Is taking average the right method to find individual answers?

So over here I need to find out the dp values which are like d2, d3, d4, and so on will dN (N being the number of data points). So to solve this I open the brakets and take the terms to LHS and then divide that by the (u(k-i)^p) term which gives me dp value. But as there are multiple loops going on i get multiple values for each P th term. So what i did was to take the average after those loops get over to get only 1 term per each p loop.

%to get only one values
load('prbs165samples.mat') %dataset loaded
y = a(:,2); % output dataset (provided)
u = a(:,1); %input datast (provided)
na = 2; 
nb = 2;
N = 161;
A = [0.2867 -0.2428];
B = [0.6607 -0.2732];
dp = [];
d = [];
tot = 0;
for p = 2:N+1
    for i = 1:na 
        for j = 1:nb 
            for k = 3:N
                    LHS = y(k) + A(i)*y(k-i) - B(j)*u(k-j);
                    deno = (u(k-j).^p);
                    dp = [dp, LHS/deno];
                    tot = tot +(LHS/deno);
            end
        end
    end
    d = [d, tot/(na*nb*(N+1-3))];
    tot = 0;
end
dp;
d

[John D'Errico]

Huh? I'm lost as to what you are asking.

https://www.mathworks.com/matlabcentral/answers/6200-tutorial-how-to-ask-a-question-on-answers-and-get-a-fast-answer

[Ankitha Pai]

The format is followed. Thank you stopping by the question though. Appretiate it :)

[Ankitha Pai]

I'm asking to obtain a single value is it right to take the average of multiple values.

As there are summations and lopps going on for the pth term i get na*nb*(N+1-3) times values. Required is only one term

Answer 1

[Image Analyst]

No that's not right. You need two for loops. One to compute the "a" sum, then a separate loop (not nested) to compute the "b" sum. Then add the two sums together. yk = asumk + bsumk. Try something like this:

% Initialize variables to random numbers or zero
% or else use whatever values you have gotten in some other way.
N = 20;
u = rand(1, N);
d = rand(1, N);
na = 5
nb = 7
a = rand(1, na);
b = rand(1, nb);
k = 18;
y(k) = 0; % Initialize y

% Compute the first term, which is only one sum.
asum = 0;
for i = 1 : na
	asum = asum + a(i) * y(k-1);
end
fprintf('The first sum = %f.\n', asum);

% Now compute the second term, which has two sums.
bsum = 0;
for i = 1 : nb
	% First compute the third summation over p
	udsum = 0;
	for p = 2 : N
		if (k-i) <= 0 || (k-i) > length(u)
			% Cannot do this index value so exit the loop.
			break;
		end
		udsum = udsum + d(p) * u(k-i) ^ p;
	end
	% Now add that third sum to u(k-i) and multiply by b(i):
	%fprintf('(k-i) = %d.\n', (k-i));
	if (k-i) >= 1 && (k-i) < length(u)
		% If the indexes are ok, so the sum.
		bsum = bsum + b(i) * (u(k-i) + udsum);
	end
end
fprintf('The second sum = %f.\n', bsum);

% Now compute y(k) as the sum of the two summations:
y(k) = -asum + bsum
fprintf('Done running %s.m\n', mfilename);

[Ankitha Pai]

OH MY GOD! thank you so much. It makes sense. I understand the mistake now.

Definetly the real mvp!

Thank you so much.