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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 $(1-\alpha)^3$ ?

alvaro_95
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  • $C \neq (1 - \alpha)^3$ in general. For example, assume that $\theta = (0,0,0)$ or $\theta = (1,1,1)$ with 50% chance for each to occur. In both parameter models the observation has $\beta$ chance to be A. You observe A. Then $\theta_i \in ]-0.5 , 0.5[$ with 50% chance; however, $\theta \in ]-0.5 , 0.5[ \times ]-0.5 , 0.5[ \times ]-0.5 , 0.5[$ also with 50% chance. If you impose more conditions, you could get a stricter answer. – Alexander Geldhof May 13 '19 at 09:39
  • So the chance is $ \le (1-\alpha)^3$, that sounds reasonable. What is the chance at minimum i don't get any results there.. – alvaro_95 May 13 '19 at 09:42
  • Since you multiply the individual probabilities to get $(1-\alpha)^3$, note that $P(A_1\cap A_2\cap A_3)=P(A_1)P(A_2)P(A_3)$ does not necessarily hold (unless of course $A_i$'s are mutually independent events). – StubbornAtom May 13 '19 at 10:09
  • The chance is NOT smaller than $(1 - \alpha)^3$. The example I gave disproves that. – Alexander Geldhof May 13 '19 at 12:42

1 Answers1

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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

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

StubbornAtom
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