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I am trying to solve the equation below for array d. I have used the snippet below:
channel_size = 9e-3;
d = [11e-3, 12e-3];
sigma = 0.49;
ee = 727/9806.65;
alpha = d-channel_size;
sym('p',[1 2])
for i = 1:2
eqn = alpha == (4.3^(1/7))*(p^(3/11))*(((1-(sigma^2))/ee)^(7/5))/(d.^(1/6))
S = solve(eqn, p)*0.015;
vpa(S/13e-12)
end
In fact, I should get two number corresponding to d(1) and d(2), but it does not work and this error appears:
Error using mupadmex
Error in MuPAD command: Operands are invalid. [linalg::matlinsolve]
Error in sym/privBinaryOp (line 1693)
Csym = mupadmex(op,args{1}.s, args{2}.s, varargin{:});
Error in sym/mrdivide (line 232)
X = privBinaryOp(A, B, 'symobj::mrdivide');
Declaring (d.^(1/6)) is wrong and instead (d(i)^(1/6)) should be used. In addition, as alpha = d-channel_size, in the equation, I should also declare alpha(i) instead of simple alpha.
I have constrained ODE optimization problem to be solved using Matlab, I started by solving the ODE using ode15s which working will this initial values, also I had very good results without constraints using fminsearch, the problem started when I used fmincon it gave me these warnings:
Warning: Length of lower bounds is < length(x); filling in missing lower bounds with -Inf.
> In checkbounds (line 33)
In fmincon (line 318)
In Optimization (line 64)
Warning: Length of upper bounds is < length(x); filling in missing upper bounds with +Inf.
> In checkbounds (line 47)
In fmincon (line 318)
In Optimization (line 64)
Warning: Failure at t=2.340250e+01. Unable to meet integration tolerances without reducing the step size below the smallest value
allowed (5.684342e-14) at time t.
> In ode15s (line 668)
In ObjFun (line 6)
In barrier
In fmincon (line 813)
In Optimization (line 64)
I tried to remove constraints, but nothing changed...
y1_max = 5; y2_max = 2.3;
y01 = 1; y02 = 1.6;
%Control function parameter
Umin = -0.3; Umax = -0.1;
%Creation the structure of model parameters
params.T0=T0; params.Tf=Tf;
params.y1d=y1d; params.y2d=y2d;
params.y01=y01; params.y02=y02;
params.y2_max=y2_max;
n=2;
U0=-0.2*ones(1,n);
params.n=n;
params.U=U0; params.Umax=Umax; params.Umin=Umin;
params.dt=(Tf-T0)/n;
%for initial value of optimization parameters
options = odeset('RelTol',1e-7,'AbsTol',1e-7);
U=U0;
[t,y]=ode15s(#ODEsolver,[T0,Tf],[y01 y02],options,params);
%adding the ode solution as input prameters
params.t=t; params.y1= y(:,1); params.y2= y(:,2);
U0=-0.2*ones(1,n);
params.n=n; params.U=U0;
params.dt=(Tf-T0)/n;
A = [];
B = [];
Aq = []; Bq = [];
options1 = optimset('MaxIter',5000);
[x,fval,exitflag,output] = fmincon(#ObjFun,U0,A,B,Aq,Bq,Umin,Umax,IneqConst(U0,params),options1,params)
function y = ODEsolver(t,y,params)
dt = params.dt;
nu = floor(t/dt)+1;
nu = min(nu,params.n-1);
t1 = (nu-1)*dt;
Ut = params.U(nu) + ((t-t1)/dt)*(params.U(nu+1) - params.U(nu));
dy1a = ((0.63*y(1))/(1 + y(1)));
dy1b = (Ut*y(1)*y(2))/(1+0.6*y(1));
dy1 = dy1a + dy1b;
dy2a = (0.15*y(2))/(0.05-0.2*y(2));
dy2 = -0.2*y(2) + dy1b * dy2a;
y = [dy1; dy2];
end
function ObjFun1 = ObjFun(u,params)
% Calculating the value of the optimization criteria
params.U=u;
options = odeset('RelTol',1e-8,'AbsTol',1e-8);
[t,y]=ode15s(#ODEsolver,[0,100],[params.y01,
params.y02],options,params);
ObjFun1 =trapz(t,(y(:,1)-params.y1d).^2);
end
function [c,Const] = IneqConst(u, params)
params.U=u;
options = odeset('RelTol',1e-8,'AbsTol',1e-8);
[t,y]=ode15s(#ODEsolver,[0,100],[params.y01, params.y02],options,params);
c =[];
yCon = y(:,2)-params.y2_max;
Const = trapz(t,(abs(yCon)+yCon).^2);
end
The bug in this code is that the dimension of the bounds (Umin, Umax) should be an array with the same size of the control function U
so if the control function parameters n = 2 then Umin = [-0.3 -0.3] & Umax = [-0.1 -0.1]
I want to calculate the part-expectation of log-normal distribution via:
m = 1;
v = 2;
mu = log((m^2)/sqrt(v+m^2));
sigma = sqrt(log(v/(m^2)+1));
syms x;
d = x*lognpdf(x,mu,sigma);
int(d, x, 0, 10);
However, MATLAB says:
Error using symfun>validateArgNames (line 211) Second input must be a
scalar or vector of unique symbolic variables.
