Posteriors, conjugacy, and exponential families
for completely random measures
Tamara Broderick, Ashia C. Wilson, Michael I. Jordan
MIT Berkeley Berkeley
Posteriors, conjugacy, and exponential families for completely - - PowerPoint PPT Presentation
Posteriors, conjugacy, and exponential families for completely random measures Tamara Broderick, Ashia C. Wilson, Michael I. Jordan MIT Berkeley Berkeley Models Beta process, Bernoulli process (IBP) Gamma process, Poisson likelihood
Tamara Broderick, Ashia C. Wilson, Michael I. Jordan
MIT Berkeley Berkeley
p(θ|x) ∝ θα+x(1 + θ)−(α+x)−(β−x+1) x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0 α > 0, β > 0 = BetaPrime(θ|α, β)
1
p(θ|x) ∝ θα+x(1 + θ)−(α+x)−(β−x+1) x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0 α > 0, β > 0 = BetaPrime(θ|α, β)
1
p(θ|x) ∝ θα+x(1 + θ)−(α+x)−(β−x+1) x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0 α > 0, β > 0 = BetaPrime(θ|α, β)
1
p(θ|x) ∝ θα+x(1 + θ)−(α+x)−(β−x+1) x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0 α > 0, β > 0 = BetaPrime(θ|α, β)
1
p(θ|x) ∝ θα+x(1 + θ)−(α+x)−(β−x+1) x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0 α > 0, β > 0 = BetaPrime(θ|α, β)
1
p(θ|x) ∝ θα+x(1 + θ)−(α+x)−(β−x+1) x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0 α > 0, β > 0 = BetaPrime(θ|α, β)
1
p(θ|x) ∝ θα+x(1 + θ)−(α+x)−(β−x+1) x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0 α > 0, β > 0 = BetaPrime(θ|α, β)
1
p(θ|x) ∝ θα+x(1 + θ)−(α+x)−(β−x+1) x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0 α > 0, β > 0 = BetaPrime(θ|α, β)
1
p(θ|x) ∝ θα+x(1 + θ)−(α+x)−(β−x+1) x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0 α > 0, β > 0 = BetaPrime(θ|α, β)
1
p(θ|x) ∝ θα+x(1 + θ)−(α+x)−(β−x+1) x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0 α > 0, β > 0 = BetaPrime(θ|α, β)
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Arts Document 1 Econ Sports Health Technology Document 2 Document 3 Document 4 Document 5 Document 6 Document 7
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Arts Document 1 Econ Health Technology Document 2 Document 3 Document 4 Document 5 Document 6 Document 7
Sports
5
K+
n = Poisson
✓ γ β β + n − 1 ◆ Sn−1,k β + n − 1 For n = 1, 2, ..., N
feature k that has occurred times with probability
point n: Sn−1,k
n = 1 2 N ... k = 1 2 ...
[Griffiths & Ghahramani 2006]
6
K+
n = Poisson
✓ γ β β + n − 1 ◆ Sn−1,k β + n − 1 For n = 1, 2, ..., N
feature k that has occurred times with probability
point n: Sn−1,k
n = 1 2 N ... k = 1 2 ...
[Griffiths & Ghahramani 2006]
6
K+
n = Poisson
✓ γ β β + n − 1 ◆ Sn−1,k β + n − 1 For n = 1, 2, ..., N
feature k that has occurred times with probability
point n: Sn−1,k
n = 1 2 N ... k = 1 2 ...
[Griffiths & Ghahramani 2006]
6
K+
n = Poisson
✓ γ β β + n − 1 ◆ Sn−1,k β + n − 1 For n = 1, 2, ..., N
feature k that has occurred times with probability
point n: Sn−1,k
n = 1 2 N ... k = 1 2 ...
[Griffiths & Ghahramani 2006]
6
K+
n = Poisson
✓ γ β β + n − 1 ◆ Sn−1,k β + n − 1 For n = 1, 2, ..., N
feature k that has occurred times with probability
point n: Sn−1,k
n = 1 2 N ... k = 1 2 ...
[Griffiths & Ghahramani 2006]
6
K+
n = Poisson
✓ γ β β + n − 1 ◆ Sn−1,k β + n − 1 For n = 1, 2, ..., N
feature k that has occurred times with probability
point n: Sn−1,k
n = 1 2 N ... k = 1 2 ...
