# Anyone seen this parametrization of Weibull?

My lecturer uses a parametrization of Weibull that I can't find any where else so I'm wondering are they mistaken. Can anyone confirm if this is legitimate pdf of a Weibull?

$$\lambda\theta y^{\lambda-1}exp(-\theta y^\lambda)$$ I'm having trouble showing this is in fact a legitimate pdf, that is that it integrates to 1 when taken over positive real line. Can anyone shed some light. I have seen the usual parametrizations of Weibull and how to show they are a pdf. My questions relates only to this specific parametrization .

## 1 Answer

It's certainly a Weibull, just a simple reparametrization of the form on the Wikipedia page for the Weibull distribution:

$$f(y;\lambda,k) =\frac{k}{\lambda}\left(\frac{y}{\lambda}\right)^{k-1}e^{-(y/\lambda)^{k}}\, I_{y>0}$$

Let $\theta=\lambda^{-k}$. Then you get

$$f(y;\theta,k) =k\theta y^{k-1}e^{-\theta y^{k}}\, I_{y>0}$$

Which is of the same form as yours (you simply have $\lambda$ where I have $k$)

To show the form in your question integrates to 1, just use substitution ($x=y^k$).

• Thanks. Would you be able to show how the original is a legitimate pdf? that is it integrates to 1? I thought it looked similar to gamma integral but couldn't get it out. – user24907 May 18 '18 at 13:38
• In the integral, do the obvious substitution $x=(y/\lambda)^k$. Should take you two or three lines. In the original equation from your question, just use $x=y^k$. Both reduce to the integral of an exponential. – Glen_b May 18 '18 at 23:03