confidence interval with MLE estimator

$$f_{\theta}(x) = 2 \theta x e^{- \theta x^2}$$ on the interval $$(0, \infty)$$. T is the MLE estimator of $$\theta$$. We construct the confidence interval of $$\theta$$ $$(aT, bT)$$, where $$a$$ and $$b$$ are constants such that: $$P_{\theta}(\theta < aT) = P_{\theta}(\theta > bT) = 0.1$$

I showed that the MLE estimator is: $$T = \frac{n}{\sum_{i = 1}^n x_i^2}$$ but how I can construct this confidence interval? We have:

$$P(\theta < a \frac{n}{\sum_{i=1}^n x_i^2})$$ and I showed that $$X^2$$ has an exponential distribution with parameter $$theta$$. We know that sum of that random variables is $$\Gamma (n, \theta)$$. What we can do now?

Multiply both parts by sum of squares and then multiply by two: $$\mathbb P\left(\theta Note that the l.h.s. $$2\theta \sum_{i=1}^n x_i^2$$ has Gamma distribution $$\Gamma\bigl(n,\frac12\bigr)=\chi^2_{2n}$$ since $$2\theta x_1^2$$ has an exponential distribution with parameter $$\frac12$$. Next you can use $$0.1$$-quantile of chi-square distribution with $$2n$$ degrees of freedom: $$\mathbb P(\chi^2_{2n} and take $$2an=h_{0.1}$$, $$a=\frac{h_{0.1}}{2n}$$.
For $$b$$, $$0.9$$-quantile of chi-square distribution with $$2n$$ degrees of freedom is needed: $$\mathbb P(\chi^2_{2n}>h_{0.9})=0.1$$ and take $$2bn=h_{0.9}$$, $$b=\frac{h_{0.9}}{2n}$$.
Finally, confidence interval will be $$\left(\frac{h_{0.1}}{2n}T,\,\frac{h_{0.9}}{2n}T\right)=\left(\frac{h_{0.1}}{2\sum_{i=1}^n x_i^2},\,\frac{h_{0.9}}{2\sum_{i=1}^n x_i^2}\right)$$