I am stuck at an exercise on a specific likelihood ratio test, I cannot figure out how the size of the test would be $\chi^2_1$ distributed.
Problem description
We consider a sample $X_1,...,X_n$ from the distribution with probability density $p_\theta(x)=2\theta x e^{-\theta x^2}\cdot1_{(0,\infty)}(x)$.
We want to test $H_0:\theta=\theta_0$ against $H_1:\theta\neq\theta_0$.
First, I could show that the likelihood ratio test-statistic was $\lambda_n=(\theta_0\overline{X^2})^{-n}e^{n(\theta_0\overline{X^2}-1)}$ (which is also the right answer according to my book). I have to show that the critical region for the likelihood ratio test at level $\alpha_0$ is $K=\{(X_1,...,X_n):2\log\lambda_n\geq\chi^2_{1,1-\alpha_0}\}$ where $\chi^2_{1,1-\alpha_0}$ is the $1-\alpha_0$ quantile of a $\chi^2$ distribution with 1 df.
Attempt at a solution
This is how far I get. First, we reject $H_0$ whenever the size is smaller than the level of the test: $$\alpha\leq\alpha_0$$ $$\sup_{\theta\in\Theta_0}P(\lambda_n\geq c_{\alpha_0})\leq\alpha_0$$ $$\sup_{\theta=\theta_0}P\left(2\log((\theta_0\overline{X^2})^{-n}e^{n(\theta_0\overline{X^2}-1)})\geq 2\log(c_{\alpha_0})\right)\leq\alpha_0$$ $$1-P\left(2\log((\theta\overline{X^2})^{-n}e^{n(\theta\overline{X^2}-1)})\leq 2\log(c_{\alpha_0})\right)\leq\alpha_0$$ $$1-P\left(-2n\log(\overline{X^2}\theta)+2n(\overline{X^2}\theta-1)\leq 2\log(c_{\alpha_0})\right)\leq\alpha_0$$ $$1-\alpha_0\leq P\left(-2n\log(\overline{X^2}\theta)+2n(\overline{X^2}\theta-1)\leq 2\log(c_{\alpha_0})\right)$$.
The answer would suggest that the random variable $-2n\log(\overline{X^2}\theta)+2n(\overline{X^2}\theta-1)$ is (almost) $\chi^2_1$-distributed. But I absolutely can't see how...
I'd be happy to get some help on this.