Weibull Distribution, what is $R^2$?

Question

Given a Weibull Distribution $f_R$, how do I transform $R\to R^2$, and what is the distribution for $R^2$?

Attempt: Since $f_R$ is distributed with parameter $k$ and $h$ as a function of $x$, so . $$f_R=(k/h)*(x/h)^{k-1}*exp(-{x/h}^k)$$

Using the random variable transformation: $g(y) = -f\left[r^{-1}(y)\right] \frac{d}{dy} r^{-1}(y)$,

then $R^2 $ has a distribution of $$k/h*(\sqrt(y)/h)^{k-1}*exp(-(\sqrt(y)/h)^k)*1/\sqrt(y)$$

Answer 1

This is mechanized in Maple: $$with(Statistics): R := RandomVariable(Weibull(k, lambda)): simplify(PDF(R^2, t)); $$ $$ \cases{0&$t\leq 0$\cr 1/2\,{t}^{-1+\lambda/2}\lambda\,{k}^{-\lambda}{{\rm e}^{-{t}^{\lambda/2} \left( {k}^{-1} \right) ^{\lambda}}}&$0<t$\cr} $$ and $$CDF(R^2, t) $$ outputs $$\cases{0&$t\leq 0$\cr -{{\rm e}^{-{t}^{\lambda/2}{k}^{-\lambda}}}+1&$0<t$\cr} $$

Answer 2

Let the Weibull $R$ have parameters $k$ and $\lambda$. Then the cumulative distribution function $F_R(x)$ of $R$ is given by $$F_R(x)=1-e^{-(x/\lambda)^k}$$ (for $x\gt 0$).

Let $Y=R^2$. Then $$F_Y(y)=\Pr(Y\le y)=\Pr(R^2\le y)=\Pr(R\le \sqrt{y})=1-e^{-(\sqrt{y}/\lambda)^k}$$ for $y\gt 0$. For completeness, note that $F_Y(y)=0$ when $y\le 0$. To find the density function $f_Y(y)$, find the derivative of $F_Y(y)$.

Remark: One can also get the density function directly, by the Method of Transformations. We took the cdf approach because (i) we think of it as more fundamental and (ii) the cdf of the Weibull is "nicer" than the density.