Say we have a beta random variable with pdf, $f_X (x) = e^{-x/\theta}/\theta$ for positive $x $. Find the exact confidence interval of $\theta$ with 95% confidence.
A solution to this says
$$\mathbb{P}(a \leq \frac{X}{\theta} \leq b) \geq 0.95$$ where $c_i $ is real.
Then the answer falls out after when the cdf of $X/\theta $ is found.
My question is why that inequality came about? Why does it consider a transformed random variable? I thought it would be
$$\mathbb{P}(c \leq \theta \leq d) \geq 0.95$$ where $c $ and $d $ are functions with respect to $X $.
Thanks
Typographical errors: Your random variable has an exponential (not beta) distribution with mean $\theta.$ Also, I don't see the relevance of $c_i.$
Confidence interval of exponential data
Then $Y = X/\theta$ has an exponential distribution with mean 1, and CDF $F_Y(y) = P(Y \le y) = 1 - e^{-y},$ for $y > 0.$ This makes it easy to find numerical values for $a$ and $b.$
Then you are correct, that you need to solve (some books say 'pivot') to isolate $\theta$ between the two confidence limits: $$P(a \le X/\theta \le b) = P\left(\frac{1}{b} \le \frac{\theta}{X} \le \frac{1}{a}\right) = P\left(\frac{X}{b} \le \theta \le \frac{X}{a}\right) = 0.95.$$
Note: A more useful confidence interval (CI) is for data $X_1, X_2, \dots, X_n$, where the $X_i$ are iid $Exp(mean = \theta).$ Then $\bar X$ has a gamma distribution, which can be used to find $a$ and $b,$ and similar manipulation can give a CI for $\theta$ in terms of $\bar X.$