Error in symfun (line 45)
y.vars = validateArgNames(inputs);
Error in sym/subsasgn (line 771)
C = symfun(B,[inds{:}]);
Error in lognpdf (line 36) x(x <= 0) = Inf;
Error in untitled (line 7) d = x*lognpdf(x,mu,sigma);
I even tried to just calculate the integral of the pdf by:
m = 1;
v = 2;
mu = log((m^2)/sqrt(v+m^2));
sigma = sqrt(log(v/(m^2)+1));
syms x;
d = lognpdf(x,mu,sigma);
int(d, x, 0, 10);
But there are still errors, and MATLAB says:
Error using symfun>validateArgNames (line 211) Second input must be a
scalar or vector of unique symbolic variables.
Error in symfun (line 45)
y.vars = validateArgNames(inputs);
Error in sym/subsasgn (line 771)
C = symfun(B,[inds{:}]);
Error in lognpdf (line 36) x(x <= 0) = Inf;
Error in untitled (line 7) d = lognpdf(x,mu,sigma);
I really don't know what happened. Should the integral of the pdf be the cdf?
Similar to an answer several months ago, the Statistics Toolbox doesn't support the Symbolic Toolbox currently.
Therefore, you can proceed by hard coding the PDF itself and integrating it:
d = exp(-(log(x)-mu)^2/(2*sigma^2))/(x*sigma*sqrt(2*pi));
int(d, x, 0, 10);
Or you can use the logncdf function, which may be cleaner.
I am implementing the expression given in the image which is the log-likelihood for AR(p) model.
In this case, p=2. I am using fmincon as the optimization tool. I checked the documentation and other examples over internet regarding the syntax of this command. Still, I am unable to mitigate the problem. Can somebody please help in eliminating the problem?
The following is the error
Warning: Options LargeScale = 'off' and Algorithm = 'trust-region-reflective' conflict.
Ignoring Algorithm and running active-set algorithm. To run trust-region-reflective, set
LargeScale = 'on'. To run active-set without this warning, use Algorithm = 'active-set'.
> In fmincon at 456
In MLE_AR2 at 20
Error using ll_AR2 (line 6)
Not enough input arguments.
Error in fmincon (line 601)
initVals.f = feval(funfcn{3},X,varargin{:});
Error in MLE_AR2 (line 20)
[theta_hat,likelihood] =
fmincon(#ll_AR2,theta0,[],[],[],[],low_theta,up_theta,[],opts);
Caused by:
Failure in initial user-supplied objective function evaluation. FMINCON cannot
continue.
The vector of unknown parameters,
theta_hat = [c, theta0, theta1, theta2] where c = intercept in the original model which is zero ; theta0 = phi1 = 0.195 ; theta1 = -0.95; theta2 = variance of the noise sigma2_epsilon.
The CODE:
clc
clear all
global ERS
var_eps = 1;
epsilon = sqrt(var_eps)*randn(5000,1); % Gaussian signal exciting the AR model
theta0 = ones(4,1); %Initial values of the parameters
low_theta = zeros(4,1); %Lower bound of the parameters
up_theta = 100*ones(4,1); %upper bound of the parameters
opts=optimset('DerivativeCheck','off','Display','off','TolX',1e-6,'TolFun',1e-6,...
'Diagnostics','off','MaxIter', 200, 'LargeScale','off');
ERS(1) = 0.0;
ERS(2) = 0.0;
for t= 3:5000
ERS(t)= 0.1950*ERS(t-1) -0.9500*ERS(t-2)+ epsilon(t); %AR(2) model y
end
[theta_hat,likelihood,exit1] = fmincon(#ll_AR2,theta0,[],[],[],[],low_theta,up_theta,[],opts);
exit(1,1)=exit1;
format long;disp(num2str([theta_hat],5))
function L = ll_AR2(theta,Y)
rho0 = theta(1); %c
rho1 = theta(2); %phi1
rho2 = theta(3); %phi2
sigma2_epsilon = theta(4);
T= size(Y,1);
p=2;
mu_p = rho0./(1-rho1-rho2); %mean of Y for the first p samples
%changed sign of the log likelihood expression
cov_p = xcov(Y);
L1 = (Y(3:end) - rho0 - rho1.*Y(1:end-1) - rho2.*Y(1:end-2)).^2;
L = (p/2).*(log(2*pi)) + (p/2).*log(sigma2_epsilon) - 0.5*log(det(inv(cov_p))) + 0.5*(sigma2_epsilon^-1).*(Y(p) - mu_p)'.*inv(cov_p).*(Y(p) - mu_p)+...
