Social Network Analytics Review for Test 2 Part 2

2 𝑐 = (𝑝 + ∑ 𝑝 𝑝 ) , 𝑖 ≠ 𝑞 ≠ 𝑗 𝑖𝑗 𝑖𝑗 𝑖𝑞 𝑞𝑗 𝑞 Take A to be the Ego: 𝑐𝐴𝐵 = (𝑝𝐴𝐵 + 𝑝𝐴𝐺𝑝𝐺𝐵) 2 = (1/4 + 1/4 × 1/6) 7 = ( 24 ) ● 2 2 = (6/24 + 1/24) 2 2 = (0. 29167) = 0. 0851 5 types of brokerage: ○ Liaison: a brokerage relation in which all 3 actors occupy different groups ■ Ex: the mediator or arbitrator in inter-organizational bargaining relations such as union-management negotiations ○ Representative: created when one member of a subgroup takes upon itself or is given the role of communicating information to, or negotiating exchanges with, outsiders. ■ Ex: in the context of inter-organizational relations, if the leading firm in an industry were to act as a spokesman for the rest of the industry in discussions with government regulatory agencies ○ Gatekeeper: when an actor screens or gathers resources from the outside and distributes them to members of his or her own subgroup ■ Because they perform “information processing” and “external representation” functions, have clear relevance for research on “boundary-spanning” role ○ Cosmopolitan or Itinerant: when 2 principals may belong to the same subgroup while the intermediary belongs to a different group. ○ Local broker or coordinator: this is a “null” type in the sense that it describes any brokerage relation in which no partition is imposed on the actors; but it also describes mediation between actors within a specified subgroup Week 12a: Statistical Models for Networks ● Classical approach: ● ○ Permutation Tests ○ ○ ● The permutation test essentially calculated all ways that the experiment would have come out given that the scores were actually independent of height, and counts the proportion of random assignments yielding correlations as large as the one actually observed. ○ The proportion is the p-value Steps for QAP correlation ○ Reshape two adjacency matrices into two columns ○ Compute the correlation coefficient ○ Calculate significance: ■ Compare observed correlation to correlation between thousands of pairs of matrices known to be independent of each other ■ Compute proportion of correlations that were as large as observed value and divide by total number ○ All this is done automatically by the QAP Correlation in R ■ qaptest {sna} ○ Usage ■ qaptest(dat, FUN, reps=1000, ...) ○ Example: correlation between ‘likes’ and ‘spending time with’ ■ Out of 5000 permutations, how many had >= or <= correlation than the observed value? ■ ■ ● Output shows percentile of simulated statistics where the observed calue occurs, in this case 97.5% of the simulated values were below the observed value, so this corresponds to a one sided p-value of .025 QAP Regression ○ We need to identify at least one predictor for the dependent variable ○ Result will not be that different from QAP correlation ○ Command to use: netlm(y,x,…) Week 12b: ● NetLogit: 𝑝 𝑙𝑛[ 1−𝑝𝑖 ] = 𝑏 + 𝑏 𝑋 0 1 𝑖 𝑖 ● Computing Probabilities 𝑝= 𝑒 𝑏 +𝑏 𝑥 0 1+𝑒 1 𝑏 +𝑏 𝑥 0 1 where e is the exponential constant approximately equal to 2.72 Answers for Question 2 on the Sample test Structural Hole: Packages: ● Igraph - plots networks ● concoR - used for blockmodeling ● Statnet - used to plot blockmodeling ● Assortnet - used to find assortativity on attributes ● Matrix - used to visualize assortativity for degree ● kableExtra - extends the basic functionality of tables produced ○ Features the pipe operator, %>% How to use kableExtra package: df <- data.frame(cnst,btwn) df %>% kbl() %>% kable_paper("hover", full_width = F, position = "left") Functions: ● as.matrix(df1) -- creates a matrix (need this for blockmodeling) ● concor_hca(list(data), p=2) ○ This creates the blocks for the data ○ Must use list() within concor_hca ○ P = 2 means make 2 divisions with 4 blocks ○ P = 3 creates 8 blocks ● blockmodel(list(data), blks$block, glabels = names(data), plabels = rownames(data)) ○ This creates the block membership with also showing the reduced form blockmodel ● Bimage <- blk_mod$block.model (blk_mod is the name of our blockmodel) ○ This creates the binary image by showing the percentages of how connected each block is ● gplot(bimage, diag=TRUE, edge.lwd=bimage*7, vertex.cex=sqrt(table(blk_mod$block.membership))/.75, gmode = "digraph", vertex.sides=50, vertex.col=gray(1-diag(bimage)), label = colnames(bimage)) ○ This shows the binary image and how each block interacts with each other ● graph_from_adjacency_matrix(m2, mode=’undirected’) ○ Creates a graph out of the data ● sort(constraint(g2)) ○ Finds the constraint of each node ○ Sort allows you to see the lowest and highest constraint ● cluster_walktrap(g2, steps=20) ○ Creates clusters using walktrap (when there are clusters of nodes but there are 2 nodes that are connected that connect the clusters) ● get.adjacency(g2) ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ○ Creates an adjacency matrix assortment.discrete(adjacency matrix variable, V(g2)$major) ○ This gets the assortativity of a discrete variable like major or sex assortment.continuous(adjacency matrix variable, V(g2)$age) ○ This gets the assortativity of a continuous