Consider the two parameters Weibull distribution $W(\alpha, \beta)$ and its cumulative distribution function: $$F(x;\alpha,\beta) = 1 - \exp[-(x/\beta)^{\alpha}]$$
Consider the random variable $T \sim W(\alpha, \beta) $ and the linearly derived random variable $Z = \gamma T$ with $\gamma \in \mathbb{R}_+^{*}$.
What is the probability distribution of the random variable $Z$ ?
I tried this : as $Z = \gamma T$ then $Pr(Z \le x) = Pr(\gamma T \le x) = Pr(T \le x/\gamma ) $ (because $\gamma > 0$) then we get $Pr(Z \le x) = 1 - exp [ -(x/(\gamma \beta))^{\alpha} ]$.
So the cumulative distribution of $Z$ would be : $1 - exp [ -(x/(\gamma \beta))^{\alpha} ]$ ?
Conclusion : $Z \sim W(\alpha, \gamma \beta)$ ?