive abilities, and consequently maintaining a link in layer k has a cost sk . We do not distinguish between these two costs, because their effective outcome is the same: there is a cost to forming relationships, and this cost limits the number of relationships that can be formed. These two limitations (limited time or cognitive abilities and limited number of links) have to be incorporated in the links distribution. To do so, we can assume that the mean time (or cognitive ability per individual, S ) as well as the mean number of links per individual, L , are known, i.e. r

r

∑ E(`k ) = L ,

∑ sk E(`k ) = S .

k=1

k=1

[2]

C. Posterior Distribution. The way to incorporate this information into a posterior distribution is through the maximum entropy principle (14). In other words, the posterior P(``|S , L , N) is obtained through maximisation of ( ) r r P(``|S , L , N) S [P] = ∑ − log − τ − γ ∑ `k − µ ∑ sk `k P(``|S , L , N), [3] P0 (``|N) ` k=1 k=1

where τ, γ and µ are Lagrange multipliers associated, respectively, to the normalisation of P(``|S , L , N) and to the two constraints Eq. (2), which are to be determined a posteriori by enforcing P(``|S , L , N) to satisfy those same constraints. The posterior distribution obtained through this method is ( ) r N −1 −1 P(``|S , L , N) = Z(γ, µ, N) exp ∑ (−γ − µsk )`k , [4] ` k=1 ( ) r N −1 Z(γ, µ, N) = ∑ exp ∑ (−γ − µsk )`k . [5] ` k=1 ` In terms of the “partition function” Z(γ, µ, N), the two constraints Eq. (2) become −

∂ log Z(γ, µ, N) = L , ∂γ

−

∂ log Z(γ, µ, N) = S . ∂µ

[6]

Besides, if we extend this partition function with new variables β = (β1 , β2 , . . . , βr ) as ( ) r N −1 Z(γ, µ, N, β ) = ∑ exp ∑ (−γ − µsk − βk )`k , ` k=1 `

[7]

we can obtain E(`k ) through E(`k ) = −

∂ log Z(γ, µ, N, β ) . ∂βk β =0

[8]

These expressions to compute L , S , and E(`k ) are particularly simple once we realise that r

N − 1 r −γ−µsk −βk `k Z(γ, µ, N, β ) = ∑ = ∏ e ` k=1 ` for then

L=

(N − 1) ∑k e−γ−µsk , 1 + ∑k e−γ−µsk

Dividing S and E(`k ) by L we obtain

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S=

1+

E(`k )

L

=

∑e

,

[9]

k=1

(N − 1) ∑k sr e−γ−µsk , 1 + ∑k e−γ−µsk

S ∑k sr e−µsk , = L ∑k e−µsk

!N−1 −γ−µsk −βk

E(`k ) =

e−µsk . ∑k e−µsk

(N − 1)e−γ−µsk . 1 + ∑k e−γ−µsk

[10]

[11]


The first of these two equations allows us to determine µ as a function of S /L , and then we can use this value to obtain γ from the first of Eqs. Eq. (10) as e−γ =

L (N − 1 − L ) ∑k e−µsk

.

[12]

As a matter of fact, we can use identity Eq. (12) to simplify the posterior distribution Eq. (4). To begin with, !N−1 Z(γ, µ, N) =

1 + e−γ

r

∑ e−µs

= 1+

k

k=1

N−1

L

= 1−

N −1−L

L

−N+1

N −1

.

[13]

Thus, denoting L ≡ ∑k `k , N−1 N − 1 −γL −µ ∑k sk `k e e N −1 ` N−1−L L −µ ∑ sk `k k L N −1 L e , = 1− −µs N −1 ` N −1 (∑k e k )L

L

P(``|S , L , N) = 1 −

[14]

and using N −1 N −1 L = . ` L `

[15]

P(``|S , L , N) = B(L, L /(N − 1), N − 1)P(``|S , L, N), −µ ∑ sk `k k L e P(``|S , L, N) = , ` (∑k e−µsk )L

[16]

we arrive at

with B(x, p, n) =

n x n−x x p (1 − p)

Remark on small populations. If

the binomial distribution, and where the constraint L = ∑k `k 6 N − 1 is to be understood in P(``|S , L, N).

