Plotting a Weibull density function
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Hello,
i've created a Maximum-Likelihood function and estimated the parameters as you can see right below:
syms A B b y0 real
%Faktorstufen A%
x11=-1;x21=-1;x31=1;x41=1;
x1=[x11;x21;x31;x41];
%Faktorstufen B%
x12=-1;x22=1;x32=-1;x42=1;
x2=[x12;x22;x32;x42];
%Ausfallzeiten%
%A low, B low% %A low, B high% %A high, B low% % A high, B high%
t1=27; t2=50; t3=25; t4=55;
Versuchsdaten=[t1;t2;t3;t4];
%Log.Likelihood-Funktion%
logL=0;
for i=1:length(Versuchsdaten)
logL=logL+log(b/(exp(y0+A*x1(i)+B*x2(i))))+log(Versuchsdaten(i)/(exp(y0+A*x1(i)+B*x2(i))))^(b-1)-(Versuchsdaten(i)/exp(y0+A*x1(i)+B*x2(i)))^(b);
end
%differentielle Ableitungen%
dLdy0=diff(logL,y0)
dLdb=diff(logL,b)
%dLdt0=diff(logL,t0)
dLdA=diff(logL,A)
dLdB=diff(logL,B)
%dLdAB=diff(logL,AB)
%range=[0 Inf; 0 Inf; -1 1; -1 1];
[y0_hat, b_hat, A_hat, B_hat] = vpasolve([dLdy0==0, dLdb==0, dLdA==0, dLdB==0],[y0, b, A, B])
The result for the parameters are:
y0_hat =
2.0121168178642018928072071428139
b_hat =
0.38450126565678579374223747169347
A_hat =
0.0045872846670491337649337697413908
B_hat =
0.35116087494702178847354983047048
Then im calculating the parameter T_hat:
if A_hat<1
x1=-1
else
x1=1
end
if B_hat<1
x2=-1
else
x2=1
end
T_hat=exp(y0_hat+A_hat*x1+B_hat*x2)
With the result:
T_hat =
5.2402471237931471278213840718739
The problem right now: When trying to plot a Weibull density function with the calculated parameters, i'm getting the following error:
x=0:0.1:130;
f=wblpdf(x,T_hat,b_hat)
plot(f);
Error using symengine
Division by zero.
Error in sym/privBinaryOp (line 973)
Csym = mupadmex(op,args{1}.s, args{2}.s, varargin{:});
Error in .^ (line 324)
B = privBinaryOp(A, p, 'symobj::zip', '_power');
Error in wblpdf (line 50)
y = z.^(B-1) .* w .* B ./ A;
Error in zwei_parametrig (line 66)
f=wblpdf(x,T_hat,b_hat)
I don't really understand where the problem is. I took the 2 solutions for the parameters b_hat and T_hat and tried to plot a density function in an extra Matlab file. It worked.
Can someone tell me, where the problem is/could be?
Sincerly yours, Ronald
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