Asymptotic distribution of MLE

Question

Let $X$ be a random variable such that it's density is given by
$$
f_X(x) =c(\theta)I_{\{|x|\leq \theta\}}\left(1-e^{-|x|}\right)
$$
Find the $c(\theta)$ that makes this function a density, find the maximum likelihood estimator of $\theta$ and the asymptotic distribution of $n(\hat{\theta}-\theta)$.

Im having trouble with the last item in the question.
My whole attempt goes like this

$$
c(\theta)\int I_{\{|x|\leq \theta\}}\left(1-e^{-|x|}\right)dx =1 \\
c(\theta)\int_{-\theta}^{\theta} \left(1-e^{-|x|}\right)dx = 1
$$
Using the fact that the function inside the integral is even, $f(x)=f(-x)$, we get
$$
c(\theta)2\int_{0}^{\theta} \left(1-e^{-x}\right)dx = 1 \\
c(\theta)2(x+e^{-x})|_{0}^{\theta}=1 \\
c(\theta) = 2^{-1}(\theta+e^{-\theta}-1)^{-1}
$$

So we get the following likelihood
$$
2^{-n}(\theta+e^{-\theta}-1)^{-n} I_{\{\max_{i \in \{1,\dots,n\}}|x|\leq \theta\}}\prod_{i=1}^{n}\left(1-e^{-|x|}\right)
$$
We can see that the likelihood is a decreasing function of $\theta$ as
$$
\frac{d}{d\theta}(\theta+e^{-\theta}-1)= 1-e^{-\theta}
$$
This is positive for values greater than $0$ so the function is increasing and thus the likelihood is decreasing in $\theta$ making $\max_{i \in \{1,\dots,n\}}|x|$ the MLE of $\theta$.

Finally for the last part, I used the fact that
$$
P(|X|<x)=P(-x<X<x)
$$
and that the distribution of the maximum is
$$
P(X_{(n)}<x)=P^{n}(X<x)
$$
Getting the following CDF
$$
F_{\hat{\theta}}(x)=\left(\frac{x-1+ e^{-x}}{\theta+e^{-\theta}-1}\right)^n
$$
I tried calculating this probability to see if I notice something about it, however it led me nowhere
$$
P(n(\hat{\theta}-\theta)<x) = P(\hat{\theta} < xn^{-1}+\theta) =\left(\frac{xn^{-1}+\theta-1 +\exp(xn^{-1}+\theta)}{\theta + e^{-\theta}-1}\right)^n
$$
I would appreciate help.

Answer 1

You can use your cdf $F_{\hat{\theta}}$ to get the pdf for $\hat{\theta}$ (which you should notice is supported on $[0,\theta]$) by taking a derivative: $$f_{\hat{\theta}}(x)=n\Bigg(\frac{x+e^{-x}-1}{\theta+e^{-\theta}-1}\Bigg)^{n-1}\Bigg(\frac{1-e^{-x}}{\theta+e^{-\theta}-1}\Bigg)$$ Since $\hat{\theta}$ is supported on $[0,\theta]$ we have that $n(\hat{\theta}-\theta)$ is supported on $[-n\theta,0]$. Using the fact that $f_{n(\hat{\theta}-\theta)}(x)=\frac{1}{n}f_{\hat{\theta}}\big(\frac{x}{n}+\theta\big)$ we get $$f_{n(\hat{\theta}-\theta)}(x)=\Bigg(\frac{\frac{x}{n}+\theta-e^{-\big(\frac{x}{n}+\theta\big)}-1}{\theta+e^{-\theta}-1}\Bigg)^n\Bigg(\frac{1-e^{-\big(\frac{x}{n}+\theta\big)}}{\frac{x}{n}+\theta+e^{-\big(\frac{x}{n}+\theta\big)}-1}\Bigg)$$ The support of $f_{n(\hat{\theta}-\theta)}$ encompasses the entire negative real axis as $n \longrightarrow \infty$. The right hand term in the above expression approaches $\frac{1-e^{-\theta}}{\theta+e^{-\theta}-1}$ as $n$ gets large. Using the substitution $m=x/n$ along with L'Hopital's rule, you can easily show that $$\Bigg(\frac{\frac{x}{n}+\theta-e^{-\big(\frac{x}{n}+\theta\big)}-1}{\theta+e^{-\theta}-1}\Bigg)^n \longrightarrow \exp \Bigg\{\frac{x\big(1-e^{-\theta}\big)}{\theta+e^{-\theta}-1}\Bigg\}$$ Finally the limiting distribution of $\hat{\theta}$ is the pdf $f$ defined by $$f(x)=\frac{1-e^{-\theta}}{\theta+e^{-\theta}-1}\exp \Bigg\{\frac{x\big(1-e^{-\theta}\big)}{\theta+e^{-\theta}-1}\Bigg\} $$ which is supported on $(-\infty,0]$. In other words, the limiting distribution of $\hat{\theta}$ is $-\text{Exponential}\Bigg(\frac{1-e^{-\theta}}{\theta+e^{-\theta}-1}\Bigg)$.