N is so small that everyone knows everybody else, then L → N − 1. In this limit B(L, L /(N − 1), N − 1) →

δL,N−1 , therefore lim P(``|S , L , N) =

L →N−1

N −1 e−µ ∑k sk `k , ` (∑k e−µsk )N−1

r

∑ `k = N − 1.

[17]

k=1

D. Linear decrease in cost. Let us consider the special case in which the cost decreases linearly with the layer, sk = s1 − (s1 − sr )(k − 1)/(r − 1), with s1 > sr > 0. In that case r

Ω(µ) =

∑ e−µs

k

= e−µs1

σ≡

∑

eµ(s1 −sr )/(r−1)

k

= e−µs1

k=0

k=1

and

r−1

S ∂ s1 − sr = − log Ω(µ) = s1 + L ∂µ r−1

eµ(s1 −sr )r/(r−1) − 1 , eµ(s1 −sr )/(r−1) − 1

eµˆ rerµˆ − , eµˆ − 1 erµˆ − 1

[18]

µˆ ≡ µ(s1 − sr )/(r − 1).

[19]

Notice that the choice |∆sk | = 1 implies that µˆ = µ, and that (s1 − sr )(k − 1)/(r − 1) > 0, so in the main text we abuse a bit of notation and denote µˆ ≡ µ. However, for full clarity, we keep the µˆ notation here. Then s1 − σ (r − 1)erµˆ − re(r−1)ˆµ + 1 = f (ˆµ) ≡ eµˆ . s1 − sr (r − 1)(erµˆ − 1)(eµˆ − 1)

[20]

Function f (ˆµ) monotonically increases from its smallest value lim f (ˆµ) = 0 (corresponding to σ = s1 ) up to lim f (ˆµ) = 1 (corresponding to µˆ →−∞

µˆ →∞

σ = sr ). As sr < σ < s1 , µˆ can take any value −∞ < µˆ < ∞. Since 1 lim f (ˆµ) = , 2

[21]

µˆ →0

whenever sr < σ < (s1 + sr )/2 the solution µˆ > 0 and whenever (s1 + sr )/2 < σ < s1 the solution µˆ < 0. As for E(`k ), we obtain E(`k ) ekµˆ − e(k−1)ˆµ εk ≡ = . L erµˆ − 1 In particular, the fraction of links in circle k (the union of all layers from 1 to k) will be χk =

1 erµˆ − 1

k

ekµˆ − 1

∑ (e jµˆ − e( j−1)ˆµ ) = erµˆ − 1 .

[22]

[23]

j=1


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If µˆ > 0 (standard Dunbar’s circles) then 1 − e−kµˆ ∼ e(k−r)ˆµ , [24] 1 − e−rµˆ providing an exponential increase with k and an approximately constant scaling ratio χk+1 /χk ≈ eµˆ . On the contrary, if µˆ < 0 (inverse Dunbar’s circles) then 1 − ekµˆ χk = , [25] 1 − erµˆ which quickly approaches 1 as k increases —implying that most links are within the first circle. χk = e(k−r)ˆµ

E. Bayesian estimate of the parameter. The likelihood for ` (hence also for L = ∑k `k ) is given by

eµˆ − 1 P(``|L , µˆ , N) = B(L, L /(N − 1), N − 1) µˆ r e −1

( ) L r−1 L exp µˆ ∑ k`k+1 . ` k=0

[26]

Our goal is to determine P(L , µˆ |``, N) using Bayesian inference. To this aim we choose a neutral uniform prior for the parameters µˆ and L . Thus, N P(L , µˆ |``, N) = P(``|L , µˆ , N), [27] (N − 1) L` Ξ(``) with Ξ(``) =

−1 Z N−1 Z ∞ L dL d µˆ P(``|L , µˆ , N). ` 0 −∞

[28]

The factor N in the numerator of Eq. (27) arises from Z N−1 0

B(L, L /(N − 1), N − 1) d L =

N −1 . N

[29]

