Cartesian product of confidence intervals

Question

I got three exact CIs $I_1,I_2,I_3$ for parameters $\theta_1,\theta_2,\theta_3$ with confidence $1-\alpha$. Furthermore let $C=I_1 \times I_2 \times I_3$ and the parameter $\theta = (\theta_1,\theta_2,\theta_3)$. What is the confidence of $C$ containing $\theta$ at least respectively at most? Do i get it wrong or is this kind of a trick question because all the three CIs are exact so the confidence should be $(1-\alpha)^3$ ?

Answer 1

Recall the inequality $P\left(\bigcap\limits_{i=1}^n A_i\right)\ge \sum\limits_{i=1}^n P(A_i)-(n-1)$.

By the same logic,

\begin{align} P(\theta\in C)&=P(\theta_1\in I_1,\theta_2\in I_2,\theta_3\in I_3) \\& \ge P(\theta_1\in I_1)+P(\theta_2\in I_2)+P(\theta_3\in I_3)-2 \\&=3(1-\alpha)-2 \\&=1-3\alpha \end{align}

So $C$ is a confidence interval for $\theta$ with confidence coefficient at least $1-3\alpha$.

And since $P\left(\bigcap\limits_{i=1}^n A_i\right)\le \min\{P(A_1),P(A_2),\ldots,P(A_n)\}$, you have

$$P(\theta\in C)=P(\theta_1\in I_1,\theta_2\in I_2,\theta_3\in I_3)\le 1-\alpha$$