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3.5
Probabilistic Bisimulation
With a solid base of the lifting operation, we can proceed to define a probabilistic
version of bisimulation.
Let s and t be two states in a pLTS. We say t can simulate the behaviour of s if
whenever the latter can exhibit action a and lead to distribution ʔ then the former can
also perform a and lead to a distribution, say ʘ , which then in turn can mimic ʔ in
successor states. We are interested in a relation between two states, but it is expressed
by invoking a relation between two distributions. To formalise the mimicking of one
distribution by the other, we make use of the lifting operation investigated in Sect. 3.3 .
Definition 3.5
A relation
R
S
×
S is a probabilistic simulation if s
R
t implies
a
−ₒ
a
−ₒ
• f s
ʔ then there exists some ʘ such that t
ʘ and ʔ
R
ʘ .
R 1
If both
R
and
are probabilistic simulations, then
R
is a probabilistic bisimu-
lation . The largest probabilistic bisimulation, denoted by
, is called probabilistic
bisimilarity .
As in the nonprobabilistic setting, probabilistic bisimilarity can be approximated
by a family of inductively defined relations.
Definition 3.6
Let S be the state set of a pLTS. We define:
0 :
=
S
×
S
s
n + 1 t , for n
0, if
a
−ₒ
a
−ₒ
n ) ʘ ;
1. whenever s
ʔ , there exists some ʘ such that t
ʘ and ʔ (
a
−ₒ
a
−ₒ
n ) ʘ .
2. whenever t
ʘ , there exists some ʔ such that s
ʔ and ʔ (
= n 0 n
In general,
ˉ :
ˉ . However, the two relations coin-
cide when limited to image-finite pLTSs where for any state s and action a , the set
{
is a strictly finer relation than
a
−ₒ
ʔ
D
( S )
|
s
ʔ
}
is finite.
Proposition 3.8
On image-finite pLTSs,
ˉ coincides with
.
Proof
It is trivial to show by induction that s
t implies s
n t for all n
0, and
thus that s
ˉ t .
Now we show that
a
−ₒ
ˉ is a bisimulation. Suppose s
ˉ t and s
ʔ .By
a
−ₒ
n ) ʘ n .
Since we are considering image-finite pLTSs, there are only finitely many different
ʘ n 's. Then for at least one of them, say ʘ ,wehave ʔ (
assumption for all n
0 there exists some ʘ n with t
ʘ n and ʔ (
n ) ʘ for infinitely many
different n 's. By a straightforward induction we can show that s n t implies s m t
for all m , n
n ) ʘ for all n
0 with n>m . It follows that ʔ (
0. We now claim
that
ˉ ) ʘ (3.12)
To see this, let A be any subset of the whole state space that may be countable. For any
n
ʔ (
n ) ʘ we know from Theorem 3.5 (1) that ʔ ( A )
0, since ʔ (
ʘ (
n ( A )) .
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