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We have given a sample $X_1,...,X_n$ which is distributed with respect to $f(x;\theta)=e^{\theta-x} \Bbb{1}_{\theta\leq x<\infty}$. We want to use the statistic $T:=\min\{X_1,...,X_n\}$ to find a confidence interval for $\theta$ at level $1-\alpha$.

I first computed the distribution of $T$ and got that $$F_T(t;\theta)=1-e^{n(\theta-t)}, t\geq \theta$$ and $$F_T(t;\theta)=0, t<\theta$$We know that from class that $F_T(T,\theta)$ is a pivot. So we can find some quantiles $q_1,q_2$ s.t. $$\Bbb{P}(q_1\leq F_T(T,\theta)\leq q_2)=1-\alpha$$

Now I wanted to take $q_1=q_{\alpha/2}$ and $q_2=q_{1-\alpha/2}$ to be the $\alpha/2$- respectivly $1-\alpha/2$ quantile of $F_T(T,\theta)$. So I wanted to solve $$\begin{align}\Bbb{P}(T\leq q_{\alpha/2})&=\alpha/2\\\Bbb{P}(T\leq q_{1-\alpha/2})&=1-\alpha/2 \end{align}$$ to get this quantiles. But I then get something compleatly different than we got in class. I got for example $$q_{\alpha/2}=\theta-\frac{1}{n}\log(1-\alpha/2)$$

Where is my error?

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  • $\begingroup$ I don't know what your specific error is, but you're trying to solve for $q_{\alpha/2}$ and $q_{1-\alpha/2}$. If they don't appear in your solution, that doesn't seem right. Check if you've accidentally cancelled something out? $\endgroup$
    – Alex J
    Commented Jun 7, 2023 at 3:50
  • $\begingroup$ You've changed from the pivot $F_T(T,\theta)$ to $T$ in the quantiles: you wanted to calculate quantiles for $F_T(T, \theta)$, $\mathbb{P}(q_1 \leq F_T(T, \theta) \leq q_2) = 1 - \alpha$, but then you switch to $T$ for the next formula, $\mathbb{P}(T \leq q_{\alpha/2}) = \alpha/2$ $\endgroup$
    – Alex
    Commented Jun 7, 2023 at 4:45
  • $\begingroup$ @AlexJ sorry there was a tipo in the question. $\endgroup$ Commented Jun 7, 2023 at 5:51
  • $\begingroup$ @Alex Sorry I don't get it do I need to $\Bbb{P}(1-e^{n(\theta-T)}\leq q_{\alpha/2})$? Because this always gives exaclty $q_{\alpha/2}$. But then I mean that this does not help me further. $\endgroup$ Commented Jun 7, 2023 at 5:53
  • $\begingroup$ dup. of stats.stackexchange.com/q/99337/56940 ? $\endgroup$
    – utobi
    Commented Jun 7, 2023 at 10:15

1 Answer 1

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$F_T(T;\theta)$ is a pivot, and it can take values in the range $[0, 1]$. For $u \in [0, 1]$ its CDF is $$\mathbb{P}(F_T(T, \theta) \leq u) = \mathbb{P}\left(T \leq \frac{n\theta - \log(1-u)}{n}\right),$$ so $$\mathbb{P}(F_T(T, \theta) \leq u) =u,$$ as you said. And this is useful because it means that $F_T(T, \theta) \sim U(0, 1)$. For the uniform distribution, the quantile $q_x$ is just equal to $x$ (for $x \in [0, 1]$).

Therefore \begin{align*} &\mathbb{P}\left(\alpha/2 \leq 1 - e^{n(\theta - T)} \leq 1 - \alpha/2\right) = 1 - \alpha\\ &\mathbb{P}\left(\alpha/2 - 1 \leq - e^{n(\theta - T)} \leq - \alpha/2\right) = 1 - \alpha\\ &\mathbb{P}\left(\alpha/2 - 1 \leq - e^{n(\theta - T)} \leq - \alpha/2\right) = 1 - \alpha\\ &\mathbb{P}\left(\alpha/2 \leq e^{n(\theta - T)} \leq 1 - \alpha/2 \right) = 1 - \alpha\\ &\mathbb{P}\left(\log(\alpha/2) \leq n(\theta - T) \leq \log(1 - \alpha/2) \right) = 1 - \alpha\\ &\mathbb{P}\left(\log(\alpha/2) + nT \leq n\theta \leq \log(1 - \alpha/2) + nT \right) = 1 - \alpha\\ &\mathbb{P}\left(\frac{\log(\alpha/2) + nT}{n} \leq \theta \leq \frac{\log(1 - \alpha/2) + nT}{n} \right) = 1 - \alpha.\\ \end{align*} Therefore a $1-\alpha$ confidence interval for $\theta$ is $$\left[\frac{\log(\alpha/2)}{n} + T, \frac{\log(1 - \alpha/2)}{n} + T \right].$$ Note that both ends of this interval are below $T$, which is what we would expect.

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    $\begingroup$ Ah I see thanks a lot! $\endgroup$ Commented Jun 7, 2023 at 11:11

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