[R] Hessian matrix issue
Yes, numDeriv::hessian is very accurate. So, I normally take the output from
the optimizer, *if it is a local optimum*, and then apply numDeriv::hessian to
it. I, then, compute the standard errors from it.
However, it is important to know that you have obtained a local optimum. In
fact, you need the gradient and Hessian to actually verify this - the first and
second order KKT conditions. However, this is tricky in *constrained*
optimization problems. If you have constraints, and at least one of the
constraints is active at the best parameter estimate, then the gradient of the
original objective function need not be close to zero and the hessian is also
incorrect.
If you employ an augmented Lagrangian approach (see alabama::auglag), then you
can obtain the correct gradient and Hessian at best parameter estimate. These
gradients and Hessian correspond to the modified objective function that
includes a Lagrangian term augmented by a quadratic penalty term. They can be
used to check the KKT conditions.
Here is a simple example:
require(alabama)
fr - function(x) { ## Rosenbrock Banana function
x1 - x[1]
x2 - x[2]
100 * (x2 - x1 * x1)^2 + (1 - x1)^2
}
grr - function(x) { ## Gradient of 'fr'
x1 - x[1]
x2 - x[2]
c(-400 * x1 * (x2 - x1 * x1) - 2 * (1 - x1),
200 * (x2 - x1 * x1))
}
p0 - c(0, 0)
ans1 - optim(par=p0, fn=fr, gr=grr, upper=c(0.9, Inf), method=L-BFGS-B,
hessian=TRUE)
grr(ans1$par) # this will not be close to zero
ans1$hessian
# Using an augmented Lagrangian optimizer
hcon - function(x) 0.9 - x[1]
hcon.jac - function(x) matrix(c(-1, 0) , 1, 2)
ans2 - auglag(par=p0, fn=fr, gr=grr, hin=hcon, hin.jac=hcon.jac)
ans2$gradient # this will be close to zero
ans2$hessian
However, it is not known whether the standard errors obtained from this Hessian
are asymptotically valid.
Best,
Ravi.
---
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins
University
Ph. (410) 502-2619
email: rvarad...@jhmi.edumailto:rvarad...@jhmi.edu
[[alternative HTML version deleted]]
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Re: [R] Hessian matrix issue
However, it is not known whether the standard errors obtained from this
Hessian are asymptotically valid.
Let me rephrase this. I think that as a measure of dispersion, the standard
error obtained using the augmented Lagrangian algorithm is correct. However,
what is *not known* is the asymptotic distribution of the parameter estimates
when constraints are active. This is a non-regular situation where MLEs
might have strange asymptotic behavior. We cannot generally assume normality
and use the standard error estimates to construct confidence intervals or
calculate significance levels.
Best,
Ravi.
---
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins
University
Ph. (410) 502-2619
email: rvarad...@jhmi.edumailto:rvarad...@jhmi.edu
From: Ravi Varadhan
Sent: Wednesday, September 07, 2011 1:37 PM
To: 'gpet...@uark.edu'; 'nas...@uottawa.ca'; 'r-help@r-project.org'
Subject: Hessian matrix issue
Yes, numDeriv::hessian is very accurate. So, I normally take the output from
the optimizer, *if it is a local optimum*, and then apply numDeriv::hessian to
it. I, then, compute the standard errors from it.
However, it is important to know that you have obtained a local optimum. In
fact, you need the gradient and Hessian to actually verify this - the first and
second order KKT conditions. However, this is tricky in *constrained*
optimization problems. If you have constraints, and at least one of the
constraints is active at the best parameter estimate, then the gradient of the
original objective function need not be close to zero and the hessian is also
incorrect.
If you employ an augmented Lagrangian approach (see alabama::auglag), then you
can obtain the correct gradient and Hessian at best parameter estimate. These
gradients and Hessian correspond to the modified objective function that
includes a Lagrangian term augmented by a quadratic penalty term. They can be
used to check the KKT conditions.
