This is equivalent to saying that conditionally given that the first split is into subtrees with n
1
≥ . . . ≥ n
i
≥ . . . ≥ n
k
≥ 1 leaves and that leaf 1 is in a subtree with n
i
leaves, the delabelled subtrees S
◦ 1
, . . . , S
◦ k
of the common ancestor are independent and distributed as T
◦ n
j
respectively, j ∈ [k]. Since this conditional distribution does not depend on i, we have established the Markov branching
property of T
◦ n
. b Notice that if
γ = 1−α, the alpha-gamma model is the model related to stable trees, the labelling of which is known to be exchangeable, see Section 3.4.
On the other hand, if γ 6= 1 − α, let us turn to look at the distribution of T
3
.
¡ ¡
¡ ¡
¡ ¡
¡ ¡
❅ ❅
❅ ❅
1 2
3 1
3 2
Probability:
γ 2−
α
Probability:
1− α
2− α
We can see the probabilities of the two labelled trees in the above picture are different although they have the same unlabelled tree. So if
γ 6= 1 − α, T
n
is not exchangeable.
2.5 Sampling consistency and strong sampling consistency
Recall that an unlabelled Markov branching tree T
◦ n
, n ≥ 2 has the property of sampling consistency, if when we select a leaf uniformly and delete it together with the adjacent branch point if its
degree is reduced to 2, then the new tree, denoted by T
◦ n,−1
, is distributed as T
◦ n−1
. Denote by d : D
n
→ D
n−1
the induced deletion operator on the space D
n
of probability measures on T
◦ n
, so that for the distribution P
n
of T
◦ n
, we define dP
n
as the distribution of T
◦ n,−1
. Sampling consistency is equivalent to dP
n
= P
n−1
. This property is also called deletion stability in [12].
Proposition 12. The unlabelled alpha-gamma trees for 0 ≤ α ≤ 1 and 0 ≤ γ ≤ α are sampling
consistent. Proof. The sampling consistency formula 14 in [16] states that dP
n
= P
n−1
is equivalent to qn
1
, . . . , n
k
=
k
X
i=1
n
i
+ 1m
n
i
+1
+ 1 n + 1m
n
i
qn
1
, . . . , n
i
+ 1, . . . , n
k ↓
+ m
1
+ 1 n + 1
qn
1
, . . . , n
k
, 1 + 1
n + 1 qn, 1qn
1
, . . . , n
k
8 for all n
1
≥ . . . ≥ n
k
≥ 1 with n
1
+ . . . + n
k
= n ≥ 2, where m
j
is the number of n
i
, i ∈ [k], that equal j, and where q is the splitting rule of T
◦ n
∼ P
n
. In terms of EPPFs 1, formula 8 is equivalent to 1 − pn, 1
pn
1
, . . . , n
k
=
k
X
i=1
pn
1
, . . . , n
i
+ 1, . . . , n
k
+ pn
1
, . . . , n
k
, 1. 9
Now according to Proposition 10, the EPPF of the alpha-gamma model with α 1 is
p
seq α,γ
n
1
, . . . , n
k
= Z
n
Γ
α
n
γ + 1 − α − γ 1
nn − 1 X
u6=v
n
u
n
v
p
PD
∗
α,−α−γ
n
1
, . . . , n
k
, 10
414
where Γ
α
n = Γn − αΓ1 − α. Therefore, we can write p
seq α,γ
n
1
, . . . , n
i
+ 1, . . . , n
k
using 2 Z
n+1
Γ
α
n + 1
γ + 1 − α − γ 1
n + 1n
X
u6=v
n
u
n
v
+ 2n − n
i
a
k
Z
n+1
Y
j: j6=i
w
n
j
w
n
i
+1
= p
seq α,γ
n
1
, . . . , n
k
+ 21 − α − γ n − 1n − n
i
− P
u6=v
n
u
n
v
n + 1nn − 1 Z
n
Γ
α
n p
PD
∗
α,−α−γ
n
1
, . . . , n
k
× n
i
− α n −
α and p
seq α,γ
n
1
, . . . , n
k
, 1 as Z
n+1
Γ
α
n + 1
γ + 1 − α − γ 1
n + 1n
X
u6=v
n
u
n
v
+ 2n
a
k+1
Z
n+1
k
Y
j=1
w
n
j
w
1
= p
seq α,γ
n
1
, . . . , n
k
+ 21 − α − γ n − 1n −
P
u6=v
n
u
n
v
n + 1nn − 1 Z
n
Γ
α
n p
PD
∗
α,−α−γ
n
1
, . . . , n
k
× k − 1α − γ
n − α
. Sum over the above formulas, then the right-hand side of 9 is
1 − 1
n − α
γ + 2
n + 1 1 − α − γ
p
seq α,γ
n
1
, . . . , n
k
. Notice that the factor is indeed p
seq α,γ
n, 1. Hence, the splitting rules of the alpha-gamma model satisfy 9, which implies sampling consistency for
α 1. The case α = 1 is postponed to Section 3.2.
Moreover, sampling consistency can be enhanced to strong sampling consistency [16] by requiring that T
◦ n−1
, T
◦ n
has the same distribution as T
◦ n,−1
, T
◦ n
.
Proposition 13. The alpha-gamma model is strongly sampling consistent if and only if γ = 1 − α.
Proof. For γ = 1 − α, the model is known to be strongly sampling consistent, cf. Section 3.4.
¡ ¡
¡ ¡
¡ ¡
¡ ¡
❅ ❅
❅ ❅
❅ ❅
t
◦ 3
t
◦ 4
If γ 6= 1 − α, consider the above two deterministic unlabelled trees.
P T
◦ 4
= t
◦ 4
= q
seq α,γ
2, 1, 1q
seq α,γ
1, 1 = α − γ5 − 5α + γ2 − α3 − α. Then we delete one of the two leaves at the first branch point of
t
◦ 4
to get t
◦ 3
. Therefore P
T
◦ 4,−1
, T
◦ 4
= t
◦ 3
, t
◦ 4
= 1
2 P
T
◦ 4
= t
◦ 4
= α − γ5 − 5α + γ
22 − α3 − α
. 415
On the other hand, if T
◦ 3
= t
◦ 3
, we have to add the new leaf to the first branch point to get t
◦ 4
. Thus P
T
◦ 3
, T
◦ 4
= t
◦ 3
, t
◦ 4
= α − γ
3 − α
P T
◦ 3
= t
◦ 3
= α − γ2 − 2α + γ
2 − α3 − α .
It is easy to check that PT
◦ 4,−1
, T
◦ 4
= t
◦ 3
, t
◦ 4
6= PT
◦ 3
, T
◦ 4
= t
◦ 3
, t
◦ 4
if γ 6= 1 − α, which means that the alpha-gamma model is then not strongly sampling consistent.
3 Dislocation measures and asymptotics of alpha-gamma trees
3.1 Dislocation measures associated with the alpha-gamma-splitting rules
