Determine $\alpha$ that satisfies $P(|X|<1) = P(|X|>1)$

Question

Suppose that $X$ is uniformly distributed $[-\alpha, +\alpha]$, where $\alpha > 0$. Whenever possible, determine $\alpha$ so that the following are satisfied. $$P(|X| < 1) = P(|X| > 1)$$

Answer: $\alpha = 2$.

This is the question 4.20 from (Paul Meyer's "Introductory probability and Statistics", 2nd ed.)

What I've managed to do was find the density function, which is: $\frac{1}{2\alpha}$. And from there I applied in the given equation using the absolute value definition, obtaining: $$P(-1<X<1) = P(X>1) + P(X<-1)$$

$$\int_{-1}^{1}\frac{1}{2\alpha}d\alpha = 1 - P(X≤1) + P(X<-1)$$ But the outcome of this integral it's not a real value. What was my mistake here?

Answer 1

Hint

The issue is that you are mislead with variable names.

The density of probability map is $f_\alpha(x)=\frac{1}{2\alpha}$, which is constant, i.e. independent of $x$. $\alpha$ is not the probability variable, but a parameter of the problem. Then you have

$$P(-1 \lt X \lt 1) = \int_{-1}^1 \frac{1}{2\alpha} \ dx= \frac{1}{\alpha}$$ providing that $\alpha \ge 1$.

Then you can follow the same path for $P(\lvert X \rvert \gt 1)$.

Answer 2

Just draw a picture to get $$2\cdot\frac{1}{2\alpha}=2(\alpha-1)\cdot\frac{1}{2\alpha},$$ i.e., $\alpha=2$.