Conditