Assuming we have a random directed weighted graph with $n$ nodes. Furthermore let us assume the nodes are divided into two categories:
- A node $i$ is of category C if there are only outgoing edges or only ingoing edges from/to $i$
- Otherwise a node $i$ is of category D
Let the number of nodes in category C be $c$,then the number of nodes in category D is simply $n-c$.
Let $p_1$ be the probability that there is an edge between two nodes of category $C$ (and we pick each direction with $p_1/2$) and let $p_2$ be the probability of an edge between two nodes $i$ and $j$ for all other cases (again picking each direction with probability $p_2/2$). We denote the weight going from a node $i$ to a node $j$ by $e_{ij}$, where the $e_{ij}$'s follow three different distributions (with means $\mu_1, \mu_2, \mu_3$ respectively) depending on following cases:
- $(i,j)\in D\times D$
- $(i,j)\in D\times C$ or $(i,j)\in C\times D$
- $(i,j)\in C\times C$
I am interested in calculating the expected value of $\sum_{j=1}^{n-1} e_{ij}-e_{ji}$. For the cases where $i$, $j$ are both in the same category, the values will simply be zero by symmetry. But how can I calculate the other cases?
My idea was to first calculate the expected value of $\sum_{j=1}^{n-1} e_{ij}$ for the two cases of $i$:
i.e. if $i \in C$, then $$\sum_{j=1}^{n-1} e_{ij}=\sum_{j\in C} e_{ij} +\sum_{j\in D} e_{ij}=\frac{ \mu_3(c-1)p_1}{2} +\frac{ \mu_2(n-c)p_2}{2}$$
And if if $i \in D$, then $$\sum_{j=1}^{n-1} e_{ij}=\sum_{j\in C} e_{ij} +\sum_{j\in D} e_{ij}=\frac{ \mu_2 c p_2}{2} +\frac{ \mu_1(n-c-1)p_2}{2}$$
Then since if a node $i$ is in C it only has arrows in one direction, the expected 'net position' would be $\frac{ \mu_3(c-1)p_1}{2} +\frac{ \mu_2(n-c)p_2}{2}$
And for the other case, if $i$ is in D the expected 'net position' would be simply $0$ by symmetry.
Is this correct until now or am I missing something?
Thanks for any help!
Best Answer
Approved by my supervisor all calculations are correct and this is the solution to the expected net-positions: