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I am presently reading and studying a paper (page 11 of this document) regarding epidemiology and disease spread over a network of contacts. I am having problems understanding a passage.

Some background

So consider having a graph where nodes are people and connections are "contacts" among them. When two individuals have a contact, there is the probability that one infected subject might infect the other one (if le latter has not yet been infected so far). The paper so introduces $r_{i,j}$ as the probability that individual $x_i$ infects individual $x_j$ in one time slot (step, the model works using a discrete time scheme). That being said, the author also introduces $T_{i,j}^{(\tau)}$ as the probability that individual $x_i$ infects $x_j$ after exactly $\tau$ time steps. It is easy to understand that:

$$T_{i,j}^{(\tau)} = 1 - (1 - r_{i,j})^\tau$$

The previous holds because $x_i$ is supposed to remain infectious for all $\tau$ steps. In these $\tau$ steps $1 - r_{i,j}$ is the probability that $x_j$ is not infected. So far so good! My problems start now!

The problem

Right after introducing and formulating an expression for $T_{i,j}^{(\tau)}$, the author says that the set $\{r_{i,j}\}_{i,j \in N}$ is assumed to be iid. Because of that he also states that $T_{i,j}^{(\tau)}$ (I am reporting here the symbols he used) is also an iid random variable and that we can work with its mean by introducing the following:

\begin{equation} T = \langle T_{i,j}^{(\tau)}\rangle = 1 - \int_0^\infty [1 - P(r)(1 - r)^\tau] \, dr \end{equation}

Where $P(r)$ is the probability density function for the distribution of $r_{i,j}$.

How does he get to this expression? What


Following are my questions:

  • How does the author get to the final expression?
  • What does he mean by "$P(r)$ is the probability density function for the distribution of $r_{i,j}$"?

I am not understanding because the author never introduces the random variables. In this case I expected the author to introduce $R_{i,j}$ as the r.v. "$i$ infects $j$ in $1$ time step" which can have values true or false $(0,1)$. So $R_{i,j}$ is a discrete r.v. and its pdf (density func.) should consist in two separate deltas. The same goes for $T_{i,j}$ which should be the r.v. "$i$ infects $j$ after $\tau$ time slots", again this is a r.v. whose values can be true or false and whose pdf $f_{T(i,j)}(x)$ consists in two deltas where $f_{T(i,j)}(x = 1) = 1 - (1 - r_{i,j})^\tau$ and $f_{T(i,j)}(x = 1) = (1 - r_{i,j})^\tau$. If I want to take the mean of this random variable I do not get the expression reported above.

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I changed $< T_{i,j}^{(\tau)} >$ to $\langle T_{i,j}^{(\tau)}\rangle$. That is standard usage. – Michael Hardy Jan 27 '13 at 14:33
Thankyou... I was typing fast and did not realize I used gt and lt... – Andry Jan 27 '13 at 14:35
Are you sure he didn't write $\displaystyle 1 - \int_0^\infty P(r)(1-r)^\tau\,dr$? – Michael Hardy Jan 27 '13 at 14:37
ABSOLUTELY sure about this... But if you think this is wrong, please tell me... – Andry Jan 27 '13 at 14:42
I am providing a link to the paper:… Please refer to page 11 and you'll find it. – Andry Jan 27 '13 at 14:43
up vote 0 down vote accepted

It appears to me that Seilheimer simply made a clumsy mistake. If $r_{ij}$ is a random variable whose probability density function is $P$, and the value of $P$ is $0$ on $(-\infty,0]$, then the expected value of $1-(1-r_{ij})^\tau$ is $$ \int_0^\infty (1-(1-r)^\tau) P(r) \, dr = 1 - \int_0^\infty (1-r)^\tau P(r) \, dr. $$

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You know I have a problem here... Sure you are right, but what about $P(r)$? What does it mean? It is a pdf, well, but here the mean is considered on a continuos r.v. However I would say that $P(r)$ is the number of edges whose $r_{i,j}$ is $r$ (say) on the total number of edges... I cannot figure what P(r) represents... – Andry Jan 27 '13 at 15:19
OK, now I think I understand... he is treating $r$ which is a probability, like a random variable... – Andry Jan 27 '13 at 16:29

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