Proof of theorem A1. (⇒) If the static estimator exists but isn’t strongly connected, then there are players i and j such that there are no -paths in . In that case, if we define as the set of players reachable from i in , and to be the set of players not reachable from I. By construction, in a game and , if we would have an edge between and , which contradicts the definition of I and J, so .
Note that
and
so the sets aren’t empty, and since
is weakly connected, there must be a game between a player in
J to another in
I, let’s say w.l.o.g. that
, and in that case, by lemma A2,
since
for any
, and that’s not possible.
(⇐) On the other hand, suppose is strongly connected. Then we show that exists by proving a stronger proposition: if we fix , there are such that for every , we have . To do this, we show by induction in m that there exist continuous functions for such that for any and , and also, that every component of is non-decreasing in each of its arguments (that is, ). For brevity, we write instead of .
First, we note that since
is weakly connected,
is strictly decreasing in
, but non-decreasing in any other rating
for
:
since one of the sums is non-empty and
is strictly increasing. From the same expression it’s easy to see that
For the base case of induction,
, it’s easy to show that the increasing function
has a unique zero, because
(since
is strongly connected, player
n scores less than 1 in some match
k) and similarly
and
f is obviously continuous (
is continuous), so just applying the intermediate value theorem allows us to define
as the only zero of
f. To see that
is non-decreasing in every coordinate of its argument, we check that
and the last implication is a consequence of
being strictly decreasing in
.
To see that
is continuous, we note that for any convergent sequence
with
, we have
and since
is strictly monotone and continuous, and therefore bijective,
.
Now for the induction step: suppose exists and is non-decreasing in and with respect to every coordinate, and . For fixed , we will show that the continuous function , which inherits continuity from and , is strictly decreasing.
Suppose
. Then, by our induction hypothesis
. Let’s define
and
, and note that
since
, and
. Note that
or
, and in the latter case
by definition of
. Using that fact, and then lemmas A1 and A2,
where
. Therefore,
since at least one of the sums is non-empty (
is connected) and
for any game between
and
.
We also check that
approaches a positive number as
x goes to infinity. Since
is non-decreasing in
x, its coordinates either have a limit as
x goes to infinity, or they also go to infinity (by the monotone convergence theorem), and we can define
for which again,
and
, and as before,
since
is connected and therefore one of the sums is non-empty. An analogous argument shows that as
, and thus, for any given
,
has a unique zero, which we denote by
, and with that we can define
.
Now, we can show that
is non-decreasing, because if
, then by induction hypothesis,
, and hence
which implies (by the fact that
is strictly decreasing and continuous, i.e. bijective) that
.
This in turn implies that
is non-decreasing by composition of
g and
, and to conclude the induction, we only need to check that
g is continuous. This is similar to the base case, using that for any convergent sequence
with
, we have
and invoking that
is bijective again.
Our induction works for , and to complete the proof (for ) we consider . By definition of , we know that , and by lemma A1, , as we wanted. □