For the pdf $f(x,\theta)=\frac{1}{2\theta}e^\frac{-|x|}{\theta}, -\infty<x<\infty, \theta>0$ derive $Y_i=|X_i|$. This is what I've done:
$Y_i=|X_i| => X_i = \pm Y_i$
Then |J| = 1 where J is the Jacobian
Then I end up with $f(y)=\frac{1}{2\theta}e^\frac{-|y|}{\theta}$
This does not seem right.
For the confidence interval, I found a pivot $\frac{\bar{X}}{2\theta} \sim \chi^2_{2n}$. I am trying to find the confidence interval for $e^\frac{-1}{\theta}$.
I found a confidence interval for $\theta$. It would be $\chi^2_{2n,1-\frac{\alpha}{2}}<\frac{\bar{X}}{2\theta}<\chi^2_{2n,\frac{\alpha}{2}}$, and then rearranging for $\theta$ I get $\frac{\bar{X}}{2\chi^@_{2n,\frac{\alpha}{2}}} < \theta<\frac{\bar{X}}{2\chi^2_{2n,1-\frac{\alpha}{2}}}$
I also need to find a confidence interval for $e^\frac{-1}{\theta}$ and I'm not sure how.
The cumulative probability function of $Y$ is found by $$P(Y\leq y)=P(|X|\leq y)=P(-y\leq X \leq y)=\int_{-y}^y\frac{1}{2\theta}e^{-\frac{|x|}{\theta}}dx=$$ $$=\frac{1}{2\theta}\bigg(\int_0^ye^{-\frac{x}{\theta}}dx+\int_{-y}^0e^{\frac{x}{\theta}}dx\bigg)=\frac{1}{2\theta}\bigg({\theta}(1-e^{-\frac{y}{\theta}})+{\theta}(1-e^{-\frac{y}{\theta}})\bigg)=1-e^{-\frac{y}{\theta}}$$ for $y \in [0,\infty)$. The probability density function of $Y$ is $$\frac{d}{dy}P(Y \leq y)=\frac{1}{\theta}e^{-\frac{y}{\theta}} \ \ \ \ \ y \in[0,\infty)$$ This is an exponential distribution with parameter $\lambda=\frac{1}{\theta}$.
Addendum
As in OP, the confidence interval for $\bar{X}/(2\theta)$ is $$P\bigg(\chi^2_{2n,1-0.5\alpha}<\frac{\bar{X}}{2\theta}<\chi^2_{2n,0.5\alpha}\bigg)$$ By rearranging an then changing sign $$P\bigg(\frac{2}{\bar{X}}\chi^2_{2n,1-0.5\alpha}<\frac{1}{\theta}<\frac{2}{\bar{X}}\chi^2_{2n,0.5\alpha}\bigg)$$ $$P\bigg(-\frac{2}{\bar{X}}\chi^2_{2n,0.5\alpha}<-\frac{1}{\theta}<-\frac{2}{\bar{X}}\chi^2_{2n,1-0.5\alpha}\bigg)$$ Also $-\theta^{-1}\in (-\infty,0)$ so $$P\bigg(\exp\bigg\{-\frac{2}{\bar{X}}\chi^2_{2n,0.5\alpha}\bigg\}<\exp\bigg\{-\frac{1}{\theta}\bigg\}<\bigg\{-\frac{2}{\bar{X}}\chi^2_{2n,1-0.5\alpha}\bigg\}\bigg)$$