$ P (| X |> 1) = P (| X | <1) $

Question

Let $X$ be a random variable following a uniform distribution over the interval $[-a, a]$ where $a$ is a constant greater than 1.

If we have $ P (| X |> 1) = P (| X | <1) $, then a must be equal to a value that is in the interval:

1. $[0, 1.5[$
2. $[1.5, 2[$
3. $[2, 2.5[$
4. $[2.5, 3[$
5. None of the preceding intervals contains $a$.

I have computed that $a=1$, but I think my answer is wrong. I think the answer is rather the choice 5. Am I wrong?

Answer 1

We can immediately dismiss the possibility that $a = 1$, because if that were true, then $X \in [-1,1]$, from which it would follow that $\Pr[|X| > 1] = 0$.

Let's translate this into something a bit more intuitive. $X$ is a number chosen at random between $-a$ and $a$, inclusive, and all values in this range are equally likely to be chosen.

If $a$ is less than $1$, then its impossible to get a value of $X$ whose absolute value will be greater than $1$, because the resulting interval is too small.

If $a$ is very large, say $1000$, then the outcome $|X| < 1$ is extremely unlikely because the range of possible values is $[-1000, 1000]$ of which $[-1,1]$ is only a tiny part.

Another way to understand the question is to think of a number line. Since we know that $a > 1$, we have the points $-a, -1, 0, 1, a$ in order from left to right. From $-a$ to $-1$, color the interval red. From $-1$ to $1$, color the interval blue. And from $1$ to $a$, color the interval red. The condition $\Pr[|X| < 1] = \Pr[|X| > 1]$ means that the total length of the red intervals must equal the length of the blue interval. What would $a$ need to be to make this true?