How to do a matrix from a loop for several values of a constant?

So I am trying to have a 4x10 matrix called Y by using a for loop for each value of a. I just don't know how to do it. I know I could do this manually by defining y1 with a=.5, y2 with a=.9,...y4 with a=1.1 and running the loop for each one but I want the computer do it all. This for solving a more complex problem. Thanks in advance!

e=rand(1,10);
a=[.5,.9,1,1.1]
y=ones(1,10);
 for i=2:10;
    y(i)=a*y(i-1)+e(i-1); %I want a matrix of Y which contains this loop for each value 
                            of a stated above.
end

Answer 1

[Jose Marques]

Hello Germán Loredo! Try this:

clear all;
close all;
clc;

e=rand(1,10)
a=[.5,.9,1,1.1]
y=ones(4,10);
for i=1:10;
    for k=1:4
        y(k,i)=a(k)*y(i)+e(i); %I want a matrix of Y which contains this loop for each value 
    end                   
end

[Germán Loredo]

Does not work. I tried this also. Those values are wrong because if you do it "manually" you'll have other values. Check it out!

[Jose Marques]

Can you give more information about the matrix y? what each element are?

[Jose Marques]

that " y(i-1) " is correct?

[Germán Loredo]

For example,a=.5 then y(2)=a*1-e(1) then y(3)=a*y(2)-e(2) ... y(10)=a+y(9)-e(9) All this results should be the 10 columns of the first line of Y. Then you do the same but for a=.9 and that should be the 10 columns of line 2 of Y. ... Finally, for a=1.1 you do the same and that should be the 10 columns of matrix Y.

By theory, when a>=1 y(n)should be very high when n goes very large. As when a<1 E(y(n)) should converge to a number.

[Germán Loredo]

and Y(1)=1.

[Jose Marques]

I have changed "y(k,i)=a(k)*y(i)+e(i)" to "y(k,i)=a(k)*y(i-1)-e(i-1);".

[Jose Marques]

e=rand(1,10);
a=[.5,.9,1,1.1]
y=ones(4,10);
for i=2:10;
    for k=1:4
        y(k,i)=a(k)*y(i-1)-e(i-1); %I want a matrix of Y which contains this loop for each value 
    end                   
end

[Germán Loredo]

Still wrong, it makes no sense because, for example, when a=1.1 and we know e is always positive,

y(2)=1.1*y(1)+e=1.1*1+e(1)

y(3)=1.1*y(2)+e(2)=(1.1^2)*1+1.1*e(1)+e(2)

y(4)=(1.1^2)*1+(1.1^2)*e(1)+1.1*e(2)+e(1).

As you can see Y(n) is always bigger and bigger when n increases. Your results do not show that.

But check this out, I think I just solve it (we were needed to put dimensions to some variables).

 e=rand(4,1000);
a=[.5,.9,1,1.1];
y=ones(4,1000);
for n=1:4
    for i=2:1000;
    y(n,i)=a(n)*y(n,i-1)+e(n,i-1);
    end
end

[Germán Loredo]

This result goes along with: When a>=1, E(y)should go to infinite when n goes infinite. As when a<1 E(y) converge to a number when n goes to infinity.

[Jose Marques]

Nice! = D