Q. 5.25 - A First Course in Probability

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.

Step-by-Step Solution

Answer

The above statement is proved.

Step 1 : Given information

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

Step 2 : 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 Y will follow exponential distribution with parameter$𝜆=1$