Question

Let $Y=\frac{X-𝜈}{𝛼}$.

Show that if X is a Weibull random variable with parameters ν, α, and β, then Y is an exponential random variable with parameter λ = 1 and vice versa.

Short answer

The above statement is proved.

Step-by-step solution

Given information

We are assuming thatXis a Weibull with parameters$𝜈,𝛼\mathrm{and}𝛽$

Calculation

We have

$$\begin{gathered} P\left\{{\left(\frac{X-𝜈}{𝛼}\right)}^{𝛽}\le x\right\}=P\left\{\left(\frac{X-𝜈}{𝛼}\right)\le x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$𝛽$}\right.}\right\} \\ =P\left(X\le 𝜈+𝛼x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$𝛽$}\right.}\right) \\ =1-e^{-x} \end{gathered}$$

which is nothing but the CDF of exponential distribution.

Therefore it's proved that Ywill follow exponential distribution with parameter$𝜆=1$