Asymptotic distribution of $ \sqrt{n} (θ_n - θ)$ then CI of $θ$

Question

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

Answer 1

I'm assuming that this is correct (haven't checked the stuff that leads to it): $$\Pr(-\Phi(1 - \frac{\alpha}{2}) \leq \sqrt{n}(θ_n*θ - θ^2) \leq \Phi(1 - \frac{\alpha}{2})) \xrightarrow[]{n\rightarrow \infty} 1 - \alpha$$

I won't do the dirty work, but your confidence interval has the form $$ -c\le -x^2+ax \le c, $$ with constants $a, c>0$, and you want to determine a value range of $x$ for which the above inequalities hold. $f(x)=-x^2+ax$ is a squared function that goes to $-\infty$ if $x$ goes to $\pm\infty$. $f(x)=-c$ will have two solutions, which are the bounds of your confidence interval (outside these, $f(x)$ will be smaller), let's say $x_1<x_2$. $f(x)=c$ may have no or one solution (in which case the confidence interval is in fact $[x_1,x_2]$), or it may have two solutions, $x_3<x_4$, say. In this situation, the interval between the two solutions has $f(x)$ too large and can be excluded from the bigger interval, meaning that you have a confidence region, $[x_1,x_3]\cup[x_4,x_2]$, rather than a single interval (although $[x_1,x_2]$ will still respect the confidence level and as such can be seen as valid confidence interval, but it is not as precise as it could be by containing some parameters that don't have to be there).

So now you only need to solve the quadratic equations and plug your $n, \theta_n,$ and $\alpha$ in.

Answer 2

1st method: plug-in

by theorem of Slutsky, we replace $θ$ by $θ_{n}$, so we got:

$\Pr(-\Phi(1 - \frac{\alpha}{2}) \leq \sqrt{n}(θ_n^2 - θ*θ_n) \leq \Phi(1 - \frac{\alpha}{2})) \xrightarrow[]{n\rightarrow \infty} 1 - \alpha$

Then by rearranging terms,

$\Pr(-\frac{\Phi(1 - \frac{\alpha}{2})}{\sqrt{n}θ_n} + θ_n \leq θ \leq \frac{\Phi(1 + \frac{\alpha}{2})}{\sqrt{n}θ_n} + θ_n) \xrightarrow[]{n\rightarrow \infty} 1 - \alpha$

our CI of $θ$ of level $1 - \alpha$ obtained by first method is:

$[-\frac{\Phi(1 - \frac{\alpha}{2})}{\sqrt{n}θ_n} + θ_n, \frac{\Phi(1 + \frac{\alpha}{2})}{\sqrt{n}θ_n} + θ_n]$

2nd method: brute force?

for $θ > 0$,

$\Pr(-\Phi(1 - \frac{\alpha}{2}) \leq \sqrt{n}(θ_n*θ - θ^2) \leq \Phi(1 - \frac{\alpha}{2})) \xrightarrow[]{n\rightarrow \infty} 1 - \alpha$

we note $θ_n = a, \frac{\Phi(1 - \frac{\alpha}{2})}{\sqrt{n}} = c$

first inequality: $-θ^2 + aθ + c \geq 0$, solve this we have $θ_1 = \frac{a -\sqrt{a^2+4c}}{2}$, $θ_2 = \frac{a +\sqrt{a^2+4c}}{2}$

seconde inequality: $-θ^2 + aθ - c \leq 0$, solve this we have $θ_3 = \frac{a -\sqrt{a^2-4c}}{2}$, $θ_4 = \frac{a +\sqrt{a^2-4c}}{2}$

our CI is $([θ_1, θ_3] \cup [θ_4, θ_2]) \cap (0, + \infty)$, notice that $θ_1 < 0$, so it turns to be $[0, θ_3] \cup [θ_4, θ_2]$, finally, we obtain a CI:

$(0, \frac{θ_n -\sqrt{θ_n^2-4\frac{\Phi(1 - \frac{\alpha}{2})}{\sqrt{n}}}}{2}] \cup [\frac{θ_n +\sqrt{θ_n^2-4\frac{\Phi(1 - \frac{\alpha}{2})}{\sqrt{n}}}}{2}, \frac{θ_n +\sqrt{θ_n^2+4\frac{\Phi(1 - \frac{\alpha}{2})}{\sqrt{n}}}}{2}]$