# Transform a Two-Tail Requirement into a One-Tail Requirement

## 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 $$a. Transform this Two-Tailed requirement on $$X$$ into a One-Tailed requirement.

## Hypothesis

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

## Question

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.

• 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 – Henry Feb 24 at 0:20