Pivotal quantity of Weibull distribution

Question

If I have $X_{1},\ldots,X_{n}$ a random sample from a Weibull distribution $X\sim WEI(\theta,2)$.How can I show that $Q=2\sum\limits_{i=1}^n X_{i}^2/\theta^2\sim \chi^2(2n)$.

I have not learnt any transformations for Weibull distributions. I believe that if it has a squared term is because it got to be standar normal somehow and then became a chi-squared. The pdf of Weibull is similar to the exponential one, but that did not help. I also try to use the Jacobian to make the transformation but that sum stopped me.

Answer 1

First of all, you posted this same question earlier and it was closed because you didn't show what you tried. Adding a little paragraph about your thoughts on the question doesn't really address this. You also didn't tag the question as homework.

No. The standard normal does not play a role here.

1. What is the CDF of the Weibull distribution with scale parameter $\theta$ and shape parameter $2$?
2. So if $X \sim {\rm Weibull}(\theta,2)$, what is the probability $\Pr[(X/\theta)^2 \le x]$? Note that since $X$ is already a nonnegative random variable, this is simply $\Pr[X \le \theta\sqrt{x}]$.
3. From the above, what can you conclude about the resulting distribution of each $(X_i/\theta)^2$? Does it depend on the parameter $\theta$? Do you recognize the distribution?
4. What do you remember about the sum of $n$ IID such random variables? What is the PDF of $Q/2$? Then, what is the PDF of $Q$ itself?
5. What is the PDF of the chi-squared distribution with $2n$ degrees of freedom?

When you have done this and shown your effort--that means showing us actual calculations, then you might get more of a response.