# If $Y=\left((X-\nu)/\alpha\right)^\beta$ is standard exponential then $X$ is Weibull$(\nu,\alpha,\beta)$?

Let $\displaystyle Y=\left(\frac{X-\nu}{\alpha}\right)^\beta$ where $\alpha$ and $\beta$ are positive. Show that if $Y$ is an exponential random variable with parameter $\lambda=1$, then $X$ is a Weibull random variable with parameters $\nu$, $\alpha$ and $\beta$.

My attempt:

Let $Y$ be an exponential random variable with parameter $\lambda=1$.

We need to prove that $X$ is a Weibull random variable. For this, $X$ has to be greater than $\nu$. (For a Weibull random variable, $F(\nu)=0\implies P\{X\le\nu\}=0\implies X>\nu$) Let us try to prove this.

$F_Y(0)=1-e^0=0$

$\implies P\{Y\le 0\}=0$

$\implies Y>0$

$\displaystyle\implies \left(\frac{X-\nu}{\alpha}\right)^\beta>0$

Now if $\beta$ is an even integer, $X-\nu$ need not be positive, i.e. $X$ can be less than $\nu$. But then $X$ cannot be a Weibull random variable.

Is there a flaw in my reasoning?

The issue you raise is due to non-uniqueness of square roots, for example. I think you should just interpret the equality as $X=\alpha Y^{1/\beta} + \nu$ with the convention that $y \mapsto y^{1/\beta}$ is always nonnegative.
Then, you have $X \ge \nu$ as you were trying to check. From here you can just compute the CDF of $X$. For $x \ge 0$, $$P(X \le x) = P\left(Y \le \left(\frac{x-\nu}{\alpha}\right)^\beta\right) = 1 - e^{- \left(\frac{x-\nu}{\alpha}\right)^\beta}.$$