Nonparametric Bayesian Discrete Latent Variable Models for ...

A Details of Derivations for the Stick-Breaking Representation of IBP
The µl for l ∈ Lk are independent given µ (1:k), therefore we can obtain the distribution
for µ (k+1) = max µl by taking the product of the cdf’s of µl’s:
l∈Lk
F (µ (k+1) | µ (1:k)) = µ
= µ
α
− K
(k)
K−k
−α K
(k)
µ α
K
(k+1) I(0 ≤ µ (k+1) ≤ µ (k)) + I(µ (k) < µ (k+1)) K−k
µ α K−k
K
(k+1) I(0 ≤ µ (k+1) ≤ µ (k)) + I(µ (k) < µ (k+1)).
Differentiating the above equation, the density of µ (k+1) is obtained to be,
p(µ (k+1) | µ (1:k)) = p(µ (k+1) | µ (k))
= α
K − k
K
K−k
K
µ−α
(k)
µ α K−k
K −1
(k+1)
A.2 Probability of a part of Z being inactive
I(0 ≤ µ (k+1) ≤ µ (k)).
(A.4)
(A.5)
Given µ (k), we can calculate the probability of all entries to the right of column k being
zero. We denote the set of indices after the kth largest feature presence probability with
Lk. The density of the (unordered) feature presence probabilities with index l ∈ Lk
is given in eq. (A.2). The entries zil are Bernoulli distributed with probability µl.
Marginalizing over µl, we have;
p(Z (:,.>k) = 0 | µ (k)) =
=
µ (k)
0
p(Z (:,.>k) = 0 | µ (k), µL)p(µL)dµL
α
K
α
K
µ−
(k)
α
µ K −1 (1 − µ) N K−k dµ
Applying change of variables ν = µ/µ(k) to the above integral,
=
=
µ(k)
0
1
0
1
= α
K
Using the binomial series,
116
= α
K
= α
K
1
0
α
K
α
K
0
1
N
i=0
0
α
K
ν α
K −1
α
K
µ−
(k)
µ α
K −1 (1 − µ) N dµ
µ− α
K
(k) (νµ (k)) α
K −1 (1 − νµ (k)) N µ (k)dν
ν α
K −1 (1 − ν + ν − νµ (k)) N dν
ν α
K −1 (1 − ν) + ν(1 − µ (k)) N dν.
N
i=0
N
(1 − µ
i
(k)) i
N
(1 − ν)
i
N−i (ν(1 − µ (k))) i dν
1
0
α
i+
ν K −1 (1 − ν) N−i dν.
(A.6)
(A.7)
(A.8)