Q:

Construct a symmetric $100(1-a)\%$ confidence interval for $\theta$

Suppose $Y = T - \theta$, where $T$ is a sufficient statistic for $\theta$, with distribution function:
$$
F_Y (y) = 1-e^{-yn}
$$
To construct a $100(1-a)\%$ confidence interval, we must find two constants $c_1,c_2$ such that:
$$
\begin{cases} F_Y(c_2) = \frac{a}{2} \\ 1-F_Y(c_1) = \frac{a}{2}  \end{cases}
$$
But after solving this system, I yield:
$$
c_1 = -\frac{\ln(\frac{a}{2})}{n} = c_2
$$
which indicates that the interval is of zero length. Is this possible or did I make some calculation or logical error during the process?

A:

$F_Y(c_2) = \frac{a}{2}$ gives  $1-e^{-c_2n}=\frac{a}{2}$ and so $$c_2 = -\frac1n\log_e\left(1-\frac{a}{2}\right)$$
$1-F_Y(c_1) = \frac{a}{2}$  gives $e^{-c_1n}=\frac{a}{2}$ and so $$c_1 = -\frac1n\log_e\left(\frac{a}{2}\right)$$

These are not the same unless $a=1$.  Your error was in the $c_2$ case

