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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?

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    $\begingroup$ I don't believe your easy answer because as $b$ increases your answer increases without bound, whereas that integral cannot exceed $1.$ Once you fix that, read stats.stackexchange.com/a/179909/919 and use its approach to argue that $b=c.$ $\endgroup$ – whuber Aug 17 '20 at 18:03
  • $\begingroup$ The distribution of $X$ is Laplace with median $\theta.$ The parameter $\theta$ has no distribution. // From $\theta -c \le X \le \theta + b$ you can show $X-b \le \theta \le X+c.$ To untangle your integral look at parts below $\theta$ and above $\theta$ separately, thus giving proper treatment to $|x-\theta|$ in the PDF. $\endgroup$ – BruceET Aug 17 '20 at 18:33
  • $\begingroup$ @BruceET I have fixed the integral and get $\frac{1}{2}(e^{-2\theta}-e^{-2\theta-b}+e^{-c}+1)$ Is it correct? $\endgroup$ – ohisamadaigaku Aug 17 '20 at 18:46
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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)].

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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:

$$\begin{align} \mathbb{P}(\theta \in \text{CI}(X)) &= \mathbb{P}(X-b \leqslant \theta \leqslant X+c) \\[6pt] &= \mathbb{P}(\theta-c \leqslant X \leqslant \theta+b) \\[6pt] &= \int \limits_{\theta-c}^{\theta+b} \text{Laplace}(x|\theta,1) \ dx \\[6pt] &= \frac{1}{2} \int \limits_{\theta-c}^{\theta+b} e^{-|x-\theta|} \ dx \\[6pt] &= \frac{1}{2} \Bigg[ \ \int \limits_{\theta}^{\theta+b} e^{-x+\theta} \ dx - \int \limits_{\theta}^{\theta+c} e^{-x+\theta} \ dx \Bigg] \\[6pt] &= \frac{1}{2} \Bigg[ (1-e^{-b}) - (1-e^{-c}) \Bigg] \\[6pt] &= \frac{e^{-c} - e^{-b}}{2}. \\[6pt] \end{align}$$

(Observe that, unlike your result, this approaches one when $b \rightarrow \infty$ or $c \rightarrow \infty$.) Thus, finding the optimal confidence interval of this form requires you to solve the following optimisation problem:

$$\text{Minimise } b+c \quad \text{ subject to } \quad e^{-c} - e^{-b} = 2(1-\alpha).$$

With a bit of work, it should be possible for you to show that the optima occurs when $b=c$, so that the optmial confidence interval is one with midpoint at $x$. This is unsurprising, given that the Laplace distribution is symmetric around the mean parameter $\theta$.

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