Consider the model of "Growing Random Networks" described in lecture 3.1 (so the first model where links are added to existing nodes uniformly at random), with m = 10m=10.

At date t=40t=40, what are:

(1) the expected degree for node 15 born at i=15, and

(2) the expected degree for node 30 born at i=30?

[Hint: Use the approximation for the expected degree at time tt of a node born at time ii, from the Lecture 3.1. Also, it helps to think about which of the two nodes has a higher expected degree.]

In [1]:

```
sum = 10
for i in 16:40
sum += 10/i
end
sum
```

Out[1]:

In [2]:

```
# (1)
10(1+ log(40/15))
```

Out[2]:

In [3]:

```
# (2)
10(1+ log(40/30))
```

Out[3]:

Consider the "Preferential Attachment Model" described in lecture 3.3, with m = 10.

At date t=40, what are:

(1) the expected degree for node 15 (born at time i=15), and

(2) the expected degree for node 30 (born at time i=30)?

[Hint: Use the approximation for the expected degree at time tt of a node born at time ii from the lecture 3.3.

You are encouraged to compare the results here to those in Question 1.]

In [4]:

```
#(1)
10*(40/15)^0.5
```

Out[4]:

In [5]:

```
#(2)
10*(40/30)^0.5
```

Out[5]:

Consider the "Hybrid Model" described in lecture 3.4, with m = 10. Let us compare the cases of a - 0.8, and a = 0.2

At data t = 10000, consider a node born at data i = 20 and a node born at data i = 9999.

Recall that the approximation of the expected degree at time t of a node born at time i is: $$d_i(t) = (m + 2am/(1-a))(t/i)^{(1-a)/2} - 2am/(1-a)$$

Which option makes the following statement correct?

At data t = 10000,

a node born at date i = 9999 has a **___** expected degree with a = 0.8 than with a = 0.2; and

a node born at data i = 20 has a ** __** expected degree with a = 0.8 than with a = 0.2

[Note: After this problem set is complete, there are two optional quizzes that you may follow. One will have you use some software to analyze real network data ( `Quiz: Optional: Empirical Analysis of Network Data '), and the other will have you use R to estimate and ERGM, (Quiz: Optional: Using Statnet in R to Estimate an ERGM']

In [1]:

```
function expected_degree(m, a, t, i)
ed = (m + 2*a*m/(1-a)) * (t/i)^((1-a)/2) - 2*a*m/(1-a)
return ed
end
```

Out[1]:

In [2]:

```
[expected_degree(10, 0.8, 10000, 9999), expected_degree(10, 0.2, 10000, 9999)]
```

Out[2]:

In [3]:

```
[expected_degree(10, 0.8, 10000, 20), expected_degree(10, 0.2, 10000, 20)]
```

Out[3]: