Not enough input arguments
4 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Farah Shahpoor
am 28 Mai 2019
Kommentiert: Farah Shahpoor
am 16 Jul. 2019
Dear community,
I want to calculate the loss of the following system:
%Hybrid NKM under commitment
%% Step 1 define the parameters
%same Parameters as from the script
%Parameters
gam_f = 0.5
gam_b =0.5
beta= 1
gam_x = 0.2
lambda = 0.5
phi = 0.5 % IS Parameter for interest rate
sig = 1 % IS Parameter for interest rate
AR_par = 0.8
%% Step 2
% System is: (w;v)(+1) = A*[w;v] + [ 1;0;0;0;0]*eps
A11= [AR_par 0 0;0 0 0; 0 0 0];
A12= [ 0 0; 1 0; 0 1];
A21 =[ 0 -gam_f/(beta^2*gam_b) 0; -1/(beta*gam_f) 0 -gam_b/gam_f]; %Klammern nicht vergessen
A22 =[ 1/(beta^2*gam_b) gam_x/(beta^2*lambda*gam_b); -gam_x/(beta*gam_f) 1/(beta*gam_f)];
A = [ [A11 A12] ; [A21 A22] ]
%% Step 3
% using the Schur Decomposition to solve the state equations
% solve the system
disp('Schur decomposition')
[Z, T] = schur(A, 'complex')
disp('reorder eigenvalues in increasing order along the principal diagonal')
[Z T] = ordschur(Z,T, 1:5)
if abs(sum(sum(Z*T*Z'-A))) > 0.0001 && sum(sum(Z'*Z-eye(lenght(Z)))) > 0.0001
disp('Error in Schur decomposition')
end
disp('check Blanchard-Kahn')
abs_diag_T = abs(diag(T))'
%% Calculating the solution time path for nu, x and pi using the following law of motion:
% z(+1) = E[z(+1)] + Z_11^-1 * [1;0;0] * eps
T_11 = T(1:3,1:3)
Z_11 = Z(1:3,1:3)
Z_21 = Z(4:5,1:3)
T=1000;
z_solution= zeros(3,T); % zeros because we have variables in i with t-1
w_solution= zeros(3,T); % zeros because we have variables in i with t-1
w_solution(:,1)=[ 1; 0; 0]; %initial jump
z_solution(:,1)=inv(Z_11)* w_solution(: ,1);%initial jump
v_solution= zeros(2,T);
i_solution= zeros(1,T); % nominal interest rate: IS umgestellt nach der Variable i.Hier liegt anscheinend das Problem.
for t= 2:T
z_solution(:,t)= T_11* z_solution(: ,t-1 );
w_solution(:,t)= Z_11 * z_solution(:,t);
v_solution(:,t)= Z_21 *inv(Z_11)* w_solution(:,t);
end
for t= 1:T-1
i_solution(:,t) =((1- phi)*v_solution(1,t+1)+phi*w_solution(2,t)-w_solution(2,t+1))*sig+ v_solution(2,t+1); % Jump in 1 anstatt 2. umgestellte IS-Kurve.
end;
For calculating the loss I need the v_solution that is a vector containing two variables x and pi. I need the mean of these two for my loss function.
function [ LossVal, vol_pi, vol_x ] = Loss_Fun( data, w_pi, w_x )
vol_pi = (mean(data.v_solution(2,1:T).^2));
vol_x = (mean(data.v_solution(1,1:T).^2));
LossVal = w_pi * vol_pi ...
+ w_x * vol_x;
w_pi = 0.5 ; %1;
w_x = 0.25; %0.5;
end
but something seems to be wrong here. But I dont knwo what
3 Kommentare
Akzeptierte Antwort
Guillaume
am 26 Jun. 2019
You haven't shown us how you are calling your function Loss_fun.
If you run it with the green run button, or call it without any input argument then yes, you'll get the error you see.
>>Loss_Fun
Not enough input arguments.
Error in Loss_fun (line 3)
vol_pi = (mean(data.v_solution(2,1:T).^2));
Since you're not passing the required inputs data, w_pi and w_px.
The proper way to call the function would something like:
result = Loss_fun(something, somethingelse, anothersomething); %3 inputs required
Even if you passed the required inputs, you'd still get an error: Undefined function of variable T since it's neither an input to the function nor defined inside the function. It is unclear where you thought that T would come from.
8 Kommentare
Guillaume
am 30 Jun. 2019
If T from the main script is needed by the function, then you need to pass it as an input argument. Hence your function should have 4 inputs:
function [LossVal, vol_pi, vol_x ] = Loss_Fun(data, w_pi, w_x, T) %function has now 4 inputs
Weitere Antworten (1)
KALYAN ACHARJYA
am 26 Jun. 2019
Bearbeitet: KALYAN ACHARJYA
am 26 Jun. 2019
# Check Commnet, this function file have no role in main script.
Save the function in different file as "Loss_Fun.m" (Same directory)
function [LossVal,vol_pi,vol_x]=Loss_Fun(data,w_pi,w_x)
vol_pi=(mean(data.v_solution(2,1:T).^2));
vol_x=(mean(data.v_solution(1,1:T).^2));
LossVal=w_pi*vol_pi+w_x*vol_x;
%w_pi = 0.5 ; %1;
%w_x = 0.25; %0.5;
end
Main Script (Run the main script)
%Hybrid NKM under commitment
%% Step 1 define the parameters
%same Parameters as from the script
%Parameters
gam_f = 0.5
gam_b =0.5
beta= 1
gam_x = 0.2
lambda = 0.5
phi = 0.5 % IS Parameter for interest rate
sig = 1 % IS Parameter for interest rate
AR_par = 0.8
%% Step 2
% System is: (w;v)(+1) = A*[w;v] + [ 1;0;0;0;0]*eps
A11= [AR_par 0 0;0 0 0; 0 0 0];
A12= [ 0 0; 1 0; 0 1];
A21 =[ 0 -gam_f/(beta^2*gam_b) 0; -1/(beta*gam_f) 0 -gam_b/gam_f]; %Klammern nicht vergessen
A22 =[ 1/(beta^2*gam_b) gam_x/(beta^2*lambda*gam_b); -gam_x/(beta*gam_f) 1/(beta*gam_f)];
A = [ [A11 A12] ; [A21 A22] ]
%% Step 3
% using the Schur Decomposition to solve the state equations
% solve the system
disp('Schur decomposition')
[Z, T] = schur(A, 'complex')
disp('reorder eigenvalues in increasing order along the principal diagonal')
[Z T] = ordschur(Z,T, 1:5)
if abs(sum(sum(Z*T*Z'-A))) > 0.0001 && sum(sum(Z'*Z-eye(lenght(Z)))) > 0.0001
disp('Error in Schur decomposition')
end
disp('check Blanchard-Kahn')
abs_diag_T = abs(diag(T))'
%% Calculating the solution time path for nu, x and pi using the following law of motion:
% z(+1) = E[z(+1)] + Z_11^-1 * [1;0;0] * eps
T_11 = T(1:3,1:3)
Z_11 = Z(1:3,1:3)
Z_21 = Z(4:5,1:3)
T=1000;
z_solution= zeros(3,T); % zeros because we have variables in i with t-1
w_solution= zeros(3,T); % zeros because we have variables in i with t-1
w_solution(:,1)=[ 1; 0; 0]; %initial jump
z_solution(:,1)=inv(Z_11)* w_solution(: ,1);%initial jump
v_solution= zeros(2,T);
i_solution= zeros(1,T); % nominal interest rate: IS umgestellt nach der Variable i.Hier liegt anscheinend das Problem.
for t= 2:T
z_solution(:,t)= T_11* z_solution(: ,t-1 );
w_solution(:,t)= Z_11 * z_solution(:,t);
v_solution(:,t)= Z_21 *inv(Z_11)* w_solution(:,t);
end
for t= 1:T-1
i_solution(:,t) =((1- phi)*v_solution(1,t+1)+phi*w_solution(2,t)-w_solution(2,t+1))*sig+ v_solution(2,t+1); % Jump in 1 anstatt 2. umgestellte IS-Kurve.
