Not enough input arguments
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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 个评论
采纳的回答
Guillaume
2019-6-26
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 个评论
Guillaume
2019-6-30
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
更多回答(1 个)
KALYAN ACHARJYA
2019-6-26
编辑:KALYAN ACHARJYA
2019-6-26
# 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 个评论
Guillaume
2019-6-26
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.
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