Show that C gives a confidence interval for θ

[mathmari]

Hey!

For $n \in \mathbb{N}$ we consider the discrete statistical product model $(X, (\mathbb{P}_{\theta})_{\theta\in \Theta})$ with $X=\mathbb{N}^n$, $\Theta=\mathbb{N}$ and $p_{\theta}(x_i)=\frac{1}{\theta}1_{\{1\leq x_i\leq \theta\}}$ forall $x_i\in \mathbb{N}$, $\theta\in \Theta$.

Let $\alpha\in (0,1)$.

We define $$C(x)=\left \{\theta\in \Theta: \max (x_1, \ldots , x_n)\leq \theta\leq a^{-1/n}\cdot \max (x_1, \ldots , x_n)\right \}$$

(a) Show that $C$ gives a confidence interval for $\theta$ to the confidence probability $1-\alpha$.

(b) We have observed the following sample $x$ : $$2 \ \ \ \ 17 \ \ \ \ 44 \ \ \ \ 4 \ \ \ \ 16 \ \ \ \ 24 \ \ \ \ 32 \ \ \ \ 26 \ \ \ \ 21 \ \ \ \ 1 \ \ \ \ 24 \ \ \ \ 6$$
For $x$, calculate a confidence interval for $\theta$ with a confidence probability of $0.95$.
Could you give me a hint for (a) ? :unsure: For (b) I have done the following :

Do we use the part (a) here ? We have that $\alpha=0.05$ then we get the confidence interval \begin{align*}C&=\left \{\theta\in \Theta: \max (2 , 17 , 44 , 4, 16 , 24 ,32 , 26 ,21 , 1 , 24 ,6)\leq \theta\leq 0.05^{-1/12}\cdot \max (2 , 17 , 44 , 4, 16 , 24 ,32 , 26 ,21 , 1 , 24 ,6)\right \}\\ & =\left \{\theta\in \Theta: 44\leq \theta\leq 0.05^{-1/12}\cdot 44\right \}\\ & \approx \left \{\theta\in \Theta: 44\leq \theta\leq 56.47703\right \}\end{align*} Is that correct ? :unsure:

 

[mathmari]

For (a) we have to show that $P_\theta(\theta \in C(X)) \geq 1 - \alpha$, right?
Let $Y : =\max (x_1, \ldots , x_n)$ then do we have that $P_{\theta}(Y\leq \theta\leq \alpha^{-1/n}\cdot Y)=P_{\theta}(\theta\leq \alpha^{-1/n}\cdot Y)-P_{\theta}(\theta<Y)$ ?
Does it hold that $P_{\theta}(\theta\leq \alpha^{-1/n}\cdot Y)=P_{\theta}(\theta\cdot \alpha^{1/n}\leq Y)=P_{\theta}(Y\geq \theta\cdot \alpha^{1/n})=1-P_{\theta}(Y< \theta\cdot \alpha^{1/n})=1-p_{\theta}(\theta\cdot \alpha^{1/n})$ and $P_{\theta}(\theta<Y)=P_{\theta}(Y>\theta)=1-P_{\theta}(Y\leq \theta)=1- p_{\theta}(\theta)$ ?

:unsure: