我正在尝试重现David Ruppert的“金融工程统计和数据分析”中的以下示例,该示例使学生t分布符合历史风险免费率:
library(MASS)
data(Capm, package = "Ecdat")
x <- Capm$rf
fitt <- fitdistr(x,"t", start = list(m=mean(x),s=sd(x)), df=3)
as.numeric(fitt$estimate)
0.437310595161651 0.152205764779349
输出附带以下警告信息:
警告讯息:
In log(s): NaNs producedWarning message:
In log(s): NaNs producedWarning message:
In log(s): NaNs producedWarning message:
In log(s): NaNs producedWarning message:
In log(s): NaNs producedWarning message:
In log(s): NaNs producedWarning message:
In log(s): NaNs producedWarning message:
In log(s): NaNs producedWarning message:
In log(s): NaNs produced
从R的帮助文件中可以看出MASS::fitdistr
使用最大可能性来查找最佳参数。但是,当我手动进行优化(同一本书)时,一切顺利,并且没有警告:
library(fGarch)
loglik_t <- function(beta) {sum( - dt((x - beta[1]) / beta[2],
beta[3], log = TRUE) + log(beta[2]) )}
start <- c(mean(x), sd(x), 5)
lower <- c(-1, 0.001, 1)
fit_t <- optim(start, loglik_t, hessian = T, method = "L-BFGS-B", lower = lower)
fit_t$par
0.44232633269102 0.163306955396773 4.12343777572566
拟合参数在可接受的标准误差范围内,除了均值和sd之外,我得到了df
。
有人可以告诉我:
答案 0 :(得分:2)
您没有将lower
参数传递给函数fitdistr
,这导致它在正域和负域中进行搜索。通过将lower
参数传递给函数
fitt <- fitdistr(x,"t", start = list(m=mean(x),s=sd(x)), df=3, lower=c(-1, 0.001))
您没有NaNs - 就像您在手动优化中所做的那样。
编辑:
fitt <- fitdistr(x,"t", start = list(m=mean(x),s=sd(x),df=3),lower=c(-1, 0.001,1))
返回非整数自由度结果。但是,我想,它的舍入值round(fitt$estimate['df'],0)
可以用于拟合自由度参数。