# How can I find a confidence intervall for $\theta$?

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?

• 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? Commented Jun 7, 2023 at 3:50
• 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, $\mathbb{P}(T \leq q_{\alpha/2}) = \alpha/2$
– Alex
Commented Jun 7, 2023 at 4:45
• @AlexJ sorry there was a tipo in the question. Commented Jun 7, 2023 at 5:51
• @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. Commented Jun 7, 2023 at 5:53
• Commented Jun 7, 2023 at 10:15

$$F_T(T;\theta)$$ 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.