# Why do two confidence intervals only overlap when their difference $(\theta_1 − \theta_2) \pm z(SE_1 + SE_2)$ includes 0?

I am able to check if the interval contains 0 which I can find out by just plugging numbers into the above formula but I am curious as to why this interval containing zero means that the two CI's of $$\theta_1$$ and $$\theta_2$$ overlap.

• Because the answer comes down to the meaning of subtraction, one wonders what you are really trying to ask. Could you elaborate a little?
– whuber
Mar 15, 2021 at 21:32

If they overlap, there is a number $$\color{blue}x$$ from their intersection, i. e.

$$\color{blue}x \in \theta_1 \pm z(\text{SE}_1)\\ \color{blue}x \in \theta_2 \pm z(\text{SE}_2)$$

By subtracting them, we obtain

$$\color{red}0 \in (\theta_1 - \theta_2)\pm z(\text{SE}_1 + \text{SE}_2)$$

(since $$\color{blue}x - \color{blue}x = \color{red}0$$).

Note:

I used the shortened notation; the expanded one is

\begin{aligned} \ \theta_1 - z(\text{SE}_1) \le\, &\color{blue}x \le \theta_1 + z(\text{SE}_1)\\ \theta_2 - z(\text{SE}_2) \le\, &\color{blue}x \le \theta_2 + z(\text{SE}_2)\\ \hline (\theta_1 - \theta_2) - z(\text{SE}_1 + \text{SE}_2) \le\, &\color{red}0 \le (\theta_1 - \theta_2) + z(\text{SE}_1 + \text{SE}_2) \end{aligned}