Now, if we denote Ft (R) ≡

Z t 0

1 − e−ˆµ 1 − e−ˆµr

L

e−ˆµR d µˆ ,

[30]

then Ξ(``) =

L L L Z ∞ µˆ Z ∞ µˆ Z ∞ e −1 e −1 1 − e−ˆµ µˆ L1 µˆ L1 e d µ ˆ = e d µ ˆ + e−ˆµL1 d µˆ µˆ r µˆ r −ˆµr −∞

e −1

e −1

0

0

1−e

[31]

= F∞ (L2 ) + F∞ (L1 ), where r−1

L1 ≡

r−1

∑ k`k+1 ,

L2 ≡ L(r − 1) − L1 =

k=0

∑ k`r−k .

[32]

k=0

Summarising, P(L , µˆ |``, N) = P(L |``, N)P(ˆµ|``), where N B(L, L /(N − 1), N − 1), N −1 µˆ L e −1 eµˆ L1 . P(ˆµ|``) = Ξ(``)−1 µˆ r e −1 P(L |``, N) =

[33] [34]

The maximum likelihood estimate for L is obviously L = L. That for µˆ can be obtained differentiating log P(ˆµ|`) = µˆ L1 + L log(eµˆ − 1) − L log(eµˆ r − 1) − log Ξ(``),

[35]

eµˆ L1 r−1 `k+1 rerµˆ = ∑k = µˆ − rµˆ = (r − 1) f (eµˆ ). L L e −1 e −1 k=1

[36]

which leads to

This equation is identical to Eq. (20) if we use ` as estimates for E(``). Finally, the 1 − 2δ confidence interval for µˆ , given ` , is obtained through the cumulative distribution G(t|``) =

Z t −∞

P(ˆµ|``) d µˆ .

[37]

The extremes of the confidence interval [t1 ,t2 ] are obtained by solving G(t1 |``) = δ, G(t2 |``) = 1 − δ. Notice that G(t|``) can be obtained as  F∞ (L1 ) − F−t (L1 )   , t < 0,   F∞ (L1 ) + F∞ (L2 ) G(t|``) = [38]   F∞ (L1 ) + Ft (L2 )   , t > 0. F∞ (L1 ) + F∞ (L2 ) 4 of 18


F. A continuum layers limit. Taking the limit r → ∞ in the formalism above transforms the model into one with a continuum of layers labelled

by an index t = (k − 1)/(r − 1) ranging in 0 6 t 6 1. Thus eq. Eq. (20) becomes eη smax − σ 1 = g(η) = η − , smax − smin e −1 η

[39]

where η ≡ (r − 1)ˆµ = (smax − smin )µ , and εk and χk become, respectively, εk → ε(t) dt = Notice that

1 lim g(η) = , 2

η→0

ηeηt dt, eη − 1

χk → χ(t) =

lim g(η) = 1, η→+∞

eηt − 1 . eη − 1

[40]

lim g(η) = 0, η→−∞

and g0 (η) > 0 for all −∞ < η < ∞, so there is one and only one solution η to equation Eq. (39) for each smin 6 σ 6 smax . Solutions η > 0 (small σ, i.e., large L ) correspond to the standard regime, solutions η < 0 (large σ, i.e., small L ) to the inverted one. 2. Background information on data sources