Here is a simple example:
require(alabama)
fr - function(x) { ## Rosenbrock Banana function
x1 - x[1]
x2 - x[2]
100 * (x2 - x1 * x1)^2 + (1 - x1)^2
}
grr - function(x) { ## Gradient of 'fr'
x1 - x[1]
x2 - x[2]
c(-400 * x1 * (x2 - x1 * x1) - 2 * (1 - x1),
200 * (x2 - x1 * x1))
}
p0 - c(0, 0)
ans1 - optim(par=p0, fn=fr, gr=grr, upper=c(0.9, Inf), method=L-BFGS-B,
hessian=TRUE)
grr(ans1$par) # this will not be close to zero
ans1$hessian
# Using an augmented Lagrangian optimizer
hcon - function(x) 0.9 - x[1]
hcon.jac - function(x) matrix(c(-1, 0) , 1, 2)
ans2 - auglag(par=p0, fn=fr, gr=grr, hin=hcon, hin.jac=hcon.jac)
ans2$gradient # this will be close to zero
ans2$hessian
However, it is not known whether the standard errors obtained from this Hessian
are asymptotically valid.
Best,
Ravi.
---
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins
University
Ph. (410) 502-2619
email: rvarad...@jhmi.edumailto:rvarad...@jhmi.edu
[[alternative HTML version deleted]]
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Re: [R] Hessian Matrix Issue
On 09/03/2011 08:22 PM, dave fournier wrote:
I wonder if your code is correct?
I ran your script until an error was reported. the data set
of 30 obs was
[1] 0 0 1 3 3 3 4 4 4 4 5 5 5 5 5 7 7 7 7 7 7 8
9 10 11
[26] 12 12 12 15 16
I created a tiny AD Model Builder program to do MLE on it.
DATA_SECTION
init_int nobs
init_vector y(1,nobs)
PARAMETER_SECTION
init_number log_mu
init_number log_alpha
sdreport_number mu
sdreport_number tau
objective_function_value f
PROCEDURE_SECTION
mu=exp(log_mu);
tau=1.0+exp(log_alpha);
for (int i=1;i=nobs;i++)
{
f-=log_negbinomial_density(y(i),mu,tau);
}
It converged quickly and
The eigenvalues of the Hessian were
4.71108977478.27632341
and the estimates and std devs of the parameters mu and tau were
index name value std dev
3 mu 6.6000e+00 7.7318e-01
4 tau2.7173e+00 7.8944e-01
where tau is the variance divided by the mean.
This was all so simple that I suspect your (rather difficult to read)
R code is wrong, otherwise R must really suck at this kind of problem.
I'd put my money on R!
Paul
Dave
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Global Climate Division
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Wilhelminalaan 10 | 3732 GK | De Bilt | Kamer B 3.39
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Re: [R] Hessian matrix issue
About the numerical calculation of the Hessian matrix, I have found
numDeriv:::hessian to be often more accurate than the Hessian returned
by optim.
Best,
Giovanni Petris
On Sat, 2011-09-03 at 13:39 -0400, John C Nash wrote:
Unless you are supplying analytic hessian code, you are almost certainly
getting an
approximation. Worse, if you do not provide gradients, these are the result
of two levels
of differencing, so you should expect some loss of precision in the
approximate Hessian.
Moreover, if your estimate of the optimum is a little bit off, or the
optimizer has
terminated (algorithms converge, programs terminate) to a point that is not
an optimum,
there is no reason the Hessian should be positive definite.
Package optimx() uses the Jacobian of the gradient if the analytic gradient
is available.
This drops the differencing to 1 level. Even better is to code the Hessian,
but that is
messy and tedious in most cases.
Best, JN
On 09/03/2011 06:00 AM, r-help-requ...@r-project.org wrote:
Message: 59
Date: Fri, 2 Sep 2011 15:33:13 -0400
From: tzai...@alcor.concordia.ca
To: r-help@r-project.org
Subject: [R] Hessian Matrix Issue
Message-ID:
e6dc43b4487eb4a4055e1ab485f015f0.squir...@webmail.concordia.ca
Content-Type: text/plain;charset=iso-8859-1
Dear All,
I am running a simulation to obtain coverage probability of Wald type
confidence intervals for my parameter d in a function of two parameters
(mu,d).
