Suppose we have $X_1,...,X_n$ iid with distribution:
$f(x)=xe^{−(\frac{x^2 − θ^2}{2})},x≥θ, θ > 0$
By calculated the median of $f(x)$, $X_{\frac{1}{2}}$ equals $ \sqrt{θ^2 + \log4}$ we obtain an estimator of $θ$, $θ_n = \sqrt{(X_{\frac{1}{2}}(n))^2 - \log4}$ which $X_{\frac{1}{2}}(n)$ is the empirical median of $X_1,...,X_n$
I got the asymptotic distribution of $\sqrt{n}(X_{\frac{1}{2}}(n) - X_{\frac{1}{2}}) \rightarrow N(0, \frac{1}{θ^2 + \log4})$ using the formula $\sqrt{n}(X_{p}(n) - X_{p}) \xrightarrow[n\rightarrow \infty]{d}N(0, \frac{p(1-p)}{f(X_{p})^2})$, here we got $f(X_{\frac{1}{2}}) = \frac{\sqrt{θ^2+\log4}}{2}$
by applying delta method: $: g(θ) = \sqrt{θ^2 + \log4}$, $g'(θ) = \frac{θ}{\sqrt{θ^2 + \log4}}$,
$\sqrt{n}(θ_n - θ) \xrightarrow[n\rightarrow \infty]{d} N(0, \frac{\frac{1}{θ^2 + \log4}}{g'(θ)^2})$
so:
$\sqrt{n}(θ_n - θ) \xrightarrow[n\rightarrow \infty]{d} N(0, \frac{1}{θ^2})$
Hope i didn't make mistake at my calculation...now i want to obtain an asymptotic confidence interval of $θ$ of level $1-\alpha$, by standardisation,
$\sqrt{n}(\frac{θ_n}{θ} - 1) \xrightarrow[n\rightarrow \infty]{d} N(0, \frac{1}{θ^4})$
I can't go further cause there's no such quantile of $N(0,\frac{1}{θ^4})$
Now I'm stuck, i can't do this either $\sqrt{n}(θ_n*θ - θ^2) \xrightarrow[n\rightarrow \infty]{d} N(0, 1)$ otherwise i'm gonna stuck on rearranging terms $θ_n$ and $θ$.
$\Pr(-\Phi(1 - \frac{\alpha}{2}) \leq \sqrt{n}(θ_n*θ - θ^2) \leq \Phi(1 - \frac{\alpha}{2})) \xrightarrow[]{n\rightarrow \infty} 1 - \alpha$
then
$\Pr(\frac{-\Phi(1 - \frac{\alpha}{2})}{\sqrt{n}} \leq θ_n*θ - θ^2 \leq \frac{\Phi(1 - \frac{\alpha}{2})}{\sqrt{n}}) \xrightarrow[]{n\rightarrow \infty} 1 - \alpha$
then
$\Pr(\frac{-\Phi(1 - \frac{\alpha}{2})}{\sqrt{n}} - θ_n*θ \leq - θ^2 \leq \frac{\Phi(1 - \frac{\alpha}{2})}{\sqrt{n}} - θ_n*θ) \xrightarrow[]{n\rightarrow \infty} 1 - \alpha$
then
$\Pr(\frac{-\Phi(1 - \frac{\alpha}{2})}{\sqrt{n}} + θ_n*θ \leq θ^2 \leq \frac{\Phi(1 - \frac{\alpha}{2})}{\sqrt{n}} + θ_n*θ) \xrightarrow[]{n\rightarrow \infty} 1 - \alpha$
then
$\Pr(\sqrt{\frac{-\Phi(1 - \frac{\alpha}{2})}{\sqrt{n}} + θ_n*θ} \leq θ \leq \sqrt{\frac{\Phi(1 - \frac{\alpha}{2})}{\sqrt{n}} + θ_n*θ)} \xrightarrow[]{n\rightarrow \infty} 1 - \alpha$
I can't go further cause no matter how i try, i couldn't seperate $θ_n$ and $θ$
How can i do this? I checked several times my calculation and i think i'm right till the part of $θ_n$
Update: Thanks to @Lewian, i got the answer, i will put the detail of calculation below