Pivot quantity and confidence interval

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 $$\theta$$.