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$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?

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Multiply both parts by sum of squares and then multiply by two: $$ \mathbb P\left(\theta<a\frac{n}{\sum_{i=1}^n x_i^2}\right)=\mathbb P\left(\theta \sum_{i=1}^n x_i^2 < an\right) = \mathbb P\left(2\theta \sum_{i=1}^n x_i^2 < 2an\right)=0.1 $$ 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: $$ \mathbb P(\chi^2_{2n}<h_{0.1})=0.1 $$ 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 $$\left(\frac{h_{0.1}}{2n}T,\,\frac{h_{0.9}}{2n}T\right)=\left(\frac{h_{0.1}}{2\sum_{i=1}^n x_i^2},\,\frac{h_{0.9}}{2\sum_{i=1}^n x_i^2}\right)$$

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