I am optimizing it using optim method L-BFGS-B to obtain MLE. As, I
want to invert the Hessian matrix to get Standard errors of the two
parameter estimates. However, my Hessian matrix at times becomes
non-invertible that is it is no more positive definite and I get the
following error msg:
Error in solve.default(ac$hessian) : system is computationally singular:
reciprocal condition number = 6.89585e-21
Thank you
Following is the code I am running I would really appreciate your comments
and suggestions:
#Start Code
#option to trace /recover error
#options(error = recover)
#Sample Size
n-30
mu-5
size- 2
#true values of parameter d
d.true-1+mu/size
d.true
#true value of zero inflation index phi= 1+log(d)/(1-d)
z.true-1+(log(d.true)/(1-d.true))
z.true
# Allocating space for simulation vectors and setting counters for
simulation
counter-0
iter-1
lower.d-numeric(iter)
upper.d-numeric(iter)
#set.seed(987654321)
#begining of simulation loop
for (i in 1:iter){
r.NB-rnbinom(n, mu = mu, size = size)
y-sort(r.NB)
iter.num-i
print(y)
print(iter.num)
#empirical estimates or sample moments
xbar-mean(y)
variance-(sum((y-xbar)2))/length(y)
dbar-variance/xbar
#sample estimate of proportion of zeros and zero inflation index
pbar-length(y[y==0])/length(y)
### Simplified function #
NegBin-function(th){
mu-th[1]
d-th[2]
n-length(y)
arg1-n*mean(y)*ifelse(mu = 0, log(mu),0)
#arg1-n*mean(y)*log(mu)
#arg2-n*log(d)*((mean(y))+mu/(d-1))
arg2-n*ifelse(d=0, log(d), 0)*((mean(y))+mu/ifelse((d-1)= 0, (d-1),
0.001))
aa-numeric(length(max(y)))
a-numeric(length(y))
for (i in 1:n)
{
for (j in 1:y[i]){
aa[j]-ifelse(((j-1)*(d-1))/mu 0,log(1+((j-1)*(d-1))/mu),0)
#aa[j]-log(1+((j-1)*(d-1))/mu)
#print(aa[j])
}
a[i]-sum(aa)
#print(a[i])
}
a
arg3-sum(a)
llh-arg1+arg2+arg3
if(! is.finite(llh))
llh-1e+20
-llh
}
ac-optim(NegBin,par=c(xbar,dbar),method=L-BFGS-B,hessian=TRUE,lower=
c(0,1) )
ac
print(ac$hessian)
muhat-ac$par[1]
dhat-ac$par[2]
zhat- 1+(log(dhat)/(1-dhat))
infor-solve(ac$hessian)
var.dhat-infor[2,2]
se.dhat-sqrt(var.dhat)
var.muhat-infor[1,1]
se.muhat-sqrt(var.muhat)
var.func-dhat*muhat
var.func
d.prime-cbind(dhat,muhat)
se.var.func-d.prime%*%infor%*%t(d.prime)
se.var.func
lower.d[i]-dhat-1.96*se.dhat
upper.d[i]-dhat+1.96*se.dhat
if(lower.d[i] = d.true d.true= upper.d[i])
counter -counter+1
}
counter
covg.prob-counter/iter
covg.prob
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Hessian Matrix Issue
Uwe Ligges ligges at statistik.tu-dortmund.de writes: I have not really looked into the details of the lengthy and almost unreadable code below. In any case, there are good reasons why numerics software typically uses Fisher scoring / IWLS in order to fit GLMs. And if your matrix is that singular, even the common numerical tricks may not save the day anymore. 7e-21 is very close to exact singularity! Uwe Ligges Your problem is with the strategy you use to try to deal with non-finite values, i.e. setting the negative log-likelihood to 10^20 if the calculated values are not finite. What happens is that, rather than just pushing the optimization away from a bad value, you get stuck there, which leads to a solution to the optimization, which is completely flat (because the objective function is 1e20 for any value near the solution), which leads to an uninvertible hessian. More specifically, inserting a browser() call at the point after the if (!is.finite()) call and inspecting the results when the objective function is not finite shows that when d=1 the ifelse((d-1)=0, ...) clause returns (d-1) as a denominator ... Beyond that, I can't spend any more time picking through this ... Ben Bolker __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Hessian Matrix Issue
I have not really looked into the details of the lengthy and almost
unreadable code below. In any case, there are good reasons why numerics
software typically uses Fisher scoring / IWLS in order to fit GLMs.
