The final Hessian matrix is full rank but has at least one negative eigenvalue. Second-order optima

rizayuli — Sat, 10 Dec 2016 05:35:49 GMT

please help me to explain my syntax.

data fish;
input count x1 x2 x3 x4;
cards;
0 83.74 54.98 80.53 1.96
0 76.36 18.45 51.33 0.3
3 76.29 56.16 70.95 1.93
0 79.32 40.94 75.68 1.34
2 95.77 87.68 83.12 1.36
1 69.27 76.85 46.16 0.71
0 93.02 58.56 89.01 0.87
0 94.42 50.06 91.5 1.55
1 98.93 91.2 97.6 3.71
0 84.4 62.04 79.87 1.56
1 98.71 50.72 97.11 1.07
0 93.31 93.5 86.73 0.81
0 100.06 76.08 98.36 4.14
0 80.71 76.57 77.14 2
21 100.41 79.48 97.81 21.62
12 90.09 73.14 82.95 1.66
0 100.32 67.6 74.45 4.48
0 95.82 84.82 93.94 1.3
0 41.92 24.11 39.17 0.62
5 75.03 46.2 72.53 1.79
2 81 38.01 78.63 0.4
0 83.05 46.8 80.6 0.73
0 90.44 25.15 82.14 1.07
0 75.45 45.28 70.81 0.21
0 51.9 63.07 41.18 1.55
1 10.51 56.22 10.14 0.55
1 77.22 54.04 74.28 1.12
0 71.46 81.42 63.99 0.43
0 73.56 51.57 69.7 0.78
2 44.69 72.74 42.6 3.32
0 51.02 59.02 47.2 0.94
0 20.4 36.19 18.55 2.13
1 14.95 1.04 9.63 0.03
;
data fish;
set fish;
bound=1;
if count > bound then count=bound+1; * This is probably
how you would see the data if it was actually censored;
proc nlmixed TECH=NRRIDG; *untuk variabel KN1 TT2+ KNL TT5;
parms a0=0 a1=0 a2=0 a3=0 a4=0 b0=0 b1=0 b2=0 b3=0 b4=0 alpha=0.5;
bounds alpha>0;
lin = a0 + a1*x1 + a2*x2 + a3*x3 + a4*x4;
w = exp(lin)/(1+exp(lin));
eta = b0 + b1*x1 + b2*x2 + b3*x3 + b4*x4;
mu = exp(eta);
phi=1/alpha;
pdf=(gamma(count+phi)/(gamma(count+1)*gamma(phi)))
*((1/(1+alpha*mu))**phi*(alpha*mu/(1+alpha*mu))** count);
l_1 = w;
l_2 = (1-w) * pdf / (1-(1+alpha*mu)**(-phi));
cdf=0;
do t=1 to bound;
cdf=cdf+((1-w)*((gamma(t+phi)/(gamma(t+1)*gamma(phi)))
*((1/(1+alpha*mu))**phi*(alpha*mu/(1+alpha*mu))**t)/(1-(1+alpha*mu)**(-phi))));
end;
l_3= 1-cdf;
if count = 0 then ll = log(l_1);
if 0 < count <= bound then ll = log(l_2);
if count <= bound then d=0; else d=1;
ll=(1-d)*ll+d*log(l_3);
model count~general(ll);
run;

but there is warning:

NOTE: GCONV convergence criterion satisfied.
NOTE: At least one element of the (projected) gradient is greater than 1e-3.
WARNING: The final Hessian matrix is full rank but has at least one negative eigenvalue. Second-order
optimality condition violated.
NOTE: PROCEDURE NLMIXED used (Total process time):
real time 0.28 seconds
cpu time 0.26 seconds

Re: The final Hessian matrix is full rank but has at least one negative eigenvalue. Second-order op

sld — Mon, 12 Dec 2016 05:52:29 GMT

Google can be your friend here: just search using the text of the error message.

If you Google "The final Hessian matrix is full rank but has at least one negative eigenvalue", you find these links:

http://tinyurl.com/z7s4q4g

http://tinyurl.com/gorokea

Another useful resource is this paper

https://support.sas.com/resources/papers/proceedings15/SAS1919-2015.pdf

Hopefully one of these will suggest a solution for your problem. Good luck!

Edited: Also, if you would provide a description of your study and and your model, someone on the list might be able to help you assess whether you have a correct model.

topic Re: The final Hessian matrix is full rank but has at least one negative eigenvalue. Second-order op in Statistical Procedures

The final Hessian matrix is full rank but has at least one negative eigenvalue. Second-order optima

Re: The final Hessian matrix is full rank but has at least one negative eigenvalue. Second-order op