I know that if $X \sim \operatorname{Exp}(\theta)$, and $Y=\theta X \implies Y \sim \operatorname{Exp}(1)$.
I want a confidence interval for $E(X)= \frac{1}{\theta}$ and $\operatorname{Var}(X)= \frac{1}{\theta^2}$
The confidence interval for the mean is: $P(a<Y<b)=1-\alpha \implies P(a<\theta X<b)=1-\alpha $ And then $P(\frac{X}{b}<\frac{1}{\theta}<\frac{X}{a})=1-\alpha$.
Can I do this? : $P((\frac{X}{b})^2<(\frac{1}{\theta})^2<(\frac{X}{a})^2)=1-\alpha \ $ is a confidence interval for the variance.
I'm not sure if this is correct.