I'm definitely doing the exponential distribution wrong, am I?

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

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?

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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

try with different lambda values
Is that how it's supposed to look? Looks odd. I tried 10 and got something similar but it looked odd to me
yes, it looks correct. A continuous uniform distribution looks flatter compared to exponential one, Refer or search any literature for further info
Please accept the answer if it worked for you

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% 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

I'm confused. Are you saying my code is right? Because it initially came out to something similar with 20 bins as opposed to your 100 bins
Yes. Your code is correct. I just change the number of samples and bin number to make the hist closer to the ideal pdf.

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Asked:

on 20 Sep 2022

Commented:

on 24 Oct 2022

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