The community of Bulgarians in Roses (1) is quite recent (average time of residence: 3.5 years). They typically arrived in Catalonia either on their own (to “try their luck”), following migratory family chains (some relatives established first and served as connections), or following a “migratory work chain” (using co-ethnic contacts to join a particular company recently installed in Roses). The initial settlements may also be followed by friends and other acquaintances. In terms of education profiles, they either are well educated or people or have little or no education. Although the Sikh community is quite numerous, it is also the smallest of the ones analysed in Barcelona (2, 3). Their religion, Sikhism, is a key aspect of their social lives: 15% of their contacts were made in their religious center. From a social perspective, a Sikh can be differentiated by a series of items with symbolic connotations (the “five Ks”): long and cared-for hair (Kesh), a wooden comb for the hair (Kangha), an iron dagger (Kirpan), short cotton trousers (Kachera), and an iron bracelet (Kara). Their businesses have a strong ethnic character. They are predominantly rural males with low levels of education, and a strong feeling of ethnic identity which is promoted by the use of the Punjabi dialect (2, 3). Despite being one of the largest groups of immigrants in Catalonia, the Chinese community is also one of the more dispersed (2, 3). The participants in the study worked mainly in family businesses (restaurants and bazaars). These create structured collectives including all close family members (parents and children) alongside other extended family members. They maintain the use of the Chinese language and encourage its use among children. However, most of them are bilingual, speaking both the dialect of their mother’s tongue (unintelligible to other Chinese speakers) plus the official Putonghua. The former status of Philippines as a Spanish colony has favoured migratory fluxes to Spain for a long time. In 2007, the Filipino community in Barcelona was especially concentrated in the neighbourhood of “Ciutat Vella”, and was made of young urban females with intermediate or high levels of education, working in domestic service, and with strong religious (catholic) ties (2, 3).


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µ = 0. 33, l = (4, 10, 12, 14)

µ = 1. 88, l = (0, 0, 5, 23)

µ = 0. 71, l = (2, 4, 14, 19)

µ = 1. 15, l = (1, 2, 10, 27)

µ = 1. 54, l = (4, 0, 6, 58)

µ = 0. 74, l = (6, 6, 19, 40)

µ = 1. 33, l = (3, 2, 10, 53)

µ = 1. 21, l = (0, 4, 3, 21)

µ = 0. 82, l = (5, 3, 8, 30)

µ = 1. 33, l = (4, 3, 9, 63)

µ = − 0. 5, l = (26, 15, 11, 5)

µ = 1. 26, l = (6, 0, 8, 56)

µ = 1. 07, l = (0, 8, 11, 39)

µ = 0. 77, l = (2, 2, 2, 12)

µ = 1. 1, l = (5, 1, 13, 48)

µ = 0. 89, l = (1, 9, 12, 35)

µ = 1. 53, l = (1, 3, 7, 49)

µ = 0. 35, l = (5, 0, 13, 8)

µ = 0. 91, l = (1, 7, 5, 26)

µ = 0. 77, l = (5, 2, 23, 33)

µ = 1. 41, l = (3, 0, 12, 53)

µ = 1. 06, l = (0, 8, 10, 37)

µ = 1. 71, l = (0, 1, 8, 37)

µ = 1. 01, l = (1, 8, 21, 50)

µ = 0. 93, l = (7, 3, 4, 42)

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

fraction of links

3. Supplementary Figures

1.0 0.8 0.6 0.4 0.2 0.0 1

2

circles

3

4

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circles

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Fig. S1. Complete set of figures for the fittings in the Students community (case considering four layers 1/4).

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fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

µ = 1. 75, l = (0, 1, 9, 43)

µ = 1. 01, l = (0, 1, 1, 4)

µ = 2. 1, l = (1, 1, 5, 65)

µ = 1. 45, l = (2, 4, 7, 58)

µ = 0. 34, l = (5, 8, 20, 14)

µ = 0. 85, l = (2, 3, 13, 23)

µ = 1. 18, l = (3, 3, 3, 35)

µ = 1. 35, l = (0, 3, 1, 17)

µ = 1. 35, l = (0, 3, 5, 25)

µ = 1. 06, l = (2, 3, 11, 33)

µ = 1. 14, l = (4, 3, 15, 55)

µ = 1. 31, l = (0, 1, 5, 14)

µ = 0. 55, l = (7, 8, 23, 31)

µ = 0. 06, l = (10, 8, 11, 11)

µ = 0. 59, l = (2, 5, 13, 15)

µ = 0. 22, l = (6, 9, 7, 13)

µ = 0. 94, l = (2, 9, 16, 46)

µ = 1. 39, l = (1, 4, 9, 49)

µ = 0. 49, l = (2, 3, 13, 9)

µ = 0. 66, l = (5, 1, 9, 19)