[Griffiths & Ghahramani 2006]
6
K+
n = Poisson
✓ γ β β + n − 1 ◆ Sn−1,k β + n − 1 For n = 1, 2, ..., N
feature k that has occurred times with probability
point n: Sn−1,k
n = 1 2 N ... k = 1 2 ...
[Griffiths & Ghahramani 2006]
6
K+
n = Poisson
✓ γ β β + n − 1 ◆ Sn−1,k β + n − 1 For n = 1, 2, ..., N
feature k that has occurred times with probability
point n: Sn−1,k
n = 1 2 N ... k = 1 2 ...
[Griffiths & Ghahramani 2006]
6
K+
n = Poisson
✓ γ β β + n − 1 ◆ Sn−1,k β + n − 1 For n = 1, 2, ..., N
feature k that has occurred times with probability
point n: Sn−1,k
n = 1 2 N ... k = 1 2 ...
[Griffiths & Ghahramani 2006]
6
K+
n = Poisson
✓ γ β β + n − 1 ◆ Sn−1,k β + n − 1 For n = 1, 2, ..., N
feature k that has occurred times with probability
point n: Sn−1,k
n = 1 2 N ... k = 1 2 ...
[Griffiths & Ghahramani 2006]
6
K+
n = Poisson
✓ γ β β + n − 1 ◆ Sn−1,k β + n − 1 For n = 1, 2, ..., N
feature k that has occurred times with probability
point n: Sn−1,k
n = 1 2 N ... k = 1 2 ...
[Griffiths & Ghahramani 2006]
6
θk ∼ Beta(1, β + m − 1) For m = 1, 2, ...
Draw a frequency of size
1
1, . . . , K+
m
θ1 θ2 θ3 θ4 θ5 θ6 K+
m ∼ Poisson
✓ γ β β + m − 1 ◆
[Hjort 1990; Kim 1999; Thibaux & Jordan 2007]
7
θk ∼ Beta(1, β + m − 1) For m = 1, 2, ...
Draw a frequency of size
1
1, . . . , K+
m
θ1 θ2 θ3 θ4 θ5 θ6 K+
m ∼ Poisson
✓ γ β β + m − 1 ◆
7
[Hjort 1990; Kim 1999; Thibaux & Jordan 2007]
θk ∼ Beta(1, β + m − 1) For m = 1, 2, ...
Draw a frequency of size
1
1, . . . , K+
m
θ1 θ2 θ3 θ4 θ5 θ6 K+
m ∼ Poisson
✓ γ β β + m − 1 ◆
7
[Hjort 1990; Kim 1999; Thibaux & Jordan 2007]
θk ∼ Beta(1, β + m − 1) For m = 1, 2, ...
Draw a frequency of size
1
1, . . . , K+
m
θ1 θ2 θ3 θ4 θ5 θ6 K+
m ∼ Poisson
✓ γ β β + m − 1 ◆
7
[Hjort 1990; Kim 1999; Thibaux & Jordan 2007]
θk ∼ Beta(1, β + m − 1) For m = 1, 2, ...
Draw a frequency of size
1
1, . . . , K+
m
θ1 θ2 θ3 θ4 θ5 θ6 K+
m ∼ Poisson
✓ γ β β + m − 1 ◆
7
[Hjort 1990; Kim 1999; Thibaux & Jordan 2007]
θk ∼ Beta(1, β + m − 1) For m = 1, 2, ...
Draw a frequency of size
1
1, . . . , K+
m
K+
m ∼ Poisson
✓ γ β β + m − 1 ◆
7
[Hjort 1990; Kim 1999; Thibaux & Jordan 2007]
θk ∼ Beta(1, β + m − 1) For m = 1, 2, ...
Draw a frequency of size
1
1, . . . , K+
m
θ1 θ2 θ3 K+
m ∼ Poisson
✓ γ β β + m − 1 ◆
7
[Hjort 1990; Kim 1999; Thibaux & Jordan 2007]
θk ∼ Beta(1, β + m − 1) For m = 1, 2, ...
Draw a frequency of size
1
1, . . . , K+
m
θ1 θ2 θ3 θ4 K+
m ∼ Poisson
✓ γ β β + m − 1 ◆
7
[Hjort 1990; Kim 1999; Thibaux & Jordan 2007]
θk ∼ Beta(1, β + m − 1) For m = 1, 2, ...
Draw a frequency of size
1
1, . . . , K+
m
θ1 θ2 θ3 θ4 θ5 θ6 K+
m ∼ Poisson
✓ γ β β + m − 1 ◆
7
[Hjort 1990; Kim 1999; Thibaux & Jordan 2007]
θk ∼ Beta(1, β + m − 1) For m = 1, 2, ...
