I'm trying to superimpose an exponential distribution to a uniform distribution, and it's not coming out right. Why? I was told I can transform from uniform to exponential by using the equation Y = − ln X/λ. Am I doing it wrong?
I'm definitely doing the exponential distribution wrong, am I?
5 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
N = 1e4;
a=0; b=1;
x = a+(b-a)*rand([N,1]);
lambda = 1;
Y = -log(x)/lambda;
figure(1); clf
histogram(x,20,'Normalization','pdf');
hold on;
histogram(Y,20,'Normalization' , 'pdf')
hold off
xlabel('random variable $x$','Interpreter','latex','FontSize',20)
ylabel('Probability density','Interpreter','latex','FontSize',20)
title('Continuous uniform PDF','Interpreter','latex','FontSize',20)
Antworten (2)
VBBV
am 20 Sep. 2022
Bearbeitet: VBBV
am 20 Sep. 2022
N = 1e4;
a=0; b=1;
x = (a+(b-a)*rand([N,1]));
lambda = 11; % try with different lambda values
Y = -log(x)/lambda;
figure(1); clf
histogram(x,20,'Normalization','pdf');
hold on;
histogram(Y,20,'Normalization' , 'pdf')
hold off
xlabel('random variable $x$','Interpreter','latex','FontSize',20)
ylabel('Probability density','Interpreter','latex','FontSize',20)
title('Continuous uniform PDF','Interpreter','latex','FontSize',20)
5 Kommentare
Chunru
am 20 Sep. 2022
% if X is uniform on [0,1] then −loge(X) follows an exponential distribution with parameter 1
N = 1e5;
a=0; b=1;
x = a+(b-a)*rand([N,1]);
lambda = 1;
Y = -log(x)/lambda;
figure(1); clf
histogram(x,100,'Normalization','pdf');
hold on;
histogram(Y,100,'Normalization' , 'pdf')
xx = 0:0.1:10;
plot(xx, pdf('Uniform', xx, 0, 1), 'r--', 'Linewidth', 2);
plot(xx, pdf('Exponential', xx, 1), 'b--', 'Linewidth', 2);
xlim([0 10])
hold off
xlabel('random variable $x$','Interpreter','latex','FontSize',20)
ylabel('Probability density','Interpreter','latex','FontSize',20)
title('Continuous uniform PDF','Interpreter','latex','FontSize',20)
legend('Hist-unif', 'Hist-exp', 'Unif', 'Exp')
3 Kommentare
Chunru
am 20 Sep. 2022
Yes. Your code is correct. I just change the number of samples and bin number to make the hist closer to the ideal pdf.
Siehe auch
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!