µ = 1. 5, l = (0, 7, 3, 51)

µ = 2. 79, l = (0, 0, 3, 43)

µ = 0. 83, l = (3, 6, 15, 34)

µ = 0. 83, l = (4, 1, 6, 21)

µ = 0. 99, l = (4, 2, 18, 42)

1

2

circles

3

4

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3

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3

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fraction of links fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

µ = 0. 68, l = (1, 6, 17, 19)

µ = 0. 56, l = (2, 4, 16, 13)

µ = 1. 42, l = (0, 1, 9, 26)

µ = 0. 88, l = (2, 9, 16, 41)

µ = 0. 42, l = (0, 0, 2, 0)

µ = 0. 23, l = (9, 15, 31, 17)

µ = 1. 97, l = (0, 1, 2, 22)

µ = 1. 08, l = (5, 3, 7, 46)

µ = 0. 82, l = (4, 4, 9, 29)

µ = 0. 54, l = (5, 1, 5, 14)

µ = 1. 35, l = (2, 1, 10, 41)

µ = 1. 18, l = (2, 2, 5, 28)

µ = 1. 88, l = (0, 3, 8, 68)

µ = − 0. 67, l = (3, 0, 0, 1)

µ = 1. 17, l = (1, 1, 14, 30)

µ = 0. 78, l = (6, 2, 3, 27)

µ = 0. 13, l = (6, 3, 4, 8)

µ = 1. 43, l = (0, 2, 14, 44)

µ = 0. 26, l = (6, 3, 8, 10)

µ = 1. 0, l = (5, 1, 13, 40)

µ = 0. 94, l = (3, 3, 20, 38)

µ = 0. 37, l = (7, 9, 15, 20)

µ = 0. 0, l = (1, 0, 0, 1)

µ = 1. 16, l = (2, 1, 5, 23)

µ = 0. 9, l = (4, 4, 12, 36)

1.0 0.8 0.6 0.4 0.2 0.0 1

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fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

µ = 0. 85, l = (2, 3, 10, 21)

µ = 1. 74, l = (0, 3, 4, 41)

µ = 1. 45, l = (0, 4, 10, 47)

µ = 0. 89, l = (4, 3, 16, 36)

µ = 0. 29, l = (0, 10, 4, 7)

µ = 0. 27, l = (6, 2, 5, 10)

µ = 1. 87, l = (0, 2, 3, 34)

µ = 1. 33, l = (1, 5, 4, 40)



9 of 18

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

µ = 0. 35, l = (4, 10, 12, 14, 23)

µ = 1. 22, l = (0, 0, 5, 23, 53)

µ = 0. 7, l = (2, 4, 14, 19, 42)

µ = 0. 8, l = (1, 2, 10, 27, 39)

µ = 0. 53, l = (4, 0, 6, 58, 14)

µ = 0. 27, l = (6, 6, 19, 40, 10)

µ = 0. 49, l = (3, 2, 10, 53, 14)

µ = 1. 05, l = (0, 4, 3, 21, 48)

µ = 0. 62, l = (5, 3, 8, 30, 35)

µ = 0. 36, l = (4, 3, 9, 63, 1)

µ = − 0. 46, l = (26, 15, 11, 5, 5)

µ = 0. 43, l = (6, 0, 8, 56, 10)

µ = 0. 52, l = (0, 8, 11, 39, 22)

µ = 0. 84, l = (2, 2, 2, 12, 26)

µ = 0. 4, l = (5, 1, 13, 48, 11)

µ = 0. 47, l = (1, 9, 12, 35, 22)

µ = 0. 59, l = (1, 3, 7, 49, 19)

µ = 0. 36, l = (5, 0, 13, 8, 15)

µ = 0. 76, l = (1, 7, 5, 26, 41)

µ = 0. 36, l = (5, 2, 23, 33, 17)

µ = 0. 5, l = (3, 0, 12, 53, 13)

µ = 0. 54, l = (0, 8, 10, 37, 23)

µ = 0. 84, l = (0, 1, 8, 37, 36)

µ = 0. 28, l = (1, 8, 21, 50, 2)

µ = 0. 5, l = (7, 3, 4, 42, 23)

1

2

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circles

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2

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Fig. S2. Complete set of figures for the fittings in the Students community (case considering five layers 1/4).

