Finding pivot random variable of $T\sim \operatorname{Exp}(\theta)$

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I have an exercise in statistical inference theory about finding a $95\%$-confidence interval for an unknown expectation $E(T) = \theta$ where $T \sim \operatorname{Exp}(\theta)$, $f_\theta(t) = \frac{1}{\theta} e^{-\frac{t}{\theta}} 1\{t \geq 0\}$ and we have one observation, $t=1000$.

To do this I need to find a pivot random variable $s = h(\theta, \hat{\theta})$ such that ${h_{\hat{\theta}}}^{-1}(s) = \theta$, where $\hat{\theta}$ is the estimator of $\theta$. I have begun by reasoning that "it seems reasonable that if the expected value of $T$ is $\theta$ then the expected value of $\frac{T}{\theta}$ should be $\frac{\theta}{\theta}=1$". If that is the case, then I have found my pivot random variable and I know how to find the confidence interval.

So far I have found the distribution function of $\frac{T}{\theta}$, $$F_{\frac{T}{\theta}}(t) = P\left(\frac{T}{\theta} \leq t\right) = P(T \leq \theta t) = \int_0^{\theta t} \frac{1}{\theta}e^{-\frac{x}{\theta}} \, dx = [-e^{-\frac{x}{\theta}}]\Bigr\rvert_0^{\theta t} = 1 - e^{-t},$$ for $ t \in [0, \infty)$.

Now I want to calculate,

\begin{align} & E\left(\frac{T}{\theta}\right) = \int x\, dF_{\frac{T}{\theta}}(x) = \int_0^\infty x f_{\frac{T}{\theta}}(x) \, dx \\[10pt] = {} & \int_0^\infty x e^{-x}\,dx =[-xe^{-x}]\Bigr\rvert_0^\infty - \int_0^\infty-e^{-x} \, dx = 0 + [-e^{-x}] \Bigr\rvert_0^\infty = 0-(-1) = 1. \end{align}

It does seem to work with $\frac{T}{\theta}$, however, it is not clear how I proceed from here. In the case of $X\sim N(\theta,\sigma^2)$ I can form the pivot function $h(\theta, \hat{\theta}) = \frac{\hat{\theta}-\theta}{\sigma}\sim N(0,1)$. But for $\frac{T}{\theta}$ how do I obtain my pivot random variable?

Thanks in advance,

Isak

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Your pivot is $T/\theta.$ The fact that its expected value does not depend on $\theta$ is not enough to show that it's a pivot. But for $t\ge0,$ you've shown that $\Pr(T/\theta \le t) = 1-e^{-t}.$ The fact that that does not depend on $\theta$ tells you that $T/\theta$ is a pivotal quantity. That enables you to find numbers $A,B$ such that $\Pr(T/\theta<A) = 0.05/2 = \Pr(T/\theta>B).$ Next, $T/\theta>A$ is equivalent to $T/A >\theta,$ and $T/\theta<B$ is equivalent to $T/B<\theta.$ So you have $$ \Pr\left( \frac T B <\theta< \frac T A \right) = 0.95. $$