The Setup

Let $X$ be a random variable with an unknown distribution. Let $p$ be the probability that $X$ is between $a$ and $b$, where $a<b$. Transform this Two-Tailed requirement on $X$ into a One-Tailed requirement.


$$\mathbb P(a \le X \le b)=p \ \rightarrow \ \mathbb P\biggl(\biggl | X-\frac{a+b}{2} \biggr | \le \frac{b-a}{2} \biggr)=p$$


I believe that my hypothesis is true, but I have been unable to prove it. Can you prove that this transform of inequalities is correct? Thank you.

  • $\begingroup$ That works. So too does $ \mathbb P\biggl(\left( X-\frac{a+b}{2} \right)^2 \le\left( \frac{b-a}{2}\right)^2 \biggr)=p$ which is why chi-squared rejection regions are often one-tailed $\endgroup$ – Henry Feb 24 at 0:20

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