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fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

µ = 0. 74, l = (0, 1, 9, 43, 28)

µ = 1. 43, l = (0, 1, 1, 4, 23)

µ = 0. 57, l = (1, 1, 5, 65, 10)

µ = 0. 47, l = (2, 4, 7, 58, 10)

µ = 0. 4, l = (5, 8, 20, 14, 31)

µ = 0. 71, l = (2, 3, 13, 23, 40)

µ = 0. 76, l = (3, 3, 3, 35, 37)

µ = 1. 29, l = (0, 3, 1, 17, 54)

µ = 1. 01, l = (0, 3, 5, 25, 48)

µ = 0. 64, l = (2, 3, 11, 33, 31)

µ = 0. 35, l = (4, 3, 15, 55, 5)

µ = 1. 37, l = (0, 1, 5, 14, 61)

µ = 0. 21, l = (7, 8, 23, 31, 12)

µ = 0. 05, l = (10, 8, 11, 11, 11)

µ = 0. 73, l = (2, 5, 13, 15, 45)

µ = 0. 59, l = (6, 9, 7, 13, 46)

µ = 0. 32, l = (2, 9, 16, 46, 8)

µ = 0. 54, l = (1, 4, 9, 49, 17)

µ = 0. 49, l = (2, 3, 13, 9, 20)

µ = 0. 64, l = (5, 1, 9, 19, 32)

µ = 0. 59, l = (0, 7, 3, 51, 20)

µ = 0. 94, l = (0, 0, 3, 43, 36)

µ = 0. 42, l = (3, 6, 15, 34, 19)

µ = 0. 89, l = (4, 1, 6, 21, 49)

µ = 0. 42, l = (4, 2, 18, 42, 16)

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fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

µ = 0. 61, l = (1, 6, 17, 19, 37)

µ = 0. 58, l = (2, 4, 16, 13, 31)

µ = 0. 95, l = (0, 1, 9, 26, 44)

µ = 0. 35, l = (2, 9, 16, 41, 13)

µ = 1. 07, l = (0, 0, 2, 0, 6)

µ = 0. 01, l = (9, 15, 31, 17, 9)

µ = 1. 32, l = (0, 1, 2, 22, 56)

µ = 0. 5, l = (5, 3, 7, 46, 20)

µ = 0. 61, l = (4, 4, 9, 29, 34)

µ = 0. 96, l = (5, 1, 5, 14, 56)

µ = 0. 65, l = (2, 1, 10, 41, 27)

µ = 0. 88, l = (2, 2, 5, 28, 43)

µ = 0. 48, l = (0, 3, 8, 68, 4)

µ = − 0. 08, l = (3, 0, 0, 1, 2)

µ = 0. 72, l = (1, 1, 14, 30, 35)

µ = 0. 73, l = (6, 2, 3, 27, 41)

µ = 0. 89, l = (6, 3, 4, 8, 56)

µ = 0. 59, l = (0, 2, 14, 44, 21)

µ = 0. 77, l = (6, 3, 8, 10, 51)

µ = 0. 48, l = (5, 1, 13, 40, 20)

µ = 0. 42, l = (3, 3, 20, 38, 17)

µ = 0. 31, l = (7, 9, 15, 20, 24)

µ = 0. 4, l = (1, 0, 0, 1, 2)

µ = 1. 01, l = (2, 1, 5, 23, 50)

µ = 0. 51, l = (4, 4, 12, 36, 26)

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fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

µ = 0. 82, l = (2, 3, 10, 21, 46)

µ = 0. 81, l = (0, 3, 4, 41, 33)

µ = 0. 59, l = (0, 4, 10, 47, 21)

µ = 0. 46, l = (4, 3, 16, 36, 22)

µ = 0. 44, l = (0, 10, 4, 7, 17)

