# Definition of Diffusion Centrality in Banerjee et al. (2019)

I am reading "Using gossips to spread information: Theory and evidence from two Randomised Control Trials" by Banerjee, Chandrashekhar, Duflo and Jackson, where they discuss the efficacy of gossips in spreading information regarding policies/ business offers.

In this paper, in page 21, they have introduced a generalised version of the network parameter "Diffusion Centrality", which the same authors had defined earlier in a 2013 paper.

If $$w$$ is a directed and weighted adjacency matrix for a social network, then they define a 'hearing matrix' for $$w$$ and time duration $$T \in N$$ as:

$$H(w, T) = \sum_{t=1}^T (w)^t$$

The $$ij$$-th entry of $$H$$, $$H(w, T)_{ij}$$ , is the expected number of times, within $$T$$ periods, that $$j$$ hears about a piece of information originating from $$i$$.

Then Diffusion Centrality is defined as:

$$DC(w, T) = H(w, T) \cdot 1$$

The paper says, "$$DC(w, T)_i$$ is the expected total number of times that some piece of information that originates from $$i$$ is heard by any of the members of the society during a $$T$$-period time interval".

I am not very familiar with network economics. I would like to understand why do $$H$$ and $$DC$$ match the verbal descriptions of them, that is, why is the $$ij$$-th entry of $$H$$ the expected number of times $$j$$ hears something from $$i$$ and similarly for $$DC$$. A proof directly written in the answer, or a reference to any good material to study this will help.

Let me use capital letter $$W$$ instead of small letter $$w$$ (as $$W$$ is actually a matrix).

So, let $$W$$ be the matrix that contains in row $$i$$ and column $$j$$ the probability that $$j$$ hears from $$i$$. Then average number of pieces of information that $$j$$ gets from $$i$$ in one period is given by $$W(i,j)$$.

The average number of information that $$j$$ gets from $$i$$ that takes 2 periods to travel is the information that goes first from $$i$$ to some intermediate person $$k$$ and then from $$k$$ to $$j$$, via the path $$i \to k \to j$$. The amount of information transmitted through this path is $$W(i,k) \cdot W(k,j)$$. To see this, notice that $$W(i,k)$$ is the amount of info send to $$k$$ and of this a fraction $$W(k,j)$$ is sent forward to $$j$$. Now, this has to be summed over all intermediate nodes $$k$$ (i.e. all paths of length 2). so we get: $$\sum_{k} W(i,k) \cdot W(k,j) = (W \times W)_{i,j} = (W^2)_{i,j}$$ Here $$\times$$ is matrix multiplication.

Let's now look at the information that takes 3 periods. This is the information that goes from $$i$$ to some intermediate $$k$$ then from $$k$$ to some intermediate $$\ell$$ and finally to $$j$$, so the path $$i \to k \to \ell \to j$$. The amount of information on this path is $$W(i,k) \cdot W(k,\ell) \cdot W(\ell, j)$$. If we sum over all these paths of length 3, we get: $$\sum_k \sum_\ell W(i,k) \cdot W(k,\ell) \cdot W(\ell, j) = (W \times W \times W)_{i,j} = (W^3)_{i,j}.$$ If we continue, then we can generalize this to any path of length $$T$$. The information sent from $$i$$ to $$j$$ through all such paths is given by: $$(W \times W \times \ldots W)_{i,j} = (W^T)_{i,j}.$$

Now if info goes from $$i$$ to $$j$$ either takes 1 period, 2 periods, 3 periods, ... $$T$$ periods to travel. As such, to compute the total expected pieces of info that $$j$$ receives from $$i$$ in $$T$$ periods is given by the sum over all these periods: $$H(W, T) = \sum_{t = 1}^T (W^t)_{i,j}.$$ Now the total amount of info sent from $$i$$ and heard by someone can be obtained by taking the sum over all recipients $$j$$. $$DC(W,T) = \sum_{j} (H(W,T))_{i,j} = (H(W,T)\times 1)_{i,j}$$ This amounts to adding up the rows of the matrix $$H(W,T)$$.

### An example

Let's take the example of three individuals $$1,2$$ and $$3$$. Assume that $$W$$ is given by: $$W = \begin{bmatrix} 0 & 0.3 & 0.2\\ 0.2 & 0 & 0.6\\ 0 & 0.4 & 0\end{bmatrix}$$ This implies that (for example) the probability that 2 hears from 1 in one period equals 0.3.

Then $$W^2 = \begin{bmatrix} 0.06 & 0.08 & 0.18\\ 0 & 0.3 & 0.04\\ 0.08 & 0 & 0.24\end{bmatrix}$$ Here, for example, the amount received by 2 from 1 in two periods is $$0.08 = 0.2 \times 0.4$$

Next, multiplying $$W^2$$ once more by $$W$$ gives: $$W^3 = \begin{bmatrix} 0.016 & 0.09 & 0.06\\ 0.06 & 0.016 & 0.18\\ 0 & 0.12 & 0.016\end{bmatrix}$$ Then: $$H(W, 3) = W + W^2 + W^3 = \begin{bmatrix} 0.076 & 0.47 & 0.44\\ 0.26 & 0.316 & 0.82\\ 0.08 & 0.52 & 0.256\end{bmatrix}$$ So in three periods $$2$$ receives on average $$0.47$$ pieces of info from $$1$$

The total information sent from $$1$$ to someone is computed by taking the sum of the elements in row $$1$$, which gives: $$0.076 + 0.47 + 0.44 = 0.986$$ In general: $$DC(W, 3) = \begin{bmatrix} 0.986\\1.396\\0.856\end{bmatrix}$$