(T-p).*0.5*log(2*pi) + 0.5*(T-p).*log(sigma2_epsilon) + 0.5*(sigma2_epsilon^-1).*L1;
L = sum(L);
end
You are trying to pass constant parameters to the objective function (Y) in addition to the optimization variables (theta).
The right way of doing so is using anonymous function:
Y = ...; %// define your parameter here
fmincon( #(theta) ll_AR2(theta, Y), theta0, [],[],[],[],low_theta,up_theta,[],opts);
Now the objective function, as far as fmincon concerns, depends only on theta.
For more information you can read about anonymous functions and passing const parameters.
I am using numerical integration in MATLAB, with one varibale to integrate over but the function also contains a variable number of terms depending on the dimension of my data. Right now this looks like the following for the 2-dimensional case:
for t = 1:T
fxt = #(u) exp(-0.5*(x(t,1)-theta*norminv(u,0,1)).^2) .* ...
exp(-0.5*(x(t,2) -theta*norminv(u,0,1)).^2);
f(t) = integral(fxt,1e-4,1-1e-4,'AbsTol',1e-3);
end
I would like to have this function flexible in the sense that there could be any number of data points in, each in the following term:
exp(-0.5*(x(t,i) -theta*norminv(u,0,1)).^2);
I hope this is understandable.
If x and u have a valid dimension match (vector-vector or array-scalar) for the subtraction, you can put the whole matrix x into the handle and pass it to the integral function using the name-parameter pair ('ArrayValued',true):
fxt = #(u) exp(-0.5*(x - theta*norminv(u,0,1)).^2) .* ...
exp(-0.5*(x - theta*norminv(u,0,1)).^2);
f = integral(fxt,1e-4,1-1e-4,'AbsTol',1e-3,'ArrayValued',true);
[Documentation]
You may need a loop if integral ever passes a vector u into the handle.
But in looking at how the integral function is written, the integration nodes are entered as scalars for array-valued functions, so the loop shouldn't be necessary unless some weird dimension-mismatch error is thrown.
Array-Valued Output
In response to the comments below, you could try this function handle:
fx = #(u,t,k) prod(exp(-0.5*(x(t,1:k)-theta*norminv(u,0,1)).^2),2);
Then your current loop would look like
fx = #(u,t,k) prod(exp(-0.5*(x(t,1:k)-theta*norminv(u,0,1)).^2),2);
k = 2;
for t = 1:T
f(t) = integral(#(u)fx(u,t,k),1e-4,1-1e-4,'AbsTol',1e-3,'ArrayValued',true);
end
The ArrayValued flag is needed since x and u will have a dimension mismatch.
In this form, another loop would be needed to sweep through the k indexes.
However, we can improve this function by skipping the loop altogether since each iterate of the loop is independent by using the ArrayValued mode:
fx = #(u,k) prod(exp(-0.5*(x(:,1:k)-theta*norminv(u,0,1)).^2),2);
k = 2;
f = integral(#(u)fx(u,k),1e-4,1-1e-4,'AbsTol',1e-3,'ArrayValued',true);
Vector-Valued Output
If ArrayValued is not desired, which may be the case if the integration requires a lot of subdivisions and a vector-valued u is preferable, you can also try a recursive version of the handle using cell arrays:
% x has size [T,K]
fx = cell(K,1);
fx{1} = #(u,t) exp(-0.5*(x(t,1) - theta*norminv(u,0,1)).^2);
for k = 2:K
fx{k} = #(u,t) fx{k-1}(u,t).*exp(-0.5*(x(t,k) - theta*norminv(u,0,1)).^2);
end
f(T) = 0;
k = 2;
for t = 1:T
f(t) = integral(#(u)fx{k}(u,t),1e-4,1-1e-4,'AbsTol',1e-3);
end
ThanksTroy but now I run into the follwing:
x = [0.3,0.8;1.5,-0.7];
T = size(x,1);
k = size(x,2);
theta= 1;
fx = #(u,t,k) prod(exp(-0.5*(x(t,1:k) - theta*norminv(u,0,1))^2));
for t = 1,T
f(t) = integral(#(u)fx(u,t,k),1e-4,1-1e-4,'AbsTol',1e-3);
end
Error using -
Matrix dimensions must agree.
Error in #(u,t,k)prod(exp(-0.5*(x(t,1:k)-theta*norminv(u,0,1))^2))
Error in #(u)fx(u,t,k)
Error in integralCalc/iterateScalarValued (line 314)
fx = FUN(t);
Error in integralCalc/vadapt (line 133)
[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);
Error in integralCalc (line 76)
[q,errbnd] = vadapt(#AtoBInvTransform,interval);
Error in integral (line 89)
Q = integralCalc(fun,a,b,opstruct);