r - Stan多项式回归参数估计模型回顾

Question

我有以下多项式回归模型:

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LaTeX 版本

$Y_i |\mu_i,\sigma^2\sim\text{正态}(\mu_i,\sigma^2), i = 1,\dots, n\\text{独立}$

$\mu_i =\alpha +\beta_1 x_{i1} +\beta_2 x_{i2} +\beta_3 x_{i1}^2 +\beta_4 x_{i2}^2 +\beta_5 x_{i1} x_{ i2}$

$\alpha\sim\text{一些合适的先验}$

$\beta_1,\dots,\beta_5\sim\text{一些合适的先验}$

$\sigma^2\sim\text{一些合适的先验}$

我想将样本大小和 $y_i$、$x_{i1}$ 和 $x_{i2}$ 上的观察向量作为输入。其代码如下:

data{
  int<lower=1> n;
  vector[n] x1;
  vector[n] x2;
  vector[n] y;
}

我想标准化(居中和缩放)两个输入变量以获得标准化的回归变量x1_std 和x2_std。其代码位于transformed data block 中，如下所示:

transformed data{
  real bar_x1;
  real x1_sd;
  vector[n] x1_std;
  real bar_x2;
  real x2_sd;
  vector[n] x2_std;
  real y_sd;

  bar_x1 = mean(x1);
  x1_sd = sd(x1);
  x1_std = (x1 - bar_x1)/x1_sd; // centered and scaled

  bar_x2 = mean(x2);
  x2_sd = sd(x2);
  x2_std = (x2 - bar_x2)/x2_sd; // centered and scaled

  y_sd = sd(y);
}

然后我想使用标准化回归变量和回归参数 $\alpha$、$\beta_1$ 和 $\dots、\beta_5$ 的回归估计来拟合上述多项式回归模型，两者 原始的标准化比例。

据此，如果我没记错的话，标准化参数到原始尺度的转换公式如下:

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LaTeX 版本

$\alpha =\tilde{\alpha} -\dfrac{\gamma_1}{s_1}\bar{x}_1 -\dfrac{\gamma_2}{s_2}\bar{x}_2 +\dfrac{\gamma_3}{s_1^2}\bar{x}_1^2 +\dfrac{\gamma_4}{s_2^2}\bar{x}_2^2 +\dfrac{\gamma_5}{s_1 s_2}\bar{x }_1\bar{x}_2$

$\beta_1 =\left(\dfrac{\gamma_1}{s_1} - 2\dfrac{\gamma_3}{s_1^2}\bar{x}_1 -\dfrac{\gamma_5}{s_1 s_2}\bar{x}_2\right)$

$\beta_2 =\left(\dfrac{\gamma_2}{s_2} - 2\dfrac{\gamma_4}{s_2^2}\bar{x}_2 -\dfrac{\gamma_5}{s_1 s_2}\栏{x}_1\右)$

$\beta_3 =\dfrac{\gamma_3}{s_1^2}$

$\beta_4 =\dfrac{\gamma_4}{s_2^2}$

$\beta_5 =\dfrac{\gamma_5}{s_1 s_2}$

实现它的代码包含在 generated quantities block 中，如下所示:

alpha = alpha_std - beta1_std*bar_x1/x1_sd - beta2_std*bar_x2/x2_sd
      + (beta3_std*bar_x1^2)/x1_sd^2 + (beta4_std*bar_x2^2)/x2_sd^2
      + (beta5_std*bar_x2*bar_x1)/(x1_sd*x2_sd);

beta1 = beta1_std/x1_sd - 2*beta3_std*bar_x1/x1_sd^2
      - beta5_std*bar_x2/(x1_sd*x2_sd);

beta2 = beta2_std/x2_sd - 2*beta4_std*bar_x2/x2_sd^2
      - beta5_std*bar_x1/(x1_sd*x2_sd);

beta3 = beta3_std/x1_sd^2;

beta4 = beta4_std/x2_sd^2;

beta5 = beta5_std/(x1_sd*x2_sd);