µ = 0. 9, l = (6, 2, 5, 10, 56)

µ = 0. 97, l = (0, 2, 3, 34, 41)

µ = 0. 69, l = (1, 5, 4, 40, 30)



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fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

µ = − 0. 57, l = (5, 20, 5, 0, 0)

µ = − 1. 94, l = (25, 5, 0, 0, 0)

µ = − 0. 08, l = (4, 7, 11, 6, 2)

µ = − 0. 23, l = (6, 7, 13, 5, 0)

µ = − 0. 12, l = (8, 4, 5, 13, 0)

µ = − 0. 98, l = (15, 14, 0, 1, 0)

µ = − 0. 07, l = (6, 7, 1, 15, 0)

µ = − 0. 39, l = (7, 10, 11, 2, 0)

µ = − 0. 62, l = (16, 0, 14, 0, 0)

µ = − 0. 62, l = (8, 19, 0, 3, 0)

µ = − 0. 89, l = (11, 19, 0, 0, 0)

µ = − 0. 62, l = (15, 6, 5, 4, 0)

µ = − 0. 57, l = (9, 12, 9, 0, 0)

µ = − 0. 52, l = (11, 11, 3, 5, 0)

µ = − 0. 97, l = (16, 8, 4, 0, 0)

µ = − 0. 89, l = (13, 15, 2, 0, 0)

µ = − 0. 54, l = (7, 15, 8, 0, 0)

µ = − 0. 1, l = (6, 2, 15, 6, 1)

µ = − 0. 37, l = (8, 9, 9, 4, 0)

µ = − 3. 43, l = (29, 1, 0, 0, 0)

µ = − 0. 07, l = (9, 1, 9, 7, 4)

µ = − 0. 48, l = (15, 2, 11, 4, 0)

µ = − 0. 64, l = (11, 12, 6, 1, 0)

µ = 0. 13, l = (8, 0, 3, 14, 5)

µ = − 0. 47, l = (7, 15, 5, 3, 0)

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Fig. S3. Complete set of figures for the fittings in the Bulgarian community

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1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

1.0 0.8 0.6 0.4 0.2 0.0

µ = 0. 08, l = (13, 0, 1, 1, 15)

µ = − 0. 73, l = (13, 12, 3, 2, 0)

µ = 0. 02, l = (2, 0, 23, 5, 0)

µ = − 3. 43, l = (29, 1, 0, 0, 0)

µ = − 0. 19, l = (5, 2, 22, 1, 0)

µ = − 0. 39, l = (5, 12, 13, 0, 0)

µ = − 0. 59, l = (10, 12, 7, 1, 0)

µ = − 0. 76, l = (11, 16, 2, 1, 0)

µ = − 0. 7, l = (14, 7, 9, 0, 0)

µ = − 0. 17, l = (2, 10, 14, 4, 0)

µ = 0. 05, l = (1, 0, 25, 3, 1)

µ = − 0. 39, l = (8, 7, 14, 1, 0)

µ = − 0. 24, l = (9, 3, 12, 5, 1)

µ = − 0. 28, l = (7, 7, 12, 3, 1)

µ = − 0. 29, l = (8, 7, 10, 4, 1)

µ = − 0. 76, l = (15, 7, 8, 0, 0)

µ = − 1. 23, l = (22, 4, 4, 0, 0)

µ = − 0. 82, l = (12, 15, 3, 0, 0)

µ = − 0. 37, l = (10, 8, 5, 7, 0)

µ = − 0. 59, l = (8, 15, 7, 0, 0)

µ = − 0. 29, l = (7, 10, 6, 7, 0)

µ = − 0. 02, l = (1, 3, 22, 4, 0)

µ = − 0. 45, l = (8, 9, 13, 0, 0)

µ = − 0. 89, l = (12, 17, 1, 0, 0)

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Fig. S4. Complete set of figures for the fittings in the Sikh community


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fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links


1.0 0.8 0.6 0.4 0.2 0.0

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

1.0 0.8 0.6 0.4 0.2 0.0

µ = − 0. 02, l = (3, 7, 8, 12, 0)

