Why does my loop stop after the first iteration?

Question

Joseph Weiss 2022-4-26

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1705980-why-does-my-loop-stop-after-the-first-iteration

评论： Voss 2022-4-29

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

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Answer 1

Voss 2022-4-26

1
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1705980-why-does-my-loop-stop-after-the-first-iteration#answer_951740

编辑：Voss 2022-4-26

在 MATLAB Online 中打开

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.)

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Voss 2022-4-27

在 MATLAB Online 中打开

You're welcome!

To answer your question, when any function is called (recursive or not), the function doesn't "remember" or "know" anything except what is passed to it as input arguments (an exception is if you are using global or persistent variables, which you aren't in this case - and you shouldn't need to).

I would focus on just getting Mn_A to work first, and then use analogous logic in Vn_A. Here's a description of how a recursive function to calculate the mean of a sequence might work, with a simple example:

function r = Mn_A(x1,n) % x1 is a sequence of random samples, n is the length of the sequence
    if n == 0 % if there are no samples, the mean is 0
        r = 0;
    elseif n == 1 % is there is only one sample, it is the mean
        r = x1;
    else
        
        % if there is more than one sample, determine the mean
        % somehow, in terms of the mean of a smaller sequence
        % e.g., I know your sequences are 0s and 1s, but suppose 
        %       you had the sequence {20,30,70}. Then:
        %         (step 1) mean({20,30,70}) is (70 + 2*mean({20,30}))/3, where
        %         (step 2) mean({20,30})    is (30 +   mean({20})   )/2, where
        %         (step 3) mean({20})       is 20, so
        % (back to step 2) mean({20,30})    is (30 +   20           )/2 = 25, so
        % (back to step 1) mean({20,30,70}) is (70 + 2*25           )/3 = 40
        
        r = (Mn_A(n-1) + (x1 - Mn_A(n-1)))/n; % <- does not (yet) correspond to the above
         
    end
end

You have to figure out what that line should be:

r = (Mn_A(n-1) + (x1 - Mn_A(n-1)))/n;

in order to express the mean of a sequence x1 in terms of the mean of the smaller sequence formed by removing one element from x1, as in the example I step through above.

I think part of the confusion (including confusion on my part) is that in this case, you are iterating and recursing. On each iteration, you would already have the mean of the previous sequence (i.e., without the most recent sample), so you could use that mean combined with the new sample to get the mean of the new sequence (including the most recent sample). Then a recursive function is not required, since you're doing the same thing that a recursive function would be doing, but doing it by storing and using the means of the smaller sequences that are calculated as you iterate.

I would do this either by iterating (calculating the new mean from the old mean and the new sample) or by recursing (given a sequence, let the recursive function figure out the mean), but I wouldn't combine both approaches - you end up doing the same thing over and over when you do that.

Joseph Weiss 2022-4-27