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Why does my loop stop after the first iteration?

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Joseph Weiss
Joseph Weiss el 26 de Abr. de 2022
Comentada: Voss el 29 de Abr. de 2022
Hi everyone! I am trying to generate a sample (either a 0 or 1) with certain probability, and then update a Mean formula and Variance formula, both of which are recursive. I will keep generating samples and keep updating the formulas until this condition is met:
if (n>10) && ((Mn_A(A,n)-2*sqrt(Vn_A(A,n)/n)>.05) || (Mn_A(A,n)+2*sqrt(Vn_A(A,n)/n) < 0.5))
disp(['95% confidence interval achieved for A with N = ' num2str(n) ' and width = ' num2str(WidthA)])
break
end
I am getting an error however. It goes through the loop on the n=1 iteration with no problem, but on n=2, I get this error:
Not enough input arguments.
Error in ee497ca2_1>Mn_A (line 14)
if n == 0
Error in ee497ca2_1>Mn_A (line 19)
r = (Mn_A(n-1) + (x1 - Mn_A(n-1)))/n;
Error in ee497ca2_1 (line 4)
Mn_A(A,n)
I think I am missing something on how recursion works, but I am not sure. Here is the whole code below. Thank you!
samples = 500;
for n = 1:samples
A = randsrc(1,1,[0,1;0.49,0.51]);
Mn_A(A,n)
Vn_A(A,n)
ConfidenceInterval_Min_A = Mn_A(A,n) - 2*sqrt(Vn_A(A,n)/n);
ConfidenceInterval_Max_A = Mn_A(A,n) + 2*sqrt(Vn_A(A,n)/n);
WidthA = ConfidenceInterval_Max_A - ConfidenceInterval_Min_A;
if (n>10) && ((Mn_A(A,n)-2*sqrt(Vn_A(A,n)/n)>.05) || (Mn_A(A,n)+2*sqrt(Vn_A(A,n)/n) < 0.5))
disp(['95% confidence interval achieved for A with N = ' num2str(n) ' and width = ' num2str(WidthA)])
break
end
end
function r = Mn_A(x1,n)
if n == 0
r = 0;
elseif n == 1
r = x1;
else
r = (Mn_A(n-1) + (x1 - Mn_A(n-1)))/n;
end
end
function q = Vn_A(x1,n)
if n == 0
q = 0;
elseif n == 1
q = 0;
else
q = Vn_A(n-1) - Vn_A(n-1)/(n-1) + ((x1-Mn_A(n-1))^2)/n;
end
end

Respuesta aceptada

Voss
Voss el 26 de Abr. de 2022
Editada: Voss el 26 de Abr. de 2022
Mn_A takes 2 input arguments, x1 and n:
function r = Mn_A(x1,n)
But later in the function Mn_A, you call Mn_A with one input:
r = (Mn_A(n-1) + (x1 - Mn_A(n-1)))/n;
% ^^^ ^^^ one input each time
When Mn_A is called with one input, it has an error on its first line:
function r = Mn_A(x1,n) % if only x1 is given, n is undefined, so
if n == 0 % there's an error trying to refer to n here
(Same for Vn_A.)
  4 comentarios
Joseph Weiss
Joseph Weiss el 27 de Abr. de 2022
Okay. I think I was confusing myself as well haha, but it makes sense after you explained it. Thank you again for your help it is much appreciated!
Voss
Voss el 29 de Abr. de 2022
You're welcome!

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