I am working with the Weibull model with this pdf:
The standard pdf that the some R functions perform is :
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.