Draw a frequency of size
1
1, . . . , K+
m
θ1 θ2 θ3 θ4 θ5 θ6 K+
m ∼ Poisson
✓ γ β β + m − 1 ◆
7
[Hjort 1990; Kim 1999; Thibaux & Jordan 2007]
θk ∼ Beta(1, β + m − 1) For m = 1, 2, ...
Draw a frequency of size
1
1, . . . , K+
m
θ1 θ2 θ3 θ4 θ5 θ6 K+
m ∼ Poisson
✓ γ β β + m − 1 ◆
7
[Hjort 1990; Kim 1999; Thibaux & Jordan 2007]
θk ∼ Beta(1, β + m − 1) For m = 1, 2, ...
Draw a frequency of size
1
1, . . . , K+
m
θ1 θ2 θ3 θ4 θ5 θ6 K+
m ∼ Poisson
✓ γ β β + m − 1 ◆
7
[Hjort 1990; Kim 1999; Thibaux & Jordan 2007]
θk ∼ Beta(1, β + m − 1) For m = 1, 2, ...
Draw a frequency of size
1
1, . . . , K+
m
θ1 θ2 θ3 θ4 θ5 θ6 K+
m ∼ Poisson
✓ γ β β + m − 1 ◆
7
[Hjort 1990; Kim 1999; Thibaux & Jordan 2007]
θk ∼ Beta(1, β + m − 1) For m = 1, 2, ...
Draw a frequency of size
1
1, . . . , K+
m
θ1 θ2 θ3 θ4 θ5 θ6 K+
m ∼ Poisson
✓ γ β β + m − 1 ◆
7
[Hjort 1990; Kim 1999; Thibaux & Jordan 2007]
Gibbs sampling)
...
distributions
How do we come up with these models?
n = 1 2 N ... k = 1 2 ...
8
Gibbs sampling)
...
distributions
How do we come up with these models?
n = 1 2 N ... k = 1 2 ...
8
Gibbs sampling)
...
distributions
How do we come up with these models?
n = 1 2 N ... k = 1 2 ...
8
distributions
Gibbs sampling)
...
How do we come up with these models?
n = 1 2 N ... k = 1 2 ...
8
Gibbs sampling)
...
How do we come up with these models?
n = 1 2 N ... k = 1 2 ...
distributions
8
Gibbs sampling)
...
distributions How do we come up with these models?
n = 1 2 N ... k = 1 2 ...
8
Likelihood (e.g. Bernoulli)
sequence (e.g. BP stick- breaking)
[Broderick, Wilson, Jordan 2014]
9
Likelihood (e.g. Bernoulli)
sequence (e.g. BP stick- breaking)
[Broderick, Wilson, Jordan 2014]
9
Likelihood (e.g. Bernoulli)
sequence (e.g. BP stick- breaking)
[Broderick, Wilson, Jordan 2014]
9
Likelihood (e.g. Bernoulli)
sequence (e.g. BP stick- breaking)
[Broderick, Wilson, Jordan 2014]
9
Likelihood (e.g. Bernoulli)
sequence (e.g. BP stick- breaking)
[Broderick, Wilson, Jordan 2014]
9
Likelihood (e.g. Bernoulli)
sequence (e.g. BP stick- breaking)
[Broderick, Wilson, Jordan 2014]
9
Likelihood (e.g. Bernoulli)
sequence (e.g. BP stick- breaking)
[Broderick, Wilson, Jordan 2014]
9
Likelihood (e.g. Bernoulli)
sequence (e.g. BP stick- breaking)
[Broderick, Wilson, Jordan 2014]
9
x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0
10
θ x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0
10
θ x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0
measure
process rate measure
atoms ν(dθ) ν(dθ)p(x|θ) ν(dθ) = γθα−1(1 − θ)−α−βdθ α ∈ (−1, 0], β > 0, γ > 0
10
θ x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0
measure
process rate measure
atoms ν(dθ) ν(dθ)p(x|θ) ν(dθ) = γθα−1(1 − θ)−α−βdθ α ∈ (−1, 0], β > 0, γ > 0
10
θ x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0
measure
process rate measure
atoms ν(dθ) ν(dθ)p(x|θ) ν(dθ) = γθα−1(1 − θ)−α−βdθ α ∈ (−1, 0], β > 0, γ > 0
10
θ x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0
measure
process rate measure
atoms ν(dθ) ν(dθ)p(x|θ) ν(dθ) = γθα−1(1 − θ)−α−βdθ α ∈ (−1, 0], β > 0, γ > 0 x
10
θ x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0
measure
process rate measure
atoms ν(dθ) ν(dθ)p(x|θ) ν(dθ) = γθα−1(1 − θ)−α−βdθ α ∈ (−1, 0], β > 0, γ > 0 x
10
θ x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0
measure
process rate measure
atoms ν(dθ) ν(dθ)p(x|θ) ν(dθ) = γθα−1(1 − θ)−α−βdθ α ∈ (−1, 0], β > 0, γ > 0 x
10
θ x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0
measure
process rate measure
atoms ν(dθ) ν(dθ)p(x|θ) ν(dθ) = γθα−1(1 − θ)−α−βdθ α ∈ (−1, 0], β > 0, γ > 0 x
10
θ x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0
measure
process rate measure
atoms ν(dθ) ν(dθ)p(x|θ) ν(dθ) = γθα−1(1 − θ)−α−βdθ α ∈ (−1, 0], β > 0, γ > 0 x
10
θ x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0
measure
process rate measure
atoms ν(dθ) ν(dθ)p(x|θ) ν(dθ) = γθα−1(1 − θ)−α−βdθ α ∈ (−1, 0], β > 0, γ > 0 x
10
measure
process rate measure
atoms
θ x x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0 ν(dθ) ν(dθ)p(x|θ) ν(dθ) = γθα−1(1 − θ)−α−βdθ α ∈ (−1, 0], β > 0, γ > 0
10
x x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0 θ ν(dθ) = γθα−1(1 − θ)−α−βdθ
11
x x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0 θ ν(dθ) = γθα−1(1 − θ)−α−βdθ
11
x x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0 θ ν(dθ) = γθα−1(1 − θ)−α−βdθ
11
θ x x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0 ν(dθ) = γθα−1(1 − θ)−α−βdθ
11
θ x x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0 R+ θ1 θ2 θ3 ν(dθ) = γθα−1(1 − θ)−α−βdθ
11
θ x R+ x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0 θ1 θ2 θ3 ν(dθ|x1 = 0) = γθα−1(1 − θ)−α−(β+1)
11
θ x R+ x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0 θ1 θ2 θ3 ν(dθ|x1 = 0) = γθα−1(1 − θ)−α−(β+1)
11
θ x R+ x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0 θ1 θ2 θ3 θ4 θ5 ν(dθ|x1 = 0) = γθα−1(1 − θ)−α−(β+1)
11
θ x R+ x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0 θ1 θ2 θ3 θ4 θ5 ν(dθ|x1:2 = 0) = γθα−1(1 − θ)−α−(β+2)
11
θ x
R+ x ∈ {0, 1} p(x|θ) = θx(1 + θ)−1 θ > 0 θ6 ν(dθ|x1:2 = 0) = γθα−1(1 − θ)−α−(β+2) θ1 θ2 θ3 θ4 θ5
11
For m = 1, 2, ...
Draw a rate of size α = 0 1, . . . , K+
m
R+ θ6 K+
m ∼ Poisson
✓ γ β β + m − 1 ◆ θk ∼ BetaPrime(1, β + m − 1) θ1 θ2 θ3 θ4 θ5
11
For m = 1, 2, ...
Draw a rate of size α = 0 1, . . . , K+
m
R+ θ6 K+
m ∼ Poisson
✓ γ β β + m − 1 ◆ θk ∼ BetaPrime(1, β + m − 1) θ1 θ2 θ3 θ4 θ5
11
For m = 1, 2, ...
Draw a rate of size α = 0 1, . . . , K+
m
R+ θ6 K+
m ∼ Poisson
✓ γ β β + m − 1 ◆ θk ∼ BetaPrime(1, β + m − 1) θ1 θ2 θ3 θ4 θ5
Marginal process derivation is similar
11
Exponential family likelihood
[Broderick, Wilson, Jordan 2014]
ν(dθ) = γ exp{hξ, η(θ)i + λ[A(θ)]}dθ + fixed atoms PPP rate measure f(dθ) = exp{hξk, η(θ)i + λk[A(θ)] B(ξk, λk)}dθ
K+
m ∼ Poisson
✓Z
x
γ · κ(0)m−1κ(x) · exp {B(ξ + (m − 1)φ(0) + φ(x), λ + m)} dx ◆
f(dθ) / Z
x
exp {hξ + (m 1)φ(0) + φ(x), η(θ)i + (λ + m)[A(θ)]} dx
12
Exponential family likelihood
[Broderick, Wilson, Jordan 2014]
p(dx|θ) = κ(x) exp{hη(θ), φ(x)i A(θ)} ν(dθ) = γ exp{hξ, η(θ)i + λ[A(θ)]}dθ + fixed atoms PPP rate measure dx f(dθ) = exp{hξk, η(θ)i + λk[A(θ)] B(ξk, λk)}dθ
K+
m ∼ Poisson
✓Z
x
γ · κ(0)m−1κ(x) · exp {B(ξ + (m − 1)φ(0) + φ(x), λ + m)} dx ◆
f(dθ) / Z
x
exp {hξ + (m 1)φ(0) + φ(x), η(θ)i + (λ + m)[A(θ)]} dx
12
Exponential family likelihood p(dx|θ) = κ(x) exp{hη(θ), φ(x)i A(θ)} ν(dθ) = γ exp{hξ, η(θ)i + λ[A(θ)]}dθ + fixed atoms PPP rate measure dx
K+
m ∼ Poisson
✓Z
x
γ · κ(0)m−1κ(x) · exp {B(ξ + (m − 1)φ(0) + φ(x), λ + m)} dx ◆
f(dθ) / Z
x
exp {hξ + (m 1)φ(0) + φ(x), η(θ)i + (λ + m)[A(θ)]} dx
f(dθ) = exp{hξk, η(θ)i + λk[A(θ)] B(ξk, λk)}dθ
12
[Broderick, Wilson, Jordan 2014]
Exponential family likelihood
[Broderick, Wilson, Jordan 2014]
p(dx|θ) = κ(x) exp{hη(θ), φ(x)i A(θ)}
ν(dθ) = γ exp{hξ, η(θ)i + λ[A(θ)]}dθ
+ fixed atoms PPP rate measure dx
f(dθ) = exp{hξk, η(θ)i + λk[A(θ)] B(ξk, λk)}dθ
12
K+
m ∼ Poisson
✓Z
x>0
γ · κ(0)m−1 · κ(x) · exp {B(ξ + (m − 1)φ(0) + φ(x), λ + m)} dx ◆
f(dθ) / Z
x>0
exp {hξ + (m 1)φ(0) + φ(x), η(θ)i + (λ + m)[A(θ)]} dx
K+
m as above
p(xn|x1:(n−1)) = κ(xn) exp ( −B(ξ +
n−1
X
m=1
xm, λ + n − 1) + B(ξ +
n−1
X
m=1
xm + xn, λ + n) )
Exponential family likelihood
[Broderick, Wilson, Jordan 2014]
p(dx|θ) = κ(x) exp{hη(θ), φ(x)i A(θ)}
ν(dθ) = γ exp{hξ, η(θ)i + λ[A(θ)]}dθ
+ fixed atoms PPP rate measure dx
f(dθ) = exp{hξk, η(θ)i + λk[A(θ)] B(ξk, λk)}dθ
12
K+
m ∼ Poisson
✓Z
x>0
γ · κ(0)m−1 · κ(x) · exp {B(ξ + (m − 1)φ(0) + φ(x), λ + m)} dx ◆
f(dθ) / Z
x>0
exp {hξ + (m 1)φ(0) + φ(x), η(θ)i + (λ + m)[A(θ)]} dx
K+
m as above
p(xn|x1:(n−1)) = κ(xn) exp ( −B(ξ +
n−1
X
m=1
xm, λ + n − 1) + B(ξ +
n−1
X
m=1
xm + xn, λ + n) )
Exponential family likelihood
[Broderick, Wilson, Jordan 2014]
p(dx|θ) = κ(x) exp{hη(θ), φ(x)i A(θ)}
ν(dθ) = γ exp{hξ, η(θ)i + λ[A(θ)]}dθ
+ fixed atoms PPP rate measure dx
f(dθ) = exp{hξk, η(θ)i + λk[A(θ)] B(ξk, λk)}dθ
12
K+
m ∼ Poisson
✓Z
x>0
γ · κ(0)m−1 · κ(x) · exp {B(ξ + (m − 1)φ(0) + φ(x), λ + m)} dx ◆
f(dθ) / Z
x>0
exp {hξ + (m 1)φ(0) + φ(x), η(θ)i + (λ + m)[A(θ)]} dx
K+
m as above
p(xn|x1:(n−1)) = κ(xn) exp ( −B(ξ +
n−1
X
m=1
xm, λ + n − 1) + B(ξ +
n−1
X
m=1
xm + xn, λ + n) )
Exponential family likelihood
[Broderick, Wilson, Jordan 2014]
p(dx|θ) = κ(x) exp{hη(θ), φ(x)i A(θ)}
ν(dθ) = γ exp{hξ, η(θ)i + λ[A(θ)]}dθ
+ fixed atoms PPP rate measure dx
f(dθ) = exp{hξk, η(θ)i + λk[A(θ)] B(ξk, λk)}dθ
12
K+
m ∼ Poisson
✓Z
x>0
γ · κ(0)m−1 · κ(x) · exp {B(ξ + (m − 1)φ(0) + φ(x), λ + m)} dx ◆
f(dθ) / Z
x>0
exp {hξ + (m 1)φ(0) + φ(x), η(θ)i + (λ + m)[A(θ)]} dx
K+
m as above
p(xn|x1:(n−1)) = κ(xn) exp ( −B(ξ +
n−1
X
m=1
xm, λ + n − 1) + B(ξ +
n−1
X
m=1
xm + xn, λ + n) )
Exponential family likelihood
[Broderick, Wilson, Jordan 2014]
p(dx|θ) = κ(x) exp{hη(θ), φ(x)i A(θ)}
ν(dθ) = γ exp{hξ, η(θ)i + λ[A(θ)]}dθ
+ fixed atoms PPP rate measure dx
f(dθ) = exp{hξk, η(θ)i + λk[A(θ)] B(ξk, λk)}dθ
12
K+
m ∼ Poisson
✓Z
x>0
γ · κ(0)m−1 · κ(x) · exp {B(ξ + (m − 1)φ(0) + φ(x), λ + m)} dx ◆
f(dθ) / Z
x>0
exp {hξ + (m 1)φ(0) + φ(x), η(θ)i + (λ + m)[A(θ)]} dx
as above
p(xn|x1:(n−1)) = κ(xn) exp ( −B(ξ +
n−1
X
m=1
xm, λ + n − 1) + B(ξ +
n−1
X
m=1
xm + xn, λ + n) )
K+
n
likelihood must have a point mass at 0
result of Poisson process
conjugacy at a different level
(i.e., discrete, continuous, or
A r t s Document 1 E c
H e a l t h T e c h n
y Document 2 Document 3 Document 4 Document 5 Document 6 Document 7 S p
t s
13
likelihood must have a point mass at 0
result of Poisson process
conjugacy at a different level
(i.e., discrete, continuous, or
A r t s Document 1 E c
H e a l t h T e c h n
y Document 2 Document 3 Document 4 Document 5 Document 6 Document 7 S p
t s
13
likelihood must have a point mass at 0
result of Poisson process
conjugacy at a different level
(i.e., discrete, continuous, or
A r t s Document 1 E c
H e a l t h T e c h n
y Document 2 Document 3 Document 4 Document 5 Document 6 Document 7 S p
t s
13
likelihood must have a point mass at 0
result of Poisson process
conjugacy at a different level
(i.e., discrete, continuous, or
A r t s Document 1 E c
H e a l t h T e c h n
y Document 2 Document 3 Document 4 Document 5 Document 6 Document 7 S p
t s
13
likelihood must have a point mass at 0
result of Poisson process
conjugacy at a different level
(i.e., discrete, continuous, or
A r t s Document 1 E c
H e a l t h T e c h n
y Document 2 Document 3 Document 4 Document 5 Document 6 Document 7 S p
t s
13
likelihood must have a point mass at 0
result of Poisson process
conjugacy at a different level
(i.e., discrete, continuous, or
A r t s Document 1 E c
H e a l t h T e c h n
y Document 2 Document 3 Document 4 Document 5 Document 6 Document 7 S p
t s
13
likelihood must have a point mass at 0
result of Poisson process
conjugacy at a different level
(i.e., discrete, continuous, or
A r t s Document 1 E c
H e a l t h T e c h n
y Document 2 Document 3 Document 4 Document 5 Document 6 Document 7 S p
t s
13
T Broderick, AC Wilson, and MI Jordan. Posteriors, conjugacy, and exponential families for completely random measures. Submitted, ArXiv, 2014.
Statistics, 1979.
Statistics, 1999.
life history data. Annals of Statistics, 1990.
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