How to run this code for 1000 times?

%For C1
lambda1 = [60.21, 41.58, 9.11, 8.71, 3.83, 3.74, 18.06]
r1 = poissrnd(lambda1)

%For C2
lambda2 = [41.58, 60.21, 8.71, 9.11, 3.74, 3.83, 18.06]
r2 = poissrnd(lambda2)

%Designed training sequences x1 and x2
x1 = [1,1,1,0,0,0,0,1,0,1,0,1,1,0,0,1] ;
x2 = [1,1,1,0,1,0,0,0,1,1,1,0,0,0,0,1] ;

%X[3] to X[16]
X3 = [1 1 1 1 1 1 1]' ;
X4 = [0 0 1 1 1 1 1]' ;
X5 = [0 1 0 0 1 1 1]' ;
X6 = [0 0 0 1 0 0 1]' ;
X7 = [0 0 0 0 0 1 1]' ;
X8 = [1 0 0 0 0 0 1]' ;
X9 = [0 1 1 0 0 0 1]' ;
X10 = [1 1 0 1 1 0 1]' ;
X11 = [0 1 1 1 0 1 1]' ;
X12 = [1 0 0 1 1 1 1]' ;
X13 = [1 0 1 0 0 1 1]' ;
X14 = [0 0 1 0 1 0 1]' ;
X15 = [0 0 0 0 1 0 1]' ;
X16 = [1 1 0 0 0 0 1]' ;

%X,a 7x14 matrix
X = [X3,X4,X5,X6,X7,X8,X9,X10,X11,X12,X13,X14,X15,X16];

%C, a 7x2 matrix
C = [r1; r2]' ;

%Y, a 14x2 matrix
Y = X'*C ;

%Yd = Poiss(Y) (at equation (8))
Yd = poissrnd(Y)

%y1, a 14x1 matrix ; y2, a 14x1 matrix
y1 = Y(:,1)
y2 = Y(:,2)

%Least Square Estimate of C

Cls = (inv(X*X'))*(X*Yd)

%  To set to zero all the negative entries of C
Cls1 = max(Cls,0)

%Mean square error of LS C and C
MSE = mean((C - Cls1).^2)
 

[Eric Chua]

I need to get the mean square error (MSE) for 1000 realization, how should i code it in a simpler way? Any help is appreciated.

[Subhadeep Koley]

@Eric Chua Which parameter will be changed in every iteration?

[Eric Chua]

Thanks for the reply. r1 and r2 are values that will change every single run as they are generated with poisson, and so do the C, Y, Yd.

[Eric Chua]

Cls and Cls1 are also changing for every run.

Answer 1

[KSSV]

N = 1000 ; 
MSE = zeros(N,2) ; 
for i = 1:N
  %For C1
  lambda1 = [60.21, 41.58, 9.11, 8.71, 3.83, 3.74, 18.06] ;
  r1 = poissrnd(lambda1) ;
  
  %For C2
  lambda2 = [41.58, 60.21, 8.71, 9.11, 3.74, 3.83, 18.06] ; 
  r2 = poissrnd(lambda2) ;
  
  %Designed training sequences x1 and x2
  x1 = [1,1,1,0,0,0,0,1,0,1,0,1,1,0,0,1] ;
  x2 = [1,1,1,0,1,0,0,0,1,1,1,0,0,0,0,1] ;
  
  %X[3] to X[16]
  X3 = [1 1 1 1 1 1 1]' ;
  X4 = [0 0 1 1 1 1 1]' ;
  X5 = [0 1 0 0 1 1 1]' ;
  X6 = [0 0 0 1 0 0 1]' ;
  X7 = [0 0 0 0 0 1 1]' ;
  X8 = [1 0 0 0 0 0 1]' ;
  X9 = [0 1 1 0 0 0 1]' ;
  X10 = [1 1 0 1 1 0 1]' ;
  X11 = [0 1 1 1 0 1 1]' ;
  X12 = [1 0 0 1 1 1 1]' ;
  X13 = [1 0 1 0 0 1 1]' ;
  X14 = [0 0 1 0 1 0 1]' ;
  X15 = [0 0 0 0 1 0 1]' ;
  X16 = [1 1 0 0 0 0 1]' ;  
  %X,a 7x14 matrix
  X = [X3,X4,X5,X6,X7,X8,X9,X10,X11,X12,X13,X14,X15,X16];  
  %C, a 7x2 matrix
  C = [r1; r2]' ;  
  %Y, a 14x2 matrix
  Y = X'*C ;  
  %Yd = Poiss(Y) (at equation (8))
  Yd = poissrnd(Y) ;  
  %y1, a 14x1 matrix ; y2, a 14x1 matrix
  y1 = Y(:,1) ; 
  y2 = Y(:,2) ;
  %Least Square Estimate of C  
  Cls = (inv(X*X'))*(X*Yd) ;  
  %  To set to zero all the negative entries of C
  Cls1 = max(Cls,0) ;  
  %Mean square error of LS C and C
  MSE(i,:) = mean((C - Cls1).^2) ;
end

[Eric Chua]

Thanks for the help!

[KSSV]

Thanks is accepting the asnwer.

[Eric Chua]

May I know how to check for the run to make sure it is correct? For example I need only the mse of third run.

[KSSV]

If you see MSE is vector of size N*2 i.e 1000x2. If you want MSE for 3rd iteration, use:

MSE(3,:)

[Eric Chua]

Thanks for the reply, what is the code for plotting this MSE graph against something?

[KSSV]

plot(MSE(:,1))

[Eric Chua]

Does the (: , 1) here mean first column of the MSE?

[KSSV]

Yes man...you should read this basic stuff...

MSE(:,1)   % first column 
MSE(:,2)  % second column

[Eric Chua]

Thank you very much, that help me alot.

[Eric Chua]

How do I add up all the MSE value in first column and divide by 1000 to get the average?

[KSSV]

Use the function mean.