R：glm函数，族=“二项式”和“重量”规格

我对体重与family =“ binomial”在glm中的工作方式非常困惑。在我的理解中，具有family =“ binomial”的glm的可能性指定如下： ，其中y是“观察到的成功比例”，n是已知的试验次数。y

$$f(y) = {n\choose{ny}} p^{ny} (1-p)^{n(1-y)} = \exp \left(n \left[ y \log \frac{p}{1-p} - \left(-\log (1-p)\right) \right] + \log {n \choose ny}\right)$$ $y$$n$

以我的理解，成功概率$p$由一些线性系数\ beta参数$\beta$化为$p=p(\beta)$并且glm函数带有family =“ binomial”搜索：

$$\textrm{arg}\max_{\beta} \sum_i \log f(y_i).$$ 然后可以将此优化问题简化为：

$$\textrm{arg}\max_{\beta} \sum_i \log f(y_i)= \textrm{arg}\max_{\beta} \sum_i n_i \left[ y_i \log \frac{p(\beta)}{1-p(\beta)} - \left(-\log (1-p(\beta))\right) \right] + \log {n_i \choose n_iy_i}\\ = \textrm{arg}\max_{\beta} \sum_i n_i \left[ y_i \log \frac{p(\beta)}{1-p(\beta)} - \left(-\log (1-p(\beta))\right) \right] \\ $$
因此，如果我们让$n_i^*=n_ic$所有$i=1,...,N$对于某一常数$c$，那么它也必须是真实的： $$\textrm{arg}\max_{\beta} \sum_i \log f(y_i) = \textrm{arg}\max_{\beta} \sum_i n^*_i \left[ y_i \log \frac{p(\beta)}{1-p(\beta)} - \left(-\log (1-p(\beta))\right) \right] \\ $$ 由此，我认为按比例缩放试验次数$n_i$给定成功y_i的比例，具有常数的变量不会影响\ beta的最大似然估计$\beta$$y_i$。

glm帮助文件显示：

 "For a binomial GLM prior weights are used to give the number of trials 
  when the response is the proportion of successes" 

因此，鉴于成功的响应比例，我期望权重的缩放不会影响估计的$\beta$。但是，以下两个代码返回不同的系数值：

 Y <- c(1,0,0,0) ## proportion of observed success
 w <- 1:length(Y) ## weight= the number of trials
 glm(Y~1,weights=w,family=binomial)

这产生：

 Call:  glm(formula = Y ~ 1, family = "binomial", weights = w)

 Coefficients:
 (Intercept)  
      -2.197     

而如果我将所有权重乘以1000，估计的系数就会不同：

 glm(Y~1,weights=w*1000,family=binomial)

 Call:  glm(formula = Y ~ 1, family = binomial, weights = w * 1000)

 Coefficients:
 (Intercept)  
    -3.153e+15  

即使权重适度缩放，我也看到了许多其他这样的示例。这里发生了什么？

您的示例仅导致R中的舍入误差。大的权重在中表现不佳glm。的确，w几乎任何较小的数字（如100）进行缩放都会导致与未缩放的结果相同的估算w。

如果您希望使用weights参数获得更可靠的行为，请尝试使用包中的svyglm函数survey。

看这里：

    > svyglm(Y~1, design=svydesign(ids=~1, weights=~w, data=data.frame(w=w*1000, Y=Y)), family=binomial)
Independent Sampling design (with replacement)
svydesign(ids = ~1, weights = ~w, data = data.frame(w = w * 1000, 
    Y = Y))

Call:  svyglm(formula = Y ~ 1, design = svydesign(ids = ~1, weights = ~w2, 
    data = data.frame(w2 = w * 1000, Y = Y)), family = binomial)

Coefficients:
(Intercept)  
     -2.197  

Degrees of Freedom: 3 Total (i.e. Null);  3 Residual
Null Deviance:      2.601 
Residual Deviance: 2.601    AIC: 2.843

我认为它归结到在使用的初始值glm.fit由family$initialize这使得该方法divergere。据我所知，glm.fit通过形成的QR分解解决问题，其中是设计矩阵和是如所描述的对角与平方根条目这里。也就是说，使用牛顿-拉夫森方法。X$\sqrt{W}X$$X$$\sqrt{W}$

