# Finding pivotal quantity of Weibull distribution using transformation(?)

I need to prove that the pivotal quanitity of the sample $$X_1,...,X_n$$ from Weibul distribution $$X\sim WEI(\theta,2)$$ is

$$Q=2\sum_{i=1}^n X_i^2/\theta^2\sim\chi^2_{(2n)}$$

I realized that the $$\theta$$ is a scale parameter, so I used $$Q=\theta_{mle}/\theta$$ to find Q. And I got $$\theta_{mle}^2/\theta^2$$ And I am stuck. I know that to prove that it follows Chi-Squared distriubtion, I need to somehow get to $$2n\bar X/\theta$$. However, I have $$\sum_{i=0}^n x_i^2$$ not X. Any idea? Thanks!

P.S. The answer provided by my prof.(image below) used transformation. However, I cannot see why that is necessary.

1. Start by showing (using the factorization theorem) that the minimal sufficient statistic for $\theta$ is $\sum X_i^2$.
2. As they showed, find the distribution of it. I.e., $X_i^2 \sim \mathcal{E}xp(1/\theta^2)$, thus $\sum_{i=1}^n X_i^2 \sim Gamma (n, 1/\theta^2)$.
3. Now, you have to find a quantity that contains $\theta$ but its distribution is independent of it, i.e., scale to $\chi^2$: $$\frac{2\sum X_i^2}{\theta^2} \sim \chi^2(2n).$$