2
$\begingroup$

Given $X\sim \text{Weibull}(\lambda,k)$, generate samples from the Weibull distribution using the inverse transform.

We know $F_X(x) = 1-\text{e}^{-(x/\lambda)^k}$ for $x\ge 0$ with $\lambda,k > 0$.

$\endgroup$
2
$\begingroup$

Start with the CDF, replace $F_X(x)$ with $U\sim U(0,1)$, $X$ for $x$, and solve for $X$.

$$\begin{align} U &= 1-\text{e}^{-(X/\lambda)^k} \\ 1-U &=\text{e}^{-(X/\lambda)^k} \\ -\text{ln}(1-U) &= \left(\frac{X}{\lambda}\right)^k \\ \left[-\text{ln}(1-U)\right]^{\frac{1}{k}} &= \frac{X}{\lambda} \\ X &= \lambda\left[-\text{ln}(1-U)\right]^{\frac{1}{k}} \quad \quad \square \end{align}$$

% MATLAB 2017a
% Inverse Transform for Weibull distribution
% Parameters
lambda = 1.5;
k = 2;
n = 1000;                             % number of samples to generate

% Generation
U = rand(n,1);                        % U ~ Uniform(0,1)
X = lambda*((-log(1-U)).^(1/k));      % X ~ Weibull(lambda,k)
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.