And if your matrix is that singular, even the common numerical tricks
may not save the day anymore. 7e-21 is very close to exact singularity!
Uwe Ligges
On 02.09.2011 21:33, tzai...@alcor.concordia.ca wrote:
Dear All,
I am running a simulation to obtain coverage probability of Wald type
confidence intervals for my parameter d in a function of two parameters
(mu,d).
I am optimizing it using optim method L-BFGS-B to obtain MLE. As, I
want to invert the Hessian matrix to get Standard errors of the two
parameter estimates. However, my Hessian matrix at times becomes
non-invertible that is it is no more positive definite and I get the
following error msg:
Error in solve.default(ac$hessian) : system is computationally singular:
reciprocal condition number = 6.89585e-21
Thank you
Following is the code I am running I would really appreciate your comments
and suggestions:
#Start Code
#option to trace /recover error
#options(error = recover)
#Sample Size
n-30
mu-5
size- 2
#true values of parameter d
d.true-1+mu/size
d.true
#true value of zero inflation index phi= 1+log(d)/(1-d)
z.true-1+(log(d.true)/(1-d.true))
z.true
# Allocating space for simulation vectors and setting counters for simulation
counter-0
iter-1
lower.d-numeric(iter)
upper.d-numeric(iter)
#set.seed(987654321)
#begining of simulation loop
for (i in 1:iter){
r.NB-rnbinom(n, mu = mu, size = size)
y-sort(r.NB)
iter.num-i
print(y)
print(iter.num)
#empirical estimates or sample moments
xbar-mean(y)
variance-(sum((y-xbar)^2))/length(y)
dbar-variance/xbar
#sample estimate of proportion of zeros and zero inflation index
pbar-length(y[y==0])/length(y)
### Simplified function #
NegBin-function(th){
mu-th[1]
d-th[2]
n-length(y)
arg1-n*mean(y)*ifelse(mu= 0, log(mu),0)
#arg1-n*mean(y)*log(mu)
#arg2-n*log(d)*((mean(y))+mu/(d-1))
arg2-n*ifelse(d=0, log(d), 0)*((mean(y))+mu/ifelse((d-1)= 0, (d-1),
0.001))
aa-numeric(length(max(y)))
a-numeric(length(y))
for (i in 1:n)
{
for (j in 1:y[i]){
aa[j]-ifelse(((j-1)*(d-1))/mu0,log(1+((j-1)*(d-1))/mu),0)
#aa[j]-log(1+((j-1)*(d-1))/mu)
#print(aa[j])
}
a[i]-sum(aa)
#print(a[i])
}
a
arg3-sum(a)
llh-arg1+arg2+arg3
if(! is.finite(llh))
llh-1e+20
-llh
}
ac-optim(NegBin,par=c(xbar,dbar),method=L-BFGS-B,hessian=TRUE,lower=
c(0,1) )
ac
print(ac$hessian)
muhat-ac$par[1]
dhat-ac$par[2]
zhat- 1+(log(dhat)/(1-dhat))
infor-solve(ac$hessian)
var.dhat-infor[2,2]
se.dhat-sqrt(var.dhat)
var.muhat-infor[1,1]
se.muhat-sqrt(var.muhat)
var.func-dhat*muhat
var.func
d.prime-cbind(dhat,muhat)
se.var.func-d.prime%*%infor%*%t(d.prime)
se.var.func
lower.d[i]-dhat-1.96*se.dhat
upper.d[i]-dhat+1.96*se.dhat
if(lower.d[i]= d.true d.true= upper.d[i])
counter-counter+1
}
counter
covg.prob-counter/iter
covg.prob
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Hessian matrix issue
Unless you are supplying analytic hessian code, you are almost certainly
getting an
approximation. Worse, if you do not provide gradients, these are the result of
two levels
of differencing, so you should expect some loss of precision in the approximate
Hessian.
Moreover, if your estimate of the optimum is a little bit off, or the optimizer
has
terminated (algorithms converge, programs terminate) to a point that is not an
optimum,
there is no reason the Hessian should be positive definite.
Package optimx() uses the Jacobian of the gradient if the analytic gradient is
available.
This drops the differencing to 1 level. Even better is to code the Hessian, but
that is
messy and tedious in most cases.
