ST5225: Mock Exam Nov 5 7-8pm
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ST5225: Mock Exam Nov 5 7-8pm
Instructions.
❼ Write your answers on A4-sized paper.
❼ Do not write your name.
❼ Start on a fresh sheet of paper for each question.
1. Consider the following adjacency matrix of an undirected network g
|
with n = 4 nodes.
Consider the model |
1 1 0 1 |
0(0)、． 0(1)．． . |
(1) |
P9 (g) = exp(90m(从)+s(9)1(9(才)(从)+92$2(从)) ,
where m(g) is the number of edges in graph g , T (g) the number of triangles and S2(g) the number of 2-stars. The constant κ(θ) =
dexp(θ0m(h) + θ1T (h) + θ2S2(h)) where the sum is over all possible graphs h with 4 nodes.
(a) How many terms are we summing over in the expression κ(θ)? (b) What is m(g), T (g) and S2(g)?
(c) Derive the pseudo-likelihood for the graph g .
(d) The eigenvalues and eigenvectors of matrix A are given below.
What are the eigenvector centrality of the four nodes? ✩values
[1] 2.1700865 0.3111078 -1.0000000 -1.4811943
✩vectors
[,1] [1,] -0.5227207 [2,] -0.5227207 [3,] -0.6116285 [4,] -0.2818452
[,2]
-0.3681604
-0.3681604
0.2536228
0.8152247
[,3]
7.071068e-01
-7.071068e-01
0.000000e+00
-5.551115e-16
[,4]
0.3020281
0.3020281
-0.7493905
0.5059367
2. Consider the lazega dataset which records the interactions between n = 36 lawyers in a law firm, The profile information of the 10 most senior lawyers are given below. “Years” refer to number of years with
the law firm.
|
|
Seniority |
Gender |
Office |
Years |
Age |
Practice |
School |
|
V1 |
1 |
1 |
1 |
31 |
64 |
1 |
1 |
|
V2 |
2 |
1 |
1 |
32 |
62 |
2 |
1 |
|
V3 |
3 |
1 |
2 |
13 |
67 |
1 |
1 |
|
V4 |
4 |
1 |
1 |
31 |
59 |
2 |
3 |
|
V5 |
5 |
1 |
2 |
31 |
59 |
1 |
2 |
|
V6 |
6 |
1 |
2 |
29 |
55 |
1 |
1 |
|
V7 |
7 |
1 |
2 |
29 |
63 |
2 |
3 |
|
V8 |
8 |
1 |
1 |
28 |
53 |
1 |
3 |
|
V9 |
9 |
1 |
1 |
25 |
53 |
2 |
1 |
|
V10 |
10 |
1 |
1 |
25 |
53 |
2 |
3 |
A logistic model is fitted with
puà exp(a0 +a1 (z1u +z10)+'''+a7 (z7u +z70))
where puà is the probability of interaction between nodes u and v, and x1u, . . . , x7u are the covariates of lawyer u. The output is as below.
Maximum Likelihood Results:
|
|
Estimate |
Std. Error |
z value |
Pr(> |z|) |
|
edges |
7.804489 |
4.409111 |
1.770 |
0.07671 |
|
nodecov.Senior |
-0.043430 |
0.033547 |
-1.295 |
0.19546 |
|
nodecov.Practice |
0.673640 |
0.165572 |
4.069 |
< 1e-04 |
|
nodecov.Years |
-0.009918 |
0.020557 |
-0.482 |
0.62948 |
|
nodecov.Gender |
-0.518730 |
0.331374 |
-1.565 |
0.11749 |
|
nodecov.Age |
-0.086409 |
0.028817 |
-2.999 |
0.00271 |
|
nodecov.Office |
0.058560 |
0.162184 |
0.361 |
0.71805 |
|
nodecov.School |
-0.048888 |
0.109813 |
-0.445 |
0.65618 |
Null Deviance: 873.4 on 630 degrees of freedom
Residual Deviance: 552.8 on 622 degrees of freedom
AIC: 568.8 BIC: 604.4 (Smaller is better).
A second logistic model is fitted with the ”‘Years” and “Age” covari- ates removed. The output is as below.
Maximum Likelihood Results: