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Stanによる混合ガンマ分布の推定

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混合ガウス分布の推定はよくやられているけど、混合ガンマ分布の推定はちょっと難しいみたい。

Stan で書いてみた。

gamma_mixture.stan
data {
  int N;
  real values[N];
  int cluster_num;
}
transformed data {
  real min_val = min(values);
  real max_val = max(values);
  real var_val = variance(values);
}
parameters {
  positive_ordered[cluster_num] mu;
  positive_ordered[cluster_num] k;
  simplex[cluster_num] ratio; 
}
transformed parameters {
  real<lower=min_val/var_val> rate[cluster_num];
  for (j in 1:cluster_num) {
    rate[j] = k[j] / mu[j];
  }
}
model {
  real loglikes[cluster_num];
  for(i in 1:N) {
    for (j in 1:cluster_num) {
      loglikes[j] = log(ratio[j]) + gamma_lpdf(values[i] | k[j], rate[j]);
    }
    target += log_sum_exp(loglikes);
  }
  for (j in 1:cluster_num) {
    mu[j] ~ uniform(min_val, max_val);
  }
}

一般の混合ガンマ分布の推定を書きたかったけど、難しいみたいなので次の制約を入れた。
M 個のガンマ分布に対して、それぞれの期待値を $\mu_i$ と表し、

$$
\mu_1 < \mu_2 < ... < \mu_M
$$

となるように番号を振る。このとき、形状パラメータ $k_i$ に対して

$$
k_1 < k_2 < ... < k_M
$$

が成り立つという制約を入れる。
これは尺度パラメータ $\theta_i$ の逆数 $r_i$ に対して次の制約を置いたものと等価である。

$$
r_1 < \frac{\mu_2}{\mu_1}r_2 < \frac{\mu_3}{\mu_1}r_3 < ... < \frac{\mu_M}{\mu_1}r_M
$$

したがって、この制約は $\theta_1 = \cdots = \theta_M$ を含む。

2 クラス推定

R
library(rstan)

N <- 1000

d1 <- rgamma(N, 1, 1)
d2 <- rgamma(N, 4, 1/2)
values <- c(d1, d2)

standata <- list(N = length(values), values = values, cluster_num = 2)

fit <- stan("gamma_mixture.stan", data = standata, chains = 3)

print(fit, pars=c("k", "rate"), digits_summary = 1)
結果
        mean se_mean  sd 2.5% 25% 50% 75% 97.5% n_eff Rhat
k[1]     1.0       0 0.0  0.9 1.0 1.0 1.0   1.1  1906    1
k[2]     5.1       0 0.4  4.3 4.8 5.0 5.3   5.9  1515    1
rate[1]  1.0       0 0.1  0.8 1.0 1.0 1.1   1.2  1358    1
rate[2]  0.6       0 0.0  0.5 0.6 0.6 0.7   0.7  1856    1

3 クラス推定

R
library(rstan)

N <- 1000

d1 <- rgamma(N, 1, 1)
d2 <- rgamma(N, 4, 1/2)
d3 <- rgamma(N, 8, 1/5)
values <- c(d1, d2, d3)

standata <- list(N = length(values), values = values, cluster_num = 3)

fit <- stan("gamma_mixture.stan", data = standata, chains = 3)

print(fit, pars=c("k", "rate"), digits_summary = 1)
結果
        mean se_mean  sd 2.5% 25% 50% 75% 97.5% n_eff Rhat
k[1]     1.0       0 0.1  0.9 0.9 1.0 1.0   1.1  1441    1
k[2]     4.7       0 0.6  3.6 4.3 4.7 5.1   5.9   938    1
k[3]     7.9       0 0.5  6.9 7.5 7.9 8.3   9.0  1947    1
rate[1]  0.9       0 0.1  0.7 0.8 0.9 1.0   1.1   998    1
rate[2]  0.6       0 0.1  0.4 0.5 0.6 0.6   0.7   986    1
rate[3]  0.2       0 0.0  0.2 0.2 0.2 0.2   0.2  2187    1

うまくいったみたい。
ちょっとずれているが、サンプルサイズ増やせばもっと正確に推定できるはず。

Enjoy!

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