We prove that: X is a delayed renewal sequence iff
there exists a renewal sequence W (on a probability space) such that
for all nonnegative integers r and s and sequences 
of 0's and 1's such that 
Proof of `` 
''. What is meant by a delayed renewal sequence
(however never properly stated) is a sequence X of 0's and 1's
such that for 
and 
we have either 
or Y is a renewal sequence
independent of T with respect to the measure
.
In particular, for any integer 
we have
Hence
where 
and
We will use this equality for 
. By the way,
observe that 
, 
, and
what we have on the left in (1) is exactly J.
To transform it to the right-hand side of (1),
we apply Prop. 1 to the renewal sequence
. Then we find that 
, where
(Actually, we use an obvious generalization of Prop 1 when instead of
the inequalities 0<n one writes 
which is harmless if
in this Prop. 1 we have 
).
It only remains to notice that, as before,
and replacing k+n with n
Thus, we get (1) with W=Y.
Proof of `` 
''.
We may assume that 
.
The above notice that J coincides with
the left-hand side shows that, for any 
,
provided . Choosing different numbering and remembering
that 
, we get that
for any
provided . Summing up with respect to r yields
It follows that the distributions of Y (with respect to P') and W coincide. Since W is a renewal sequence and being a renewal sequence is completely defined by these distributions (cf., for instance, Prop. 1), we conclude that Y is a renewal sequence.
Finally, replacing the last factor in (2) with the left-hand side of (3), we conclude that T and Y are independent with respect to P'.
