R函数产生NAN，诊断。STAT：：Optim
我运行这段代码是为了计算零膨胀模型中贝塔二项分布的MLE。 然而，对于一个特定的数据，我得到了一些错误。你能帮帮我吗？ 以下是R代码：R函数产生NAN，诊断。STAT：：Optim,r,diagnostics,R,Diagnostics,我运行这段代码是为了计算零膨胀模型中贝塔二项分布的MLE。 然而，对于一个特定的数据，我得到了一些错误。你能帮帮我吗？ 以下是R代码： type = "zi"; lowerbound = 0.01; upperbound = 10000; n=27.59554;alpha1=19.22183;alpha2=41.90441; x=c(9, 12, 13, 11, 12, 0, 11, 0, 11, 0, 7, 12, 0, 11, 12, 0,
type = "zi"; lowerbound = 0.01; upperbound = 10000;
n=27.59554;alpha1=19.22183;alpha2=41.90441;
x=c(9, 12, 13, 11, 12, 0, 11, 0, 11, 0, 7, 12, 0, 11, 12, 0, 6, 2, 6, 2, 0, 0,
9, 10, 10, 0, 0, 0, 10, 11)
N = length(x)
t = x[x > 0]
m = length(t)
neg.log.lik <- function(y)
{
n1 = y[1]
a1 = y[2]
b1 = y[3]
logA = lgamma(a1 + n1 + b1) + lgamma(b1)
logB = lgamma(a1 + b1) + lgamma(n1 + b1)
ans = m * log(1 - exp(logB - logA)) + m * logA - m *
lgamma(n1 + 1) - sum(lgamma(t + a1)) - sum(lgamma(n1 -
t + b1)) - m * lgamma(a1 + b1) + sum(lgamma(t + 1)) +
sum(lgamma(n1 - t + 1)) + m * lgamma(a1)
return(ans)
}
gp <- function(y)
{
#n1=27.59554;a1=19.22183;b1=41.90441;
n1 = y[1]
a1 = y[2]
b1 = y[3]
logA = lgamma(a1 + n1 + b1) + lgamma(b1)
logB = lgamma(a1 + b1) + lgamma(n1 + b1)
dn = -m * exp(logB - logA) * (digamma(n1 + b1) - digamma(a1 +
n1 + b1))/(1 - exp(logB - logA)) - m * digamma(n1 +
1) - sum(digamma(n1 + b1 - t)) + sum(digamma(n1 -
t + 1)) + m * digamma(a1 + n1 + b1)
da = -m * exp(logB - logA) * (digamma(a1 + b1) - digamma(a1 +
n1 + b1))/(1 - exp(logB - logA)) - sum(digamma(t +
a1)) - m * digamma(a1 + b1) + m * digamma(a1 + n1 +
b1) + m * digamma(a1)
db = -m * exp(logB - logA) * (digamma(a1 + b1) + digamma(n1 +
b1) - digamma(a1 + n1 + b1) - digamma(b1))/(1 - exp(logB -
logA)) + m * digamma(b1) - sum(digamma(n1 - t + b1))-
m * digamma(a1 + b1) + m * digamma(a1 + n1 + b1)
return(c(dn, da, db))
}
estimate = stats::optim(par = c(n, alpha1, alpha2), fn = neg.log.lik,
gr = gp, method = "L-BFGS-B", lower = c(max(x) - lowerbound,
lowerbound, lowerbound), upper = c(upperbound, upperbound, upperbound))
type=“zi”；lowerbound=0.01；上限=10000；
n=27.59554；alpha1=19.22183；alpha2=41.90441；
x=c（9,12,13,11,12,0,11,0,11,0,7,12,0,11,12,0,6,2,2,0,0，
9, 10, 10, 0, 0, 0, 10, 11)
N=长度（x）
t=x[x>0]
m=长度（t）
neg.log.lik问题在于，在某个点上，由于digamma（0），dn变为NaN。
一个选择是像我一样考虑到这种可能性。但你应该探索在这种情况下该怎么做。
这里有一个问题，digamma（n1+b1-t）
在某个点上，它会产生digamma（0），所以会失败
type = "zi"; lowerbound = 0.01; upperbound = 10000;
n=27.59554;alpha1=19.22183;alpha2=41.90441;
x=c(9, 12, 13, 11, 12, 0, 11, 0, 11, 0, 7, 12, 0, 11, 12, 0, 6, 2, 6, 2, 0, 0,
9, 10, 10, 0, 0, 0, 10, 11)
N = length(x)
t = x[x > 0]
m = length(t)
neg.log.lik <- function(y)
{
n1 = y[1]
a1 = y[2]
b1 = y[3]
logA = lgamma(a1 + n1 + b1) + lgamma(b1)
logB = lgamma(a1 + b1) + lgamma(n1 + b1)
ans = m * log(1 - exp(logB - logA)) + m * logA - m *
lgamma(n1 + 1) - sum(lgamma(t + a1)) - sum(lgamma(n1 -
t + b1)) - m * lgamma(a1 + b1) + sum(lgamma(t + 1)) +
sum(lgamma(n1 - t + 1)) + m * lgamma(a1)