end;
Output
gam_f =
0.5000
gam_b =
0.5000
beta =
1
gam_x =
0.2000
lambda =
0.5000
phi =
0.5000
sig =
1
AR_par =
0.8000
A =
0.8000 0 0 0 0
0 0 0 1.0000 0
0 0 0 0 1.0000
0 -1.0000 0 2.0000 0.8000
-2.0000 0 -1.0000 -0.4000 2.0000
Schur decomposition
Z =
0.0000 - 0.0000i 0.0000 - 0.0000i -0.2949 + 0.0000i -0.3320 + 0.0694i 0.0258 - 0.8929i
-0.1586 - 0.3781i 0.2772 + 0.2163i 0.7860 + 0.0000i -0.0795 + 0.1836i 0.0683 - 0.2138i
0.2674 - 0.1121i 0.0822 - 0.3647i -0.1149 + 0.0000i 0.0359 + 0.8130i 0.3023 + 0.0965i
0.0856 - 0.7009i 0.5138 - 0.1167i -0.3731 + 0.0000i 0.0328 - 0.2581i -0.0960 + 0.0882i
0.4956 + 0.0605i -0.0443 - 0.6760i 0.3779 + 0.0000i -0.0505 - 0.3353i -0.1247 - 0.1357i
T =
1.4956 + 0.8535i 0.5385 + 0.0000i 0.8527 - 1.0580i 0.9668 - 0.8355i -0.4607 + 0.3907i
0.0000 + 0.0000i 1.4956 - 0.8535i -0.7757 + 1.1632i 1.2424 - 0.1948i -1.0957 - 0.3520i
0.0000 + 0.0000i 0.0000 + 0.0000i 0.8000 + 0.0000i 0.2651 - 0.3937i -0.1464 + 0.7130i
0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.5044 + 0.2878i -0.6670 + 0.0000i
0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.5044 - 0.2878i
reorder eigenvalues in increasing order along the principal diagonal
Z =
0.0000 + 0.0000i -0.0000 - 0.0000i 0.4092 + 0.2263i 0.2994 - 0.2642i 0.1468 - 0.7748i
0.1237 - 0.6952i 0.4512 - 0.2721i 0.2129 + 0.1177i 0.0988 + 0.3509i -0.1754 + 0.0321i
0.4916 + 0.0874i -0.2543 - 0.6279i -0.3849 - 0.2129i 0.1566 + 0.1010i -0.0454 - 0.2480i
-0.1377 - 0.3862i 0.3311 + 0.1184i -0.5507 - 0.3046i -0.0533 - 0.4638i 0.2349 - 0.1977i
0.2731 - 0.0974i -0.0364 - 0.3721i 0.3187 + 0.1763i -0.3534 - 0.5745i 0.2796 + 0.3322i
T =
0.5044 - 0.2878i -0.1691 + 0.0663i 0.7809 - 0.5847i 0.4891 - 0.5818i 0.0876 + 1.3505i
0.0000 + 0.0000i 0.5044 + 0.2878i -0.8802 - 0.0511i 0.4373 - 1.3339i -0.8048 - 0.1024i
0.0000 + 0.0000i 0.0000 + 0.0000i 0.8000 + 0.0000i 0.4870 + 0.8044i -0.1559 + 0.7071i
0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 1.4956 - 0.8535i -1.2778 - 0.5945i
0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 1.4956 + 0.8535i
check Blanchard-Kahn
abs_diag_T =
0.5807 0.5807 0.8000 1.7220 1.7220
T_11 =
0.5044 - 0.2878i -0.1691 + 0.0663i 0.7809 - 0.5847i
0.0000 + 0.0000i 0.5044 + 0.2878i -0.8802 - 0.0511i
0.0000 + 0.0000i 0.0000 + 0.0000i 0.8000 + 0.0000i
Z_11 =
0.0000 + 0.0000i -0.0000 - 0.0000i 0.4092 + 0.2263i
0.1237 - 0.6952i 0.4512 - 0.2721i 0.2129 + 0.1177i
0.4916 + 0.0874i -0.2543 - 0.6279i -0.3849 - 0.2129i
Z_21 =
-0.1377 - 0.3862i 0.3311 + 0.1184i -0.5507 - 0.3046i
0.2731 - 0.0974i -0.0364 - 0.3721i 0.3187 + 0.1763i
>>
7 Kommentare
Guillaume
am 26 Jun. 2019
I'm not sure how many times I can explain that like any function, a function that you write must be called with its required inputs. A function Loss_fun with 3 input arguments cannot be called with just Loss_fun the same way that the sin function cannot be called with just sin.
Siehe auch
Kategorien
Mehr zu Matrix Computations finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!