Normal Distribution generation using acceptance rejection

Hi guys, I want to implement the following algorithm and plot the pdf graph of X at the end.

1.Y1=−ln(U1) , U1 is a uniform random number

2.Y2=−ln(U2); U2 is a uniform random number

If Y2≥(Y1−1)^2/2, set|Z|=Y1; otherwise return to Step 1.

3. Generate U. Set Z=|Z| if U≤0.5 and Z= -|Z| if U >0.5

.4. Set X=0.5 Z -2

Does anyone have a suggestion on how to do that?

[Tarek Mahjoub]

Here is my attempt. Can anyone help me fix this code?

n=100;
Z = zeros(1,n);
absZ = zeros(1,n);
X = zeros(1,n);

function exp

Y1 = -log(rand(1,n));
Y2 = -log(rand(1,n));

if (Y1 -1)^2/2 <= Y2
    absZ = Y1;
else
     exp
end
end
U = rand(1,n);
if U <= 0.5
    Z = absZ;
else
  Z = -absZ;
end
X = Z * 0.5 - 2;

Answer 1

[Yazan]

Here is a simple script to estimqte the pdf of X.

clc, clear

% number of iterations, the higher this value is, the better is the pdf 
% estimation
itt = 1000;

X = nan(itt, 1);

for n=1:itt
    
    while 1
        Y1 = -log(rand(1));
        Y2 = -log(rand(1));
        
        if Y2>=(Y1-1).^2/2
            absZ = Y1;
            break
        end
    end
    
    U = rand(1);
    if U>0.5
        Z = -absZ;
    else
        Z = absZ;
    end
    
    X(n) = 0.5*Z -2;
    
end

% plot pdf
histogram(X, 'Normalization', 'probability');
ax = gca;

ax.Title.String = 'Probability distribution';
ax.XLabel.String = 'X';
ax.YLabel.String = 'Probability';