相关$intialize代码为：

if (NCOL(y) == 1) {
    if (is.factor(y)) 
        y <- y != levels(y)[1L]
    n <- rep.int(1, nobs)
    y[weights == 0] <- 0
    if (any(y < 0 | y > 1)) 
        stop("y values must be 0 <= y <= 1")
    mustart <- (weights * y + 0.5)/(weights + 1)
    m <- weights * y
    if (any(abs(m - round(m)) > 0.001)) 
        warning("non-integer #successes in a binomial glm!")
}

这是其中的简化版本glm.fit，显示了我的观点

> #####
> # setup
> y <- matrix(c(1,0,0,0), ncol = 1)
> weights <- 1:nrow(y) * 1000
> nobs <- length(y)
> family <- binomial()
> X <- matrix(rep(1, nobs), ncol = 1) # design matrix used later
> 
> # set mu start as with family$initialize
> if (NCOL(y) == 1) {
+   n <- rep.int(1, nobs)
+   y[weights == 0] <- 0
+   mustart <- (weights * y + 0.5)/(weights + 1)
+   m <- weights * y
+   if (any(abs(m - round(m)) > 0.001)) 
+     warning("non-integer #successes in a binomial glm!")
+ }
> 
> mustart # starting value
             [,1]
[1,] 0.9995004995
[2,] 0.0002498751
[3,] 0.0001666111
[4,] 0.0001249688
> (eta <- family$linkfun(mustart))
          [,1]
[1,]  7.601402
[2,] -8.294300
[3,] -8.699681
[4,] -8.987322
> 
> #####
> # Start loop to fit
> mu <- family$linkinv(eta)
> mu_eta <- family$mu.eta(eta)
> z <- drop(eta + (y - mu) / mu_eta)
> w <- drop(sqrt(weights * mu_eta^2 / family$variance(mu = mu)))
> 
> # code is simpler here as (X^T W X) is a scalar
> X_w <- X * w
> (.coef <- drop(crossprod(X_w)^-1 * ((w * z) %*% X_w)))
[1] -5.098297
> (eta <- .coef * X)
          [,1]
[1,] -5.098297
[2,] -5.098297
[3,] -5.098297
[4,] -5.098297
> 
> # repeat a few times from "start loop to fit"

我们可以重复最后一部分两次，以查看牛顿-拉夫森方法的不同之处：

> #####
> # Start loop to fit
> mu <- family$linkinv(eta)
> mu_eta <- family$mu.eta(eta)
> z <- drop(eta + (y - mu) / mu_eta)
> w <- drop(sqrt(weights * mu_eta^2 / family$variance(mu = mu)))
> 
> # code is simpler here as (X^T W X) is a scalar
> X_w <- X * w
> (.coef <- drop(crossprod(X_w)^-1 * ((w * z) %*% X_w)))
[1] 10.47049
> (eta <- .coef * X)
         [,1]
[1,] 10.47049
[2,] 10.47049
[3,] 10.47049
[4,] 10.47049
> 
> 
> #####
> # Start loop to fit
> mu <- family$linkinv(eta)
> mu_eta <- family$mu.eta(eta)
> z <- drop(eta + (y - mu) / mu_eta)
> w <- drop(sqrt(weights * mu_eta^2 / family$variance(mu = mu)))
> 
> # code is simpler here as (X^T W X) is a scalar
> X_w <- X * w
> (.coef <- drop(crossprod(X_w)^-1 * ((w * z) %*% X_w)))
[1] -31723.76
> (eta <- .coef * X)
          [,1]
[1,] -31723.76
[2,] -31723.76
[3,] -31723.76
[4,] -31723.76

如果您以weights <- 1:nrow(y)或开头，则不会发生这种情况weights <- 1:nrow(y) * 100。

注意，可以通过设置mustart参数来避免分歧。例如做

> glm(Y ~ 1,weights = w * 1000, family = binomial, mustart = rep(0.5, 4))

Call:  glm(formula = Y ~ 1, family = binomial, weights = w * 1000, mustart = rep(0.5, 
    4))

Coefficients:
(Intercept)  
     -2.197  

Degrees of Freedom: 3 Total (i.e. Null);  3 Residual
Null Deviance:      6502 
Residual Deviance: 6502     AIC: 6504