return(ans)
}
gp <- function(y)
{
#n1=27.59554;a1=19.22183;b1=41.90441;
n1 = y[1]
a1 = y[2]
b1 = y[3]
logA = lgamma(a1 + n1 + b1) + lgamma(b1)
logB = lgamma(a1 + b1) + lgamma(n1 + b1)
dn = -m * exp(logB - logA) * (digamma(n1 + b1) - digamma(a1 +
n1 + b1))/(1 - exp(logB - logA)) - m * digamma(n1 +
1) - sum(digamma(n1 + b1 - t)) + sum(digamma(n1 -
t + 1)) + m * digamma(a1 + n1 + b1)
if(is.na(dn)){
dn=-99999999
}
da = -m * exp(logB - logA) * (digamma(a1 + b1) - digamma(a1 +
n1 + b1))/(1 - exp(logB - logA)) - sum(digamma(t +
a1)) - m * digamma(a1 + b1) + m * digamma(a1 + n1 +
b1) + m * digamma(a1)
db = -m * exp(logB - logA) * (digamma(a1 + b1) + digamma(n1 +
b1) - digamma(a1 + n1 + b1) - digamma(b1))/(1 - exp(logB -
logA)) + m * digamma(b1) - sum(digamma(n1 - t + b1))-
m * digamma(a1 + b1) + m * digamma(a1 + n1 + b1)
print("dn")
print(dn)
print("da")
print(da)
print("db")
print(db)
return(c(dn, da, db))
}
estimate = stats::optim(par = c(n, alpha1, alpha2), fn = neg.log.lik,
gr = gp, method = "L-BFGS-B", lower = c(max(x) - lowerbound,
lowerbound, lowerbound), upper = c(upperbound, upperbound, upperbound))
您需要确保digamma只通过严格的正值。该函数没有为我所做的x定义。t=x[x>0]会这样做。我想不出是什么问题。非常感谢你的回答。我感谢你的帮助。最好的。
type = "zi"; lowerbound = 0.01; upperbound = 10000;
n=27.59554;alpha1=19.22183;alpha2=41.90441;
x=c(9, 12, 13, 11, 12, 0, 11, 0, 11, 0, 7, 12, 0, 11, 12, 0, 6, 2, 6, 2, 0, 0,
9, 10, 10, 0, 0, 0, 10, 11)
N = length(x)
t = x[x > 0]
m = length(t)
neg.log.lik <- function(y)
{
n1 = y[1]
a1 = y[2]
b1 = y[3]
logA = lgamma(a1 + n1 + b1) + lgamma(b1)
logB = lgamma(a1 + b1) + lgamma(n1 + b1)
ans = m * log(1 - exp(logB - logA)) + m * logA - m *
lgamma(n1 + 1) - sum(lgamma(t + a1)) - sum(lgamma(n1 -
t + b1)) - m * lgamma(a1 + b1) + sum(lgamma(t + 1)) +
sum(lgamma(n1 - t + 1)) + m * lgamma(a1)
return(ans)
}
gp <- function(y)
{
#n1=27.59554;a1=19.22183;b1=41.90441;
n1 = y[1]
a1 = y[2]
b1 = y[3]
logA = lgamma(a1 + n1 + b1) + lgamma(b1)
logB = lgamma(a1 + b1) + lgamma(n1 + b1)
dn = -m * exp(logB - logA) * (digamma(n1 + b1) - digamma(a1 +
n1 + b1))/(1 - exp(logB - logA)) - m * digamma(n1 +
1) - sum(digamma(n1 + b1 - t)+0.001) + sum(digamma(n1 -
t + 1)) + m * digamma(a1 + n1 + b1)
da = -m * exp(logB - logA) * (digamma(a1 + b1) - digamma(a1 +
n1 + b1))/(1 - exp(logB - logA)) - sum(digamma(t +
a1)) - m * digamma(a1 + b1) + m * digamma(a1 + n1 +
b1) + m * digamma(a1)
db = -m * exp(logB - logA) * (digamma(a1 + b1) + digamma(n1 +
b1) - digamma(a1 + n1 + b1) - digamma(b1))/(1 - exp(logB -
logA)) + m * digamma(b1) - sum(digamma(n1 - t + b1))-
m * digamma(a1 + b1) + m * digamma(a1 + n1 + b1)
print("dn")
print(dn)
print("da")
print(da)
print("db")
print(db)
return(c(dn, da, db))
}
estimate = stats::optim(par = c(n, alpha1, alpha2), fn = neg.log.lik,
gr = gp, method = "L-BFGS-B", lower = c(max(x) - lowerbound,
lowerbound, lowerbound), upper = c(upperbound, upperbound, upperbound))