I am reading Networks, Crowds, and Markets Reasoning about a Highly Connected World. section 8 of Chapter 28. I am confused about the another equivalent expectation. I never see that before. Does any can help me? Thanks for your time.
To recap the discussion there, the definition for the expected value is $$ E[X_n] = 1 × Pr[X_n = 1] + 2 × Pr[X_n = 2] + 3 × Pr[X_n = 3]+· · · (21.3) $$ and an alternate but equivalent way to write the right-hand side is $$Pr[X_n ≥ 1] + Pr[X_n ≥ 2] + Pr[X_n ≥ 3]+· · · , (21.4)$$ since we observe that each term $Pr[X_n = i]$ contributes exactly i copies of itself to the sum in (21.4). Therefore, we have $$E[X_n] = Pr[X_n ≥ 1] + Pr[X_n ≥ 2] + Pr[X_n ≥ 3]+· · · . (21.5)$$
It seems probability distribution is:
$X_n$| 1 2 3 .
P | $Pr[X_n = 1]$ $Pr[X_n = 2]$ $Pr[X_n = 3]$
but why
$$Pr[X_n ≥ 1] + Pr[X_n ≥ 2] + Pr[X_n ≥ 3]+· · · , (21.4)$$
Should I give more information about this? Thanks for your consideration.
EDIT
The Expected Number of Infected Individuals. We start by considering an approach to this problem that gets us partway to a proof of the claim: considering the expected number of infected individuals at each level of the tree.
First, let’s consider the total number of individuals at each level. The number of individuals at any given level exceeds the number at the previous level by a factor of $k$, and therefore the number who are at level n is $k_n$. (This is also true at level $n = 0$; the top level consists of just the root, and $k_0 = 1$.)
Now, let $X_n$ be a random variable equal to the number of infected individuals at level n. One way to think about the expected value $E[X_n]$ to write $X_n$ as a sum of simpler random variables as follows. For each individual $j$ at level $n$, let $Y_{nj}$ be a random variable equal to $1$ if $j$ is infected, and equal to $0$ otherwise. Then $$X_n = Y_{n1} + Y_{n2} +· · ·+Y_{nm},$$ where $m = k_n$, since the right-hand side simply counts, one by one, the number of infected individuals at level n. Linearity of expectation says that the expectation of the sum of a set of random variables is equal to the sum of their expectations, and so $$E[X_n] = E [Y_{n1} + Y_{n2} +· · ·+Y_{nm}] = E[Y_{n1}] + E[Y_{n2}]+· · ·+E[Y_{nm}] . (21.1)$$ The reason to write things this way is that each expectation on the right-hand side is extremely easy to work out: $E[Y_{nj}] = 1 × Pr[Y_{nj} = 1] + 0 × Pr[Y_{nj} = 0]= Pr[Y_n = 1]$, and so the expectation of each $Y_{nj }$ is just the probability that individual $j$ gets infected.Individual $j$ at depth $n$ gets infected precisely when all of the $n$ contacts leading from the root to $j$ successfully transmit the disease, as shown in Figure 21.14. Since each contact transmits the disease independently with probability $p$, individual $j$ is infected with probability $p_n$. Therefore, $E[Y_{nj}] = p_n$.We have already concluded that there are $k_n$ individuals at level $n$ of the tree, and hence $k_n $ terms on the right-hand side of Equation (21.1). Therefore, as summed up in Figure 21.15, we conclude that $$E[X_n] = p^n k^n = (pk)^n = R^n_0. (21.2)$$ From Expected Values to Probabilities of Persistence. Equation (21.2) suggests the importance of the basic reproductive number $R_0$ in reasoning about the spread of an epidemic in the branching process model. Now let’s consider what this tells us about $q^∗$, the probability that the epidemic persists indefinitely.
First, the fact that $E[X_n] = R^n_0$ immediately establishes part (a) the claim we are trying to prove, that when $R_0 < 1$ we have $q^∗ = 0$. To see why, we go back to the definition of $E[X_n]$ and apply a fact that we also found useful in Section 20.7.To recap the discussion there, the definition for the expected value is $$ E[X_n] = 1 × Pr[X_n = 1] + 2 × Pr[X_n = 2] + 3 × Pr[X_n = 3]+· · · (21.3) $$ and an alternate but equivalent way to write the right-hand side is $$Pr[X_n ≥ 1] + Pr[X_n ≥ 2] + Pr[X_n ≥ 3]+· · · , (21.4)$$ since we observe that each term $Pr[X_n = i]$ contributes exactly i copies of itself to the sum in (21.4). Therefore, we have $$E[X_n] = Pr[X_n ≥ 1] + Pr[X_n ≥ 2] + Pr[X_n ≥ 3]+· · · . (21.5)$$
From Equation (21.5) we observe that $E[X_n]$ must be at least as large as the first term on the right-hand side, and so $E[X_n] ≥ Pr[X_n ≥ 1]$. Notice also that $Pr[X_n ≥ 1]$ is precisely the definition of $q_n$, and so $E[X_n] ≥ q_n$. But $E[X_n] = R^n_0$, which is converging to $0$ as $n$ grows; hence, $q_n$ must also be converging to $0$. This shows that $q^∗ = 0$ when $R_0 < 1$.
$$p(X\geq1)=p(X=1)+p(X=2)+p(X=3)+...+p(X=n)$$ $$p(X\geq2)=p(X=2)+p(X=3)+...p(X=n)$$
Keep writing these out and add them all up, and the result follows