Question

我从伽马密度模拟了100个观测值：

x <- rgamma(100,shape=5,rate=5)

我尝试用R中的optim函数得到最大似然估计量的渐近方差。为此，我手动计算了伽马密度的对数似然的表达式，并将其乘以-1因为optim是最低限度的。这是我的代码：

min <- function(data, par) {
    with(data, 1/par[2]*sum(x)-(par[1]-1)*sum(log(x))+n*(log(gamma(par[1]))+par[1]*log(par[2])))
}

mle <- optim(par=c(0,1),min,method='BFGS', hessian=TRUE)

AV <- 1 / mle$hessian

但是，我得到以下错误（因为第二行代码）：

Error in eval(substitute(expr), data, enclos = parent.frame()) : 
    numeric 'envir' arg not of length one

有没有办法解决这个问题？任何提示都会很感激。谢谢。

P.S：我知道我可以找到渐近方差：

library(MASS)

AV <- (fitdistr(x, "gamma", start=list(shape=1, rate=1))$sd)^2

但那不是我想要的方式。

Answer 1

Ok, let's go by parts.

First of all, your expression for the minus log likelihood is incorrect, as it is actually parameterized as a function of the shape and scale parameters. Since you sampled the data by defining the shape and rate parameters, it is simpler to maintain consistency. See https://en.wikipedia.org/wiki/Gamma_distribution

Second, the specification of the function for optim is incorrect; the documentation explicitly mentions that in the function "first argument the vector of parameters over which minimization is to take place". In the code below, I left the data as a global variable that is accessed by the function to be minimized.

Third, in such a scenario it is advisable to enforce the constraints on the two parameters being estimated, otherwise the fitting may fail in some cases, depending on the sampled data.

Finally, the calculation of the asymptotic variance is incorrect: you need to invert the Fisher's information matrix; see, e.g., https://stats.stackexchange.com/questions/68080/basic-question-about-fisher-information-matrix-and-relationship-to-hessian-and-s

Here is the code:

# Data
x <- rgamma(100, shape = 5, rate = 5)

# Minus log likelihood function
minus_log_L_fun <- function(params) {
    a <- params[1] # shape
    b <- params[2] # rate (1 / scale)

    n <- length(x) # sample size

    log_L <- (a - 1) * sum(log(x)) - b * sum(x) + n * a * log(b) - n * log(gamma(a))    
    return (-log_L)
}

# Impose constraints on the estimates of the shape and rate parameters: both being strictly positive
# Use algorithm 'L-BFGS-B', since it allows for box constraints
mle <- optim(par = c(1, 1), minus_log_L_fun, method = "L-BFGS-B", lower = c(1e-10, 1e-10), upper = c(100, 100), hessian = TRUE)

# Retrieve the point estimates
shape_fit <- mle$par[1]
rate_fit <- mle$par[2]

# Fisher Information Matrix (equal to the Hessian, since minimizing minus log likelihood)
I <- mle$hessian
# Obtain the asymptotic variances (need to invert the FIM)
var_Theta <- diag(solve(I))

cat(sprintf("Results:\n"))
cat(sprintf("Point estimates: shape = %g, rate = %g\n", shape_fit,     rate_fit))
cat(sprintf("Asymptotic ML variances: shape = %g, rate = %g\n", var_Theta[1], var_Theta[2]))

producing, for a particular run:

Results:
Point estimates: shape = 4.25661, rate = 4.08384
Asymptotic ML variances: shape = 0.336318, rate = 0.34875

Using the MASS package to confirm:

library(MASS)

res <- fitdistr(x, "gamma", start = list(shape = 1, rate = 1))
MASS_PE <- res$estimate
MASS_AV <- (res$sd)^2

cat(sprintf("From the MASS package:\n"))
cat(sprintf("Point estimates:\n"))
print(MASS_PE)
cat(sprintf("Asymptotic variances:\n"))
print(MASS_AV)

leads to:

From the MASS package:
Point estimates:
   shape     rate 
4.256613 4.083836 
Asymptotic variances:
> print(MASS_AV)
    shape      rate 
0.3363179 0.3487502

Answer 2

@tguzella的回答是非常彻底和教学的。为方便起见，您也可以这样做：

set.seed(101)
x <- rgamma(100,shape=5,rate=5)
library("bbmle")
m1 <- mle2(x~dgamma(shape,rate=rate),
           start=list(shape=2,rate=2), ## anything reasonable
           data=data.frame(x)  ## data must be specified as a data frame
      )

我们收到一条警告信息，在这种情况下无害，但我们可以通过以下方式减轻：

m2 <- update(m1,lower=c(0,0),
             method="L-BFGS-B")

现在我们可以轻松得到点估计和渐近方差 - 协方差矩阵：

coef(m2)
vcov(m2)

注意：

bbmle::mle2是stats4::mle的扩展，也应该适用于此问题（mle2有一些额外的花里胡哨，并且更加健壮），尽管你会必须将对数似然函数定义为：

 nLL <- function(shape,rate) {
     -sum(dgamma(x,shape,rate=rate,log=TRUE))
 }

一般情况下，当dgamma()适用于您的问题时，使用内置分发功能（例如http://www.example.com/file.xxx）是一个好主意;它们经过了充分测试，功能强大，并且使代码更易于阅读。

R中具有最优似然估计的渐近方差

2 个答案: