我想知道,为什么当我直接提供
q
(
看见
f2
)
uniroot()
非常好,但当我提供
Q
作为其他输入值的函数
单根()
(
看见
f1
)失败?
在代码中,所有
...1
后缀(例如,
F1
)与我间接提供
Q
. 所有的一切
...2
后缀(例如,
F2
)与我直接提供
Q
.
我的目标是解决
df2
这样的话
y = .15
(正确答案是
~ 336.3956
)(
)
alpha = c(.025, .975); df1 = 3; q = 48.05649 ; peta = .3 # input values
f1 <- function(alpha, q, df1, df2, ncp){ # Objective function (`q` indirectly)
alpha - suppressWarnings(pf(q = (peta / df1) / ((1 - peta)/df2), df1, df2,
ncp, lower.tail = FALSE))
}
f2 <- function(alpha, q, df1, df2, ncp){ # Objective function (`q` directly)
alpha - suppressWarnings(pf(q = q, df1, df2, ncp, lower.tail = FALSE))
}
ncp1 <- function(df2){ # root finding
b <- sapply(c(alpha[1], alpha[2]),
function(x) uniroot(f1, c(0, 1e7), alpha = x, q = peta, df1 = df1, df2 = df2)[[1]])
b / (b + (df2 + 4))
}
ncp2 <- function(df2){ # root finding
b <- sapply(c(alpha[1], alpha[2]),
function(x) uniroot(f2, c(0, 1e7), alpha = x, q = q, df1 = df1, df2 = df2)[[1]])
b / (b + (df2 + 4))
}
m1 <- function(df2, y){ # A Utility function
abs(abs(diff(ncp1(df2))) - y)
}
m2 <- function(df2, y){ # A Utility function
abs(abs(diff(ncp2(df2))) - y)
}
optimize(m1, c(1, 1e7), y = .15)[[1]] # Incorrect answer: 1e+07
optimize(m2, c(1, 1e7), y = .15)[[1]] # Correct answer: 336.3956
