Edit: it was a typo. I correted the question and will put my solution to it.
I need to find a confindence interval for a given parameter. Let $X_1 \ldots X_n$ be i.i.d rv.'s with a p.d.f. given by:
$$ f(x|\theta) = \frac{1}{2\theta} e^{-|x|/\theta}$$
where the parameter $\theta \in (0, \infty)$.
Show that $$ 2 \frac{\sum_{j=1}^{n} |X_j|}{\theta} \sim \chi^2_{2n} $$ has a chi-squared distribution so it can be used as a pivot quantity, and create a confidence interval from that.
Let $Z = |X|$. Then:
$$\mathbb{P}(Z\leq t) = \mathbb{P}(-t \leq X \leq t) = \int_{-t}^{t} \frac{1}{2\theta}e^{|x|/\theta}dx = 1-e^{-t/\theta} $$
So the probability density function will be given by:
$$ f_Z(z) = \frac{1}{\theta} e^{-z/\theta} \ \mathbb{I}_{(0,\infty)}(z) $$
From that, we can evaluate the MGF:
$$ \phi_Z(t) = \mathbb{E}(e^{tZ}) = \int_{0}^{\infty} \frac{1}{\theta }e^{tz-z/\theta}dz = \frac{1}{1-t\theta} $$
Which results:
$$ \phi_{\frac{2\sum_{j=1}^{n} Z_j}{\theta}}(t) = \phi_{\sum_{j=1}^{n} Z_j}\left(\frac{2t}{\theta}\right) = \left(1-2t\right)^{-n}$$
And that's precisely the MGF of a $ \chi_{2n}^2 $ distribution, and is a pivot quantity. It can then be used to create a confidence interval for the parameter $\theta$, meaning:
If we choose $a,b \in \mathbb{R}$ such that:
$$ \mathbb{P}(a \leq \chi^{2}_{2n} \leq b) = 1-\alpha $$
then:
$$ \left[\frac{2\sum_{j=1}^{n} |X_j|}{b}; \frac{2\sum_{j=1}^{n} |X_j|}{a} \right] $$
is a confidence interval for the parameter $\theta$.