Difficulty Coming up with a solution, would there be a convenient command for this ??

I am looking to implement a for loop but am not quite sure of how to do this given the for loop. I have thought really hard and it may either be due to a lack of skills but am unable to find an answer. Suppose I have a matrix and some predefined variable:

Matrix = [1; 2; 3; 4; 5; 6; 7; 8; 9; 10]
X = 0.94

What I would like is for matlab to do the following: First construct running mean as follows:

m = zeros(10,1)
for n = 1: m(1,1) = (1-0.94)*(0.94)*(1) = 0.0564
for n = 2: m(2,1) = (1-0.94)*(0.94.^2)*(1) + (1-0.94)*(0.94.^1)(2) = 0.053016
for n = 3: m(3,1) = (1-0.94)*(0.94.^3)*(1) + (1-0.94)*(0.94.^2)(2) + (1-0.94)*(0.94.^1)(3)
for n = 4: m(4,1) = (1-0.94)*(0.94.^4)*(1) + (1-0.94)*(0.94.^3)*(2) + (1-0.94)*(0.94.^2)*(3)+(1-0.94)*(0.94.^1)*(4)

m1 = [0.0564; 0.053016; ... ] %%this is my output matrix

To be precise, I am looking to calculate something similar to the formula where d is equal to X set as 0.94. The calculation above describes the process for calculating the mean but if I get the intuition behind executing it, I can perform it for the variance.

    Matrix = [1; 2; 3; 4; 5; 6; 7; 8; 9; 10];
    X = 0.94;
    r = zeros(10);
    for t = 1:10
        r(t,1) = Matrix(t,1);
        for i = 1:10
            m(i,1) = (1-X)*(X.^i)*r(t,1);
        end
    end

This is a code that I came up with but clearly, it is very wrong and I have also tried to index it from t = 10:-1:1, which is also wrong. I have been stuck on this for a very long time and it would be great if someone can help me.

Any help is much appreciated. Thank you

Answer 1

[Triveni]

I don't know what is your "m1". As i understand. following should be help you

 Matrix = [1; 2; 3; 4; 5; 6; 7; 8; 9; 10];
 X = 0.94;

 Matrix1 = [5; 6 ; 7 ; 8 ; 9 ;10 ;0 ;0 ;0 ; 0];
 Matrix2 = [4; 6 ; 7; 1; 0 ; 0; 0 ; 0 ; 0 ; 0];
 Matrix3 = [1; 2 ; 3; 0; 0 ; 0; 0 ; 0 ; 0 ; 0];
 Z = [Matrix Matrix1 Matrix2 Matrix3];
 out = zeros(1,1,10);
 m1=Matrix1; % I have taken m1=Matrix1
 out(:,:,1) = (1-X)*((X.^1)*(Matrix(1,1)-m1(1,1)).^2);
 for k=2:10

    out(:,:,k) = out((k-1)) + (1-X)*((X.^k+1)*(Matrix(k,1)-m1(k,1)).^2)

 end

[Putsandcalls]

m = zeros(10,1)
  for n = 1: m(1,1) = (1-0.94)*(0.94)*(1) = 0.0564
  for n = 2: m(2,1) = (1-0.94)*(0.94.^2)*(1) + (1-0.94)*(0.94.^1)(2) = 0.053016
  for n = 3: m(3,1) = (1-0.94)*(0.94.^3)*(1) + (1-0.94)*(0.94.^2)(2) + (1-0.94)*(0.94.^1)(3)
  for n = 4: m(4,1) = (1-0.94)*(0.94.^4)*(1) + (1-0.94)*(0.94.^3)*(2) + (1-0.94)*(0.94.^2)*(3)+(1-0.94)*(0.94.^1)*(4)
  ...
  and this repeats until 
  n = 10 just repeating some the above procedure. Giving me some output matrix m1 = [0.0564; 0.053016; ... ]

Sorry I just meant to say M1 as a matrix defined by the following calculation above based on the matrix i defined as:

Matrix = [1; 2; 3; 4; 5; 6; 7; 8; 9; 10];

Problem is how would I repeat this for my large matrix Z ? since I need to iterate the procedure again.

[Triveni]

 for n =1  (1-X)*((X.^1)*(Matrix(1,1)-m1(1,1)).^2).
what did you mean by "m1(1,1)"

[Putsandcalls]

So it is to take the element from row1,column1 subtracting the calculated matrix from the above procedure.

Iterating the process over, for n=1, it would be:

for n = 1: (1-0.94)*(0.94)*(1) = 0.0564

So for the second step, it would be:

for n =1 (1-X)*((X.^1)*(Matrix(1,1)-m1(1,1)).^2) where m(1,1) = 0.0564.

Sorry for the confusion. I redefined the above comment to make it more clear. Also, the fact that it yields a 3 dimensional matrix is quite problematic. Is there any way to make it so that it is not k-dimensional?