You could use ylim() to simply prevent the peak from being drawn. Or you could increase the bar widths, such as by decreasing the number of bars you ask for.

A=[a(1:135756).column1]; figure subplot(2,1,1); histogram(A,'Normalization','pdf'); ylabel('pdf'); subplot(2,1,2); histogram(A,'Normalization','cdf'); ylabel('cdf');

You can use the histogram function to create and plot a histogram. You can use the ksdensity function to estimate a non-parametric fit to the probability density function, and then the plot function to plot it.

Your code looks good, and you've definitely normalized the histogram correctly. However, you've got one thing that needs fixing. The random values, p, in your probability density function should be the same random values. You have generated them independently, but p only need to be randomly generated once. I've made a simple fix to your code and commented on the lines! maxiter = 10^5; x_limit = 15; x_res = 0.01; % Generate vector of uniform random numbers randVec = rand(maxiter,1); % f(p) = ln(p/(1-p)) => PDF random values utilize same p vector X = log(randVec./(1-randVec)); x = -x_limit:x_res:x_limit; [N, BIN] = histc(X, x); Nnorm = N./trapz(x,N); subplot(2,1,1) h = bar(x,Nnorm, 'hist'); hold on; plot(x, exp(-x)./(1+exp(-x)).^2,'r--') title('PDF for logistic distribution') hold off A1 = trapz(x, exp(-x)./(1+exp(-x)).^2); A2 = trapz(x,Nnorm); subplot(2,1,2) bar(x, cumsum(Nnorm*x_res)), hold on plot(x,exp(x)./(1+exp(x)),'r--') title('Cumulative PDF for logistic distribution')

|hist| function is not recommended. Please use |histogram| function instead. And by setting the 'Normalization' option, PDF can be plotted, like: AllRoll = randi(6,50000, 1); SumRoll = sum(AllRoll, 2); Edges = 0.5:1:6.5; histogram(SumRoll,Edges,'Normalization','probability'); title(sprintf('PDF')); xlabel(sprintf('1-6')); ylabel(sprintf('Probability')); grid on hold on <>

You're plotting a normal probability density function which are probabilities between 0 and 1 that sum to 1 (depending on sampling). Check out your norm data. Note the y value at the peak of the curve. % from your code figure edges = [-3:0.2:3]; norm = normpdf(edges, 0,1); plot(edges,norm) Look at the numerical integration which should sum (close to) 1. trapz(edges,norm) Now let's put this into your data and increase the curve's line width do you can see it. You'll see the curve squished down at the bottom because it only reaches a height of ~0.4 while your histogram's height is ~750. % from your code figure N = 10000; M = 30; alpha = 0.7; phi = unifrnd(-pi, pi,1,N); x = cos(phi); y = zeros(1, N); y(1) = x(1); for i = 2:N y(i) = y(i-1)*alpha + x(i); end figure edges = [-3:0.2:3]; p2 = histogram(y, edges); hold on norm = normpdf(edges, 0,1); plot(edges, norm, 'LineWidth', 4) hold off Solutions histfit It's unclear what the goal is but if you want to fit the distribution to a normal curve use histfit. Note that histfit only allows you to enter the number of bins rather than the bin edges so the histogram may differ from the one you generated if the bins are not the same. figure() hf = histfit(y, 30); % for 30 bins fitdist Alternatively you could fit the distribution with your specified bin edges and plot the curve yourself using fitdist. In this example, the amplitude of the normal curve is set to the max bin count. hcounts = histcounts(y,edges); % get bin counts pd = fitdist(y(:), 'normal'); % normal distribution params (mean, sigma) xd = linspace(min(edges),max(edges),100); yd = max(hcounts)*exp(-(((xd-pd.mean).^2)/(2*pd.sigma.^2))); % new y-values (use any x vals) figure() p2 = histogram(y, edges); hold on plot(xd, yd, 'r-','LineWidth',3)

The histogram is a good numerical estimate for the PDF sampled a the bin centers. Just scale it so that the sum over all bins is 1.

The quality of the estimate depends on number of samples and "good" choice of the bins. The latter one is a trade off: smaller bins give you better amplitude resolution and wider bins make the PDF in each band more accurate.

In your case this looks fairly uniformly distributed. The two outer bins are lower than the rest which could be real, but it could also be caused by the bins extending past the max and the min of a signal. It's important how to line up the bins with the min, mean & max of the signal.

To reiterate the comments, you’ll need to plot your projections separately from the histogram; this just isn’t a functionality that histogram2 has. Also, your Delta_x and Delta_y should be calculated with the point being plotted, not with the points in your dataset.

N=5000;
x=randn(N,1); %along-track direction [Nx1]
y=randn(N,1); %cross-track direction [Nx1]

mu_x=mean(x);
mu_y=mean(y);
Delta_x= @(x)x-mu_x;
Delta_y= @(y)y-mu_y;
rho_xy= 0; %termine di correlazione ipotizzato nullo

% STD calcolate dalla sola posizione geometrica dei punti di caduta
sigma_x= sqrt(1/N*sum((x-mu_x).^2));
sigma_y= sqrt(1/N*sum((y-mu_y).^2));

 %valutazione pdf
term1 = 1/((2 * pi* sigma_x * sigma_y )*(sqrt(1-rho_xy^2)));
term2 = -1/(2*(1-rho_xy^2));
term3 = @(x)(Delta_x(x) / sigma_x) .^ 2;
term4 = @(y)(Delta_y(y) / sigma_y) .^ 2;
term5 = @(x,y)-2 *rho_xy*Delta_x(x).*Delta_y(y)/(sigma_x*sigma_y);
Pdf = @(x,y)term1 * exp(term2 * (term3(x) + term4(y) + term5(x,y)));

h=histogram2(x,y);

ax=h.Parent;
numPts=100;
xProjected = linspace(ax.XLim(1),ax.XLim(2), numPts);
yProjected = linspace(ax.YLim(1),ax.YLim(2), numPts);

zOverX = Pdf(xProjected,mu_y);
zOverX = zOverX/max(zOverX)*max(h.Values,[],'all');
zOverY = Pdf(mu_x,yProjected);
zOverY = zOverY/max(zOverY)*max(h.Values,[],'all');

hold on
plot3(xProjected,repmat(ax.YLim(2),1, numPts),zOverX)
plot3(repmat(ax.XLim(2),1, numPts),yProjected,zOverY)
hold off