2.2.2- Calculation of the Extinction Probability and the Metastable Distribution
By making a few more approximations, we can derive expressions for the
probability of extinction as a function of time, the expected lifetime of the
epidemic, and the form of the long-lived metastable distribution
First, we shall
derive an approximate expression for
where p(I) = 0 for I;SPMlt;0 or I;SPMgt;N and
However, the survival component
is nearly a solution, and is only prevented from being one because of the slow leakage of
probability to the extinction component. We may obtain the survival component by artificially
stopping this leakage, which is achieved by setting p(I=1) to some arbitrary constant
In the more interesting intermediate case in which
Motivated by the form of the survival component in the last
two frames of Fig. 3, we match a gaussian to
Thus we find that the metastable distribution is a gaussian with mean
Having obtained the metastable distribution, we can now use it to estimate the lifetime of the metastable survival component. First, we must obtain an expression for p(0,t), the extinction probability as a function of time. Returning to the dynamical equation for the evolution of the probability distribution (Eq. 6), substituting I=0, and using Eqs. 8 (which is valid once the distribution has assumed its metastable form) and 11, we obtain:
Solving Eq. 12 for p(0,t), we find that, after the initial transient, the survival component decays exponentially with a characteristic lifetime given by
Numerical solution of Eq. 9 reveals that Eq.
13 yields a rather severe overestimate of the lifetime of the metastable phase.
This can be attributed to the fact that, while the gaussian approximation
of Eq. 11 to the survival component is very good in the vicinity of
Finally, we can obtain a good approximation to the probability that the infection will become
extinct before it reaches the metastable phase by letting
the number of nodes
which is the well-known linear birth and death process [34]. A solution can be obtained by the method of generating functions, with the result:
where
To summarize, the probabilistic analysis corroborates the deterministic result that an epidemic can
not occur if
|
.
:
.
Substituting I=0 into Eq. 
and using Eq. 
, one can show that the survival component has an
extremely short lifetime, which is consistent with the conclusion of the deterministic
analysis that no epidemic can occur if the infection rate is less than the cure rate. On the
other hand, if
, one
can show that practically all of the nodes will be infected, and the lifetime of the survival
component will be extremely long.
, p(I) attains
a maximum for some
. The value of
is determined by the condition
. Substitution of this condition into Eq.
,
, and
and normalize the sum of the extinction and survival components to unity, with the
result:
and standard
deviation
. The mean is identical to that obtained from the deterministic
approximation, and the standard deviation is virtually equal to that obtained from the
numerical solution of Eq. 
for a 100-node graph is reduced to only 888
for a 10-node graph, all other parameters being equal. The constant multiplying N in the
exponent is about half that predicted by Eq. 
.
(rendering the lifetime infinite). Then Eq.
is the number of infected nodes which are originally present in the population.
. Contrary to the findings of the deterministic analysis, even
if
, the probability that an epidemic will not occur is greater than zero, being equal to
, where
. Thus
, which lends some justification to the assumption that they
are zero in the deterministic analysis.