variable like age as.network(inf) ○ Creates network objects from data matrices network.density(as.network(inf)) modularity(wt) ○ The lower the number, then there are more nodes that are connected ○ 0.5 means that there are multiple clusters that are not connected outside of their cluster at all sizes(wt) ○ Shows the size of communities length(wt) ○ Shows the number of communities membership(wt) ○ Shows the membership of each node graph.data.frame(df, directed=F) order(degree(g)) ○ Used for visualizing assortativity for degree image(A[i,i]) network(data, matrix.type=”adjacency”) ○ Creates a network with the matrix type being an adjacency gcor(data) ○ Computes the correlation qaptest(list(frnd), gcor, g1=1,g2=2) ○ Estimates the p-values of the network data netlm(spnd, reps=1000) Sample Exam Questions part of 3, 4, and 5 3. Consider the nodes in problem 2 to be a group of students. Their attributes are provided in the p2attr.csv file. Your task a. Find communities in the network and plot showing those communities b. Analyze if there is any assortativity based on any of the attributes assortment.discrete(A, V(g2)$major) $r = -0.09, which means the graph is slightly disassortive, meaning there is a tendency for nodes with dissimilar properties to connect. This being said the value is very close to 0 meaning the network is close to being non-assortive based on major. assortment.discrete(A, V(g2)$sex) $r = 0.31, this means that there is a tendencency for nodes with the same sex to connect. assortment.continuous(A, V(g2)$age) $r = .756, this means there is strong assortavity between nodes with similar age, meaning nodes with similar ages are the most likely to connect. 4. Consider the following MRQAP output based on the model - r2 <- netlm(extent_of_trade, list(bilateral_relation,cultural_exchanges,gdpmat), reps = 1000). The dependent variable is a valued matrix: extent_of_trade (scale 1-7) Independent variables are Valued matrices: bilateral_relation,cultural_exchanges (scale 1-7) Difference matrix: based on attribute (GDP): gdpmat a. Write the linear model 𝑒𝑥𝑡𝑒𝑛𝑡 𝑜𝑓 𝑡𝑟𝑎𝑑𝑒 = 2. 794 + 0. 741 × 𝑏𝑖𝑙𝑎𝑡𝑒𝑟𝑎𝑙𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 + 0. 076 × 𝑐𝑢𝑙𝑡𝑢𝑟𝑎𝑙 𝑒𝑥𝑐ℎ𝑎𝑛𝑔𝑒𝑠 + 1. 348 × 𝑔𝑑𝑝𝑚𝑎𝑡 b. Explain the results of the regression i. Is the model significant? Why or why not? ● The model is significant because the p-value associated with the model is 0.04539. Since this is less than 0.05, we can be at least 95% confident that the model is significant ii. Which of the independent variables is a statistically significant predictor? ● Bilateral_relation and gdpmat are significant because their respective p-values based on the direction of the hypothesis are 0.048 and 0.004 respectively. They are both less than 0.05 iii. Explain the coefficients in the equation ● For every unit increase in bilateral_relation, the corresponding increase in extent_of_trade is 0.7409. For every unit increase in cultural_exchanges, the corresponding increase in extent_of_trade is 0.076. For every unit increase in the difference in GDP, the corresponding increase in extent_of_trade is 1.348. 5. Consider the following netlogit output based on the model – netlogit(trade_or_not, bilateral_relation). The dependent variable is a binary matrix: trade_or_not (scale 1= significant trade, 0nonsignificant trade) Independent variable is Valued matrix: bilateral_relation (scale 1-7) a. Write the model 𝑝(𝑡𝑟𝑎𝑑𝑒𝑜𝑟𝑛𝑜𝑡) 𝑙𝑛[ 1−𝑝(𝑡𝑟𝑎𝑑𝑒𝑜𝑟𝑛𝑜𝑡) ] = − 6. 431 + 1. 370 × 𝑏𝑖𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 b. Explain the results of the regression i. Is the model significant? Why or why not? ● ii. iii. iv. Yes the model is significant, based on the p-value associated with the Chi-Squared test of fit improvement, which is 1.101001e-07 or almost 0. Is the independent variable statistically a significant predictor? ● Yes, because the p-value associated with that estimate is 0.007 Explain the coefficients in the equation ● The estimate of interest is 1.370. The literal interpretation is that logit(p) increases by 1.37 if the bilateral_relations increases by 1. This does not make too much sense. Therefore, we use the odds ratio. If the value of bilateral_relations is 4.2, what is the probability that there is significant trade? ● 𝑝= 𝑒 𝑏0+𝑏1𝑥 1+𝑒 ● v. 𝑏0+𝑏1𝑥 Code: P_estimate = exp(-6.431+1.370*4.2)/(1+exp(-6.431+1.370*4.2)) This gives an answer of 0.3369 How do you interpret Exp(b) for the predictor variable? ● Exp(b) for bilateral_relations is 3.74289. This is the odds ratio associated with a one-unit increase in the exposure. An odds ratio of 1 indicates that the condition or event under study is equally likely to occur (that is there is a 50-50 chance). An odds ratio of 1.2 means there is a 20% increase in the odds of an outcome with a given exposure. An odds ratio of 2 means there is a 100% increase in the odds of an outcome with a given level of the independent variable. So, an odds ratio of 3.74289 implies that there is an increase of 274.289% increase in the odds of there being significant trade for a unit increase in bilateral_relations.