Determine the conditional probability and uniform distribution of a random variable.

Question

Given that $X$ is a random variable, such that $P(X=-1)=P(X=1)=\frac{1}{4}, P(X=0)=\frac{1}{2}$, determine: (a) the uniform distribution of $X^2$ and (b) the conditional probabilities of $X$ given $X^2$.

I am not sure how to go about doing these.

For (a), I think that $X^2(\omega) = \{0,1\}$ but then what do I do with the probabilities? Do I simply square them as well?

For (b), I know that: $P(X=x \ | \ X^2 = x^2) = \frac{P(X=x, X^2 = x^2)}{P(X^2 = x^2)}$. But then how do I go on from here?

Answer 1

As mentioned in the comments, $ \{X^2=1\}= \{X=-1\}\cup\{X=1\}, $ so $$ \mathbb P(X^2=1) = \mathbb P(X=-1)+\mathbb P(X=1) = \frac12. $$Moreover $\{X^2=0\}=\{X=0\}$ so, $$ \mathbb P(X^2=0)=\mathbb P(X=0)=\frac12. $$ For the conditional distribution, we use the definition of conditional probability: \begin{align} \mathbb P(X=0\mid X^2=0) &= \frac{\mathbb P(X=0,X^2=0)}{\mathbb P(X^2=0)}=\frac{\mathbb P(X^2=0)}{\mathbb P(X^2=0)} = 1\\ \mathbb P(X=1\mid X^2=1) &= \frac{\mathbb P(X=1,X^2=1)}{\mathbb P(X^2=1)} =\frac{\mathbb P(X=1)}{\mathbb P(X^2=1)} = \frac{1/4}{1/2}= \frac12\\ \mathbb P(X=-1\mid X^2=1) &= \frac{\mathbb P(X=-1,X^2=1)}{\mathbb P(X^2=1)} =\frac{\mathbb P(X=-1)}{\mathbb P(X^2=1)} =\frac{1/4}{1/2}= \frac12. \end{align}