µ = − 0. 03, l = (1, 3, 23, 3, 0)

µ = − 0. 39, l = (5, 16, 6, 2, 1)

µ = − 0. 45, l = (9, 8, 12, 1, 0)

µ = − 0. 7, l = (14, 11, 1, 4, 0)

µ = − 0. 12, l = (6, 3, 13, 8, 0)

µ = − 0. 45, l = (11, 6, 10, 3, 0)

µ = 0. 0, l = (3, 11, 2, 11, 3)

µ = − 0. 08, l = (3, 4, 18, 5, 0)

µ = − 0. 33, l = (6, 10, 11, 3, 0)

µ = − 0. 86, l = (12, 17, 0, 1, 0)

µ = − 0. 24, l = (6, 5, 16, 3, 0)

µ = − 0. 19, l = (4, 10, 9, 7, 0)

µ = − 0. 13, l = (5, 6, 12, 6, 1)

µ = − 0. 12, l = (3, 10, 8, 9, 0)

µ = − 0. 2, l = (9, 7, 2, 11, 1)

µ = 0. 35, l = (0, 5, 0, 25, 0)

µ = 0. 05, l = (3, 6, 6, 15, 0)

µ = − 0. 22, l = (4, 11, 9, 6, 0)

µ = − 0. 57, l = (4, 22, 4, 0, 0)

µ = − 0. 93, l = (15, 12, 3, 0, 0)

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Fig. S5. Complete set of figures for the fittings in the Chinese community

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1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

fraction of links

1.0 0.8 0.6 0.4 0.2 0.0

1.0 0.8 0.6 0.4 0.2 0.0

µ = − 0. 17, l = (2, 16, 2, 10, 0)

µ = − 0. 1, l = (3, 5, 17, 5, 0)

µ = 0. 05, l = (4, 2, 11, 13, 0)

µ = − 0. 02, l = (5, 8, 0, 17, 0)

µ = 0. 17, l = (0, 1, 18, 11, 0)

µ = − 0. 57, l = (9, 16, 1, 4, 0)

µ = 0. 02, l = (5, 5, 6, 12, 2)

µ = − 0. 43, l = (7, 10, 13, 0, 0)

µ = − 0. 29, l = (6, 5, 19, 0, 0)

µ = − 0. 12, l = (6, 2, 15, 7, 0)

µ = − 0. 37, l = (5, 13, 10, 2, 0)

µ = − 0. 67, l = (13, 8, 9, 0, 0)

µ = − 1. 79, l = (24, 6, 0, 0, 0)

µ = 0. 07, l = (1, 5, 13, 11, 0)

µ = 0. 08, l = (0, 7, 11, 12, 0)

µ = − 0. 43, l = (10, 4, 16, 0, 0)

µ = − 0. 13, l = (1, 17, 1, 11, 0)

µ = 0. 17, l = (2, 3, 8, 17, 0)

µ = − 0. 5, l = (13, 6, 7, 3, 1)

µ = − 0. 45, l = (11, 5, 12, 2, 0)

µ = − 0. 19, l = (5, 1, 24, 0, 0)

µ = 0. 12, l = (4, 2, 7, 17, 0)

µ = − 0. 03, l = (4, 4, 12, 10, 0)

µ = 0. 02, l = (1, 13, 0, 16, 0)

µ = − 0. 54, l = (11, 13, 0, 6, 0)

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Fig. S6. Complete set of figures for the fittings in the Filipino community


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References

1. Sílvia Gómez Mestres, Jose Luis Molina, Sarah Hoeksma, and Miranda Lubbers. Bulgarian migrants in spain: Social networks, patterns of transnationality, community dynamics and cultural change in catalonia (northeastern spain) 1. Southeastern Europe, 36(2):208–236, 2012. 2. José Luis Molina, Sören Petermann, and Andreas Herz. Defining and measuring transnational social structures. Field Methods, 27(3):223– 243, 2015. 3. José Luís Molina and Fabien Pelissier. Les xarxes socials de sikhs, xinesos i filipins a barcelona, 2010.

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