I am working with the Weibull model with this pdf:enter image description here

The standard pdf that the some R functions perform is : enter image description here

I am required to find the median, which in the standard parameterisation is $\lambda (\ln2)^{\frac{1}{k}}$. I have worked out the median of my given pdf to be $\lambda^{-\alpha}(\ln 2)^{\frac{1}{\alpha}}$.

So here the shape k is $\alpha$; let $\lambda$ be the standard scale, the scale in my pdf can be denoted as $\lambda'$. So $\lambda'= \lambda^{-k}= \lambda^{-\alpha}$. Please check my reparameterisation for me. I have obtained nonsensical results with those $\alpha, \lambda$ values.

  • $\begingroup$ if you calculated via the integral, can you share how you did it? Or did you just use reparam.? $\endgroup$
    – gunes
    Aug 30, 2020 at 7:32

1 Answer 1


Your calculation is wrong. $\lambda'$ is not a scale parameter, so the scale is not $\lambda'$.

The scale is $\lambda$ - i.e. $(\lambda')^{-1/\alpha}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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