Given a network with $G(n, p)$ with $n = 3600$ vertices and an edge probability of $p$. The set of vertices consists of three different types of vertex: $type_i$, $type_j$ and $type_k$. The proportions of those types are known: 80% of $type_i$, 15% of $type_j$ and 5% of $type_k$.
An edge is said to be of $type_{(i, j)}$ if it connects a $type_i$ vertex and a $type_j$ vertex. The graph is without self-loops and undirected, so edge $type_{(i, j)} \equiv type_{(j, i)}$.
What are the expected number of edges of each type? So: $type_{(i, j)}$, $type_{(i, k)}$, $type_{(j, k)}$, $type_{(i, i)}$, $type_{(j, j)}$ and $type_{(k, k)}$?
Here's what I got so far:
- $P(type_{(i, i)}) = 0.8^2$
- $P(type_{(j, j)}) = 0.15^2$
- $P(type_{(k, k)}) = 0.05^2$
- $P(type_{(i, j)}) = 2 \times 0.8 \times 0.15$
- $P(type_{(i, k)}) = 2 \times 0.8 \times 0.05$
- $P(type_{(j, k)}) = 2 \times 0.15 \times 0.05$
The sum of all probabilities gives 1, so that seems about right. To get the number of channels of a specific type, you would do the following for each type:
$p \times 3600 \times P(type_{(i, i)})$
My solution seems to disregard the fact that the graph doesn't self-loop, but I figured that with enough vertices that wouldn't make a difference.
Are my assumptions correct? Is the math right? Any pointer would be appreciated.
You are very close, but made a mistake in the end.
The probabilities of each type, for a given existing edge are ok.
For each potential edge, then the probability of each type is indeed $p\times P\left(type_{\cdot,\cdot}\right)$.
But when you are taking the total expected number : You do not have 3600 edges. You have 3600 vertices, hence $$\binom{3600}{2} \text{ possible edges}$$