I'm definitely doing the exponential distribution wrong, am I?
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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)
1 Comment
Kevin Nelson
on 20 Sep 2022
Answers (2)
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 Comments
VBBV
on 20 Sep 2022
try with different lambda values
Kevin Nelson
on 20 Sep 2022
Kevin Nelson
on 20 Sep 2022
VBBV
on 24 Oct 2022
Please accept the answer if it worked for you
% 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 Comments
Kevin Nelson
on 20 Sep 2022
Chunru
on 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.
Kevin Nelson
on 20 Sep 2022
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