Best, JN
On 09/03/2011 06:00 AM, r-help-requ...@r-project.org wrote:
Message: 59
Date: Fri, 2 Sep 2011 15:33:13 -0400
From: tzai...@alcor.concordia.ca
To: r-help@r-project.org
Subject: [R] Hessian Matrix Issue
Message-ID:
e6dc43b4487eb4a4055e1ab485f015f0.squir...@webmail.concordia.ca
Content-Type: text/plain;charset=iso-8859-1
Dear All,
I am running a simulation to obtain coverage probability of Wald type
confidence intervals for my parameter d in a function of two parameters
(mu,d).
I am optimizing it using optim method L-BFGS-B to obtain MLE. As, I
want to invert the Hessian matrix to get Standard errors of the two
parameter estimates. However, my Hessian matrix at times becomes
non-invertible that is it is no more positive definite and I get the
following error msg:
Error in solve.default(ac$hessian) : system is computationally singular:
reciprocal condition number = 6.89585e-21
Thank you
Following is the code I am running I would really appreciate your comments
and suggestions:
#Start Code
#option to trace /recover error
#options(error = recover)
#Sample Size
n-30
mu-5
size- 2
#true values of parameter d
d.true-1+mu/size
d.true
#true value of zero inflation index phi= 1+log(d)/(1-d)
z.true-1+(log(d.true)/(1-d.true))
z.true
# Allocating space for simulation vectors and setting counters for simulation
counter-0
iter-1
lower.d-numeric(iter)
upper.d-numeric(iter)
#set.seed(987654321)
#begining of simulation loop
for (i in 1:iter){
r.NB-rnbinom(n, mu = mu, size = size)
y-sort(r.NB)
iter.num-i
print(y)
print(iter.num)
#empirical estimates or sample moments
xbar-mean(y)
variance-(sum((y-xbar)2))/length(y)
dbar-variance/xbar
#sample estimate of proportion of zeros and zero inflation index
pbar-length(y[y==0])/length(y)
### Simplified function #
NegBin-function(th){
mu-th[1]
d-th[2]
n-length(y)
arg1-n*mean(y)*ifelse(mu = 0, log(mu),0)
#arg1-n*mean(y)*log(mu)
#arg2-n*log(d)*((mean(y))+mu/(d-1))
arg2-n*ifelse(d=0, log(d), 0)*((mean(y))+mu/ifelse((d-1)= 0, (d-1),
0.001))
aa-numeric(length(max(y)))
a-numeric(length(y))
for (i in 1:n)
{
for (j in 1:y[i]){
aa[j]-ifelse(((j-1)*(d-1))/mu 0,log(1+((j-1)*(d-1))/mu),0)
#aa[j]-log(1+((j-1)*(d-1))/mu)
#print(aa[j])
}
a[i]-sum(aa)
#print(a[i])
}
a
arg3-sum(a)
llh-arg1+arg2+arg3
if(! is.finite(llh))
llh-1e+20
-llh
}
ac-optim(NegBin,par=c(xbar,dbar),method=L-BFGS-B,hessian=TRUE,lower=
c(0,1) )
ac
print(ac$hessian)
muhat-ac$par[1]
dhat-ac$par[2]
zhat- 1+(log(dhat)/(1-dhat))
infor-solve(ac$hessian)
var.dhat-infor[2,2]
se.dhat-sqrt(var.dhat)
var.muhat-infor[1,1]
se.muhat-sqrt(var.muhat)
var.func-dhat*muhat
var.func
d.prime-cbind(dhat,muhat)
se.var.func-d.prime%*%infor%*%t(d.prime)
se.var.func
lower.d[i]-dhat-1.96*se.dhat
upper.d[i]-dhat+1.96*se.dhat
if(lower.d[i] = d.true d.true= upper.d[i])
counter -counter+1
}
counter
covg.prob-counter/iter
covg.prob
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Hessian Matrix Issue
I wonder if your code is correct?
I ran your script until an error was reported. the data set
of 30 obs was
[1] 0 0 1 3 3 3 4 4 4 4 5 5 5 5 5 7 7 7 7 7 7 8 9
10 11
[26] 12 12 12 15 16
I created a tiny AD Model Builder program to do MLE on it.
