What is the value that makes the minimum length of confidence interval? A random variable $X$ follows
$$f(x|\theta)=\frac{1}{2}e^{-|x-\theta|} \quad  -\infty<x<\infty$$
I consider of a confidence interval of $\theta$, $S(X)=[X-b,X+c]$.
When I set confidence level at $1-\alpha$, what are the values of $b$ and $c$ which makes the minimum length of the confidence interval $d=b+c$?

What I have found
The question prior to this asked about the probability of
$${\theta-c \leq X \leq \theta +b}$$
and I easily got the answer $$\int_{\theta -c }^{\theta+b } f(x|\theta) dx=\frac{1}{2}(e^b-e^c)$$
I think if I need a confidence interval of $/theta$, I need to set
$$P(X-b\leq\theta\leq X+c)>1-\alpha$$
but I don't know the PDF of $\theta$. This is where I got stuck.
Can anyone help me?
 A: Since the pdf you provided is a conditional pdf of X under given θ, it is possible to derive the confidence interval (CI) of X under given θ, but not the CI of θ.
Contrarily, if the pdf of f(θ|x) is given by the same expression, then the shortest CI of θ can be derived as S(x) = [x+ln(alfa) x-ln(alfa)].
A: There is a mistake in your probability result (which should be clear by the fact that it is unbounded).  Using the interval $\text{CI}(X) = [X-b, X+c]$ you should have the coverage probability:
$$\begin{align}
\mathbb{P}(\theta \in \text{CI}(X))
&= \mathbb{P}(X-b \leqslant \theta \leqslant X+c) \\[6pt]
&= \mathbb{P}(\theta-c \leqslant X \leqslant \theta+b) \\[6pt]
&= \int \limits_{\theta-c}^{\theta+b} \text{Laplace}(x|\theta,1) \ dx \\[6pt]
&= \frac{1}{2} \int \limits_{\theta-c}^{\theta+b} e^{-|x-\theta|} \ dx \\[6pt]
&= \frac{1}{2} \Bigg[ \ \int \limits_{\theta}^{\theta+b} e^{-x+\theta} \ dx + \int \limits_{\theta-c}^{\theta} e^{x-\theta} \ dx \Bigg] \\[6pt]
&= \frac{1}{2} \Bigg[ \ \int \limits_{0}^{b} e^{-r} \ dr + \int \limits_{-c}^{0} e^{r} \ dr \Bigg] \\[6pt]
&= \frac{1}{2} \Bigg[ (1-e^{-b}) + (1-e^{-c}) \Bigg] \\[6pt]
&= 1 - \frac{e^{-c} + e^{-b}}{2}. \\[6pt]
\end{align}$$
(Observe that, unlike your result, this approaches one when $b \rightarrow \infty$ and $c \rightarrow \infty$.)  Thus, finding the optimal confidence interval of this form requires you to solve the following optimisation problem:
$$\text{Minimise } b+c \quad \text{ subject to } \quad e^{-c} + e^{-b} \leqslant 2 \alpha.$$
With a bit of work, it should be possible for you to show that the optima occurs when $b=c$, so that the optimal confidence interval is one with midpoint at $x$.  This is unsurprising, given that the Laplace distribution is symmetric around the mean parameter $\theta$.