我的整个模型如下:

data{
  int<lower=1> n;
  vector[n] x1;
  vector[n] x2;
  vector[n] y;
}
transformed data{
  real bar_x1;
  real x1_sd;
  vector[n] x1_std;
  real bar_x2;
  real x2_sd;
  vector[n] x2_std;
  real y_sd;

  bar_x1 = mean(x1);
  x1_sd = sd(x1);
  x1_std = (x1 - bar_x1)/x1_sd; // centered and scaled

  bar_x2 = mean(x2);
  x2_sd = sd(x2);
  x2_std = (x2 - bar_x2)/x2_sd; // centered and scaled

  y_sd = sd(y);
}
parameters{
  real<lower=0> sigma;
  real alpha_std;
  real beta1_std;
  real beta2_std;
  real beta3_std;
  real beta4_std;
  real beta5_std;
}
transformed parameters {
  real mu[n];

  for(i in 1:n) {
    mu[i] = alpha_std + beta1_std*x1_std[i]
      + beta2_std*x2_std[i] + beta3_std*x1_std[i]^2
      + beta4_std*x2_std[i]^2 + beta5_std*x1_std[i]*x2_std[i];
  }
}
model{
  alpha_std ~ normal(0, 10);
  beta1_std ~ normal(0, 2.5);
  beta2_std ~ normal(0, 2.5);
  beta3_std ~ normal(0, 2.5);
  beta4_std ~ normal(0, 2.5);
  beta5_std ~ normal(0, 2.5);
  sigma ~ exponential(1 / y_sd);

  y ~ normal(mu, sigma);
}
generated quantities {
  real alpha;
  real beta1;
  real beta2;
  real beta3;
  real beta4;
  real beta5;
  
  alpha = alpha_std - beta1_std*bar_x1/x1_sd - beta2_std*bar_x2/x2_sd
      + (beta3_std*bar_x1^2)/x1_sd^2 + (beta4_std*bar_x2^2)/x2_sd^2
      + (beta5_std*bar_x2*bar_x1)/(x1_sd*x2_sd);

  beta1 = beta1_std/x1_sd - 2*beta3_std*bar_x1/x1_sd^2
      - beta5_std*bar_x2/(x1_sd*x2_sd);

  beta2 = beta2_std/x2_sd - 2*beta4_std*bar_x2/x2_sd^2
      - beta5_std*bar_x1/(x1_sd*x2_sd);

  beta3 = beta3_std/x1_sd^2;

  beta4 = beta4_std/x2_sd^2;

  beta5 = beta5_std/(x1_sd*x2_sd);
}

我正在使用 R 的 MASS 包中的 hills 数据集:

library(MASS)
hills[18, 3] <- 18.65 # Fixing transcription error
x1 <- hills$dist
x2 <- hills$climb
y <- hills$time
n <- length(x1)
data.in <- list(x1 = x1, x2 = x2, y = y, n = n)
model.fit <- sampling(example, data.in)

现在我输出标准化的 (alpha_std, beta1_std, beta2_std, beta3_std, beta4_std , beta5_std) 和原始比例 (alpha, beta1, beta2, beta3, beta4, beta5)回归参数:

print(model.fit, pars = c("alpha_std", "alpha", "beta1_std", "beta2_std", "beta3_std", "beta4_std", "beta5_std", "beta1", "beta2 ", "beta3", "beta4", "beta5", "sigma"), probs = c(0.05, 0.5, 0.95), digits = 5)

我做对了吗？我还对数学进行了两次和三次检查，所以我认为它应该是正确的。尽管如此，让我感到紧张的一件事是 beta4 是 0.00000。这是否表明我犯了错误？正如我所说，我已经检查了我所有的代码和数学，所以据我所知，一切似乎都很好。

Answer 1

好吧，我刚刚发现问题是我没有打印具有足够数字的值(5 不够)以查看该值不是 0.00000。其他一切都很好。