Probability of global epidemic

Solution 1:

This does not compute an exact value but some rigorous upper bounds of the probability $r$ that every site in $\mathbb Z^2$ is reached, eventually.

Consider the Galton-Watson branching process where the progeny of every individual has the distribution of the number of signals transmitted at time $n+1$ by any site reached at time $n$. Thus the progeny of any node in the Galton-Watson tree is $0$, $1$, $2$, $3$ and $4$ with respective probabilities $1$, $4$, $6$, $4$ and $1$ divided by $16$.

Then the generation $n$ of the process overestimates the number of sites reached at time $n$ since the signals received from different neighbors stay separated. Hence the probability $q$ that the tree becomes extinct underestimates the probability that the cloud of sites receiving a signal becomes empty. On the other hand, if the cloud survives forever, the path of any eternally transmitted signal is a simple random walk on $\mathbb Z^2$, and these are recurrent hence every site is reached by the trace of the cloud on the grid, eventually.

The extinction probability $q$ is classically the smallest root in $[0,1]$ of the one-step equation $$ 16q=1+4q+6q^2+4q^3+q^4, $$ which is $q=0.087378$. Hence every site on the grid is reached by the original transmission process with probability $r\lt91.2622\%$.

Now, the probability $r_T$ that $T=(10^5,10^5)$ is reached corresponds to the fact that generation $n=2\cdot10^5$ of the tree is not empty, whose probability $1-q_n$ is very close to $1-q$, hence $r_T\lt1-q_n\approx1-q$.

Nota: One gets a simpler bound noticing that, if the initial site transmits nothing, no other site is reached, and this happens with probability $1/16=6.25\%$, hence $r\lt93.75\%$.

Solution 2:

In the same vein as Didier's answer providing bounds on the extinction probability $q$, we can also obtain bounds on the transmission probability $p$ required for the probability of $N$ signals dying out to go to $0$ as $N\to\infty$, which is the probability required for the extinction probability $q$ not to be $1$.

Didier's equation for $q$ can be rewritten as $q=(1-(1-q)/2)^4$, which says that the signal goes extinct if all four signals die, either because they don't make it across the edge ($1/2$) or because they go extinct afterwards ($1-q$). Generalizing to $d$ dimensions and transmission probability $p$, this is

$$q=(1-p(1-q))^{2d}\;.$$

This equation always has a root at $1$. For sufficiently large $q$, it also has a second root in $[0,1]$, and the extinction probability is given by that root. The critical case in which the extinction probability becomes $1$ occurs when these two roots coincide. Differentiating with respect to $q$ and substituting $q=1$ yields the condition for $1$ to be a double root:

$$ \begin{eqnarray} 1&=&2dp(1-p(1-q))\;, \\ 1&=&2dp\;, \\ p&=&\frac1{2d}\;. \end{eqnarray} $$

This is a lower bound, since in the case of confluent signals there are fewer signals to keep the fire burning. Your experiments seem to indicate that the bound becomes better with increasing dimension, which makes sense since the signals become less likely to collide.