DATA_SECTION
init_int nobs
init_vector y(1,nobs)
PARAMETER_SECTION
init_number log_mu
init_number log_alpha
sdreport_number mu
sdreport_number tau
objective_function_value f
PROCEDURE_SECTION
mu=exp(log_mu);
tau=1.0+exp(log_alpha);
for (int i=1;i=nobs;i++)
{
f-=log_negbinomial_density(y(i),mu,tau);
}
It converged quickly and
The eigenvalues of the Hessian were
4.71108977478.27632341
and the estimates and std devs of the parameters mu and tau were
index name value std dev
3 mu 6.6000e+00 7.7318e-01
4 tau2.7173e+00 7.8944e-01
where tau is the variance divided by the mean.
This was all so simple that I suspect your (rather difficult to read)
R code is wrong, otherwise R must really suck at this kind of problem.
Dave
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Hessian Matrix Issue
Dear All,
I am running a simulation to obtain coverage probability of Wald type
confidence intervals for my parameter d in a function of two parameters
(mu,d).
I am optimizing it using optim method L-BFGS-B to obtain MLE. As, I
want to invert the Hessian matrix to get Standard errors of the two
parameter estimates. However, my Hessian matrix at times becomes
non-invertible that is it is no more positive definite and I get the
following error msg:
Error in solve.default(ac$hessian) : system is computationally singular:
reciprocal condition number = 6.89585e-21
Thank you
Following is the code I am running I would really appreciate your comments
and suggestions:
#Start Code
#option to trace /recover error
#options(error = recover)
#Sample Size
n-30
mu-5
size- 2
#true values of parameter d
d.true-1+mu/size
d.true
#true value of zero inflation index phi= 1+log(d)/(1-d)
z.true-1+(log(d.true)/(1-d.true))
z.true
# Allocating space for simulation vectors and setting counters for simulation
counter-0
iter-1
lower.d-numeric(iter)
upper.d-numeric(iter)
#set.seed(987654321)
#begining of simulation loop
for (i in 1:iter){
r.NB-rnbinom(n, mu = mu, size = size)
y-sort(r.NB)
iter.num-i
print(y)
print(iter.num)
#empirical estimates or sample moments
xbar-mean(y)
variance-(sum((y-xbar)^2))/length(y)
dbar-variance/xbar
#sample estimate of proportion of zeros and zero inflation index
pbar-length(y[y==0])/length(y)
### Simplified function #
NegBin-function(th){
mu-th[1]
d-th[2]
n-length(y)
arg1-n*mean(y)*ifelse(mu = 0, log(mu),0)
#arg1-n*mean(y)*log(mu)
#arg2-n*log(d)*((mean(y))+mu/(d-1))
arg2-n*ifelse(d=0, log(d), 0)*((mean(y))+mu/ifelse((d-1)= 0, (d-1),
0.001))
aa-numeric(length(max(y)))
a-numeric(length(y))
for (i in 1:n)
{
for (j in 1:y[i]){
aa[j]-ifelse(((j-1)*(d-1))/mu 0,log(1+((j-1)*(d-1))/mu),0)
#aa[j]-log(1+((j-1)*(d-1))/mu)
#print(aa[j])
}
a[i]-sum(aa)
#print(a[i])
}
a
arg3-sum(a)
llh-arg1+arg2+arg3
if(! is.finite(llh))
llh-1e+20
-llh
}
ac-optim(NegBin,par=c(xbar,dbar),method=L-BFGS-B,hessian=TRUE,lower=
c(0,1) )
ac
print(ac$hessian)
muhat-ac$par[1]
dhat-ac$par[2]
zhat- 1+(log(dhat)/(1-dhat))
infor-solve(ac$hessian)
var.dhat-infor[2,2]
se.dhat-sqrt(var.dhat)
var.muhat-infor[1,1]
se.muhat-sqrt(var.muhat)
var.func-dhat*muhat
var.func
d.prime-cbind(dhat,muhat)
se.var.func-d.prime%*%infor%*%t(d.prime)
se.var.func
lower.d[i]-dhat-1.96*se.dhat
upper.d[i]-dhat+1.96*se.dhat
if(lower.d[i] = d.true d.true= upper.d[i])
counter -counter+1
}
counter
covg.prob-counter/iter
covg.prob
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