# Exponentiation of a random variable $\Rightarrow X$ is not a random variable

Let $$X^2$$ be a real random variable. Is $$X$$ a random variable, too?

I figured out that $$X$$ is not a random variable.

I want to give a counter example.

So, I tried:

Let $$A \subset \mathbb{R}$$ be a non-measurable subset.

$$X(\omega)=\mathbf{1}_A(\omega)-\mathbf{1}_{A^c}(\omega),\quad\omega\in\mathbb{R}$$

So $$X$$ is not a random variable.

Is this way correct or is there another counter example for that?

• The map $x\mapsto \sqrt x$ is measurable so $X = \sqrt{X^2}$ is a random variable (here we are considering the positive square root). Nov 30, 2019 at 18:55
• Your example wouldn't work anyway as then $$X^2(\omega) = \mathsf 1_A(\omega) - 2\mathsf 1_A(\omega)\mathsf 1_{A^c}(\omega) + \mathsf 1_{A^c}(\omega) = \mathsf 1_A(\omega) + \mathsf 1_{A^c}\omega) = X(\omega),$$ since $\mathsf 1_A(\omega)\mathsf 1_{A^c}(\omega) = 0$. Nov 30, 2019 at 18:59

Yes, your example is valid. For this case $$X(\omega)$$ is not a random variable since $$X^{-1}(\{1\})=A\not\in \mathcal F$$ and $$X^2(\omega)\equiv 1$$ is measurable.
All counterexamples here should have the same structure. The only difference can be that you do not need to take $$\Omega=\mathbb R$$.
Let $$\Omega=\{-1,1\}$$ with trivial $$\sigma$$-algebra $$\mathcal F=\{\varnothing, \Omega\}$$. Let $$X(\omega)=\omega$$. It is not measurable since for Borel set $$\{1\}$$ $$X^{-1}(\{1\})=\{1\}\not\in \mathcal F.$$ And $$X^2(\omega)=1$$ is $$\mathcal F$$-measurable.
Note that it is just the same example as you provided: $$X(\omega)=\mathbf{1}_{\{1\}}(\omega)-\mathbf{1}_{\{-1\}}(\omega),\quad\omega\in\Omega$$.
So the right statement should be: if $$X^2$$ is a random variable than $$X$$ need not to be a random variable.