Let $X$ be a random variable with p.d.f.: $$f(X|\theta) = \frac{e^{x-\theta}}{(1+e^{x-\theta})^2}$$ where $-\infty<x<\infty$ and $-\infty<\theta<\infty$
Use the pivotal method to verify that if $0<\alpha_1<0.5$ and $0<\alpha_2<0.5$, then
$[Y-\log(\frac{1-\alpha_2}{\alpha_2}),Y-\log(\frac{\alpha_1}{1+\alpha_1})]$ is a confidence interval for $\theta$ with converge probability $1-(\alpha_1+\alpha_2)$ here my attempt if I set $Y = X-\theta$ I get this $$\begin{equation} \begin{aligned} F_Y(y) \equiv \mathbb{P}(Y \leqslant y) &= \mathbb{P}({X-\theta} \leqslant y) \\[6pt] &= \mathbb{P}(X \leqslant y+\theta) \\[6pt] &= \int \limits_{-\infty}^{y+\theta} \frac{e^{x-\theta}}{(1+e^{x-\theta})^2} dx \\[6pt] &= \Bigg[ \frac{-1}{1+e^{x-\theta}} \Bigg]_{x=-\infty}^{x=y+\theta} \\[6pt] &= 1-\frac{1}{1+e^{y}}. \\[6pt] \end{aligned} \end{equation}$$ from this I derived the pdf equal to $$\frac{e^y}{(1+e^y)^2}$$
from this I show that the distribution does not depend on the parameter anymore. how can I show the confidence interval for $\theta$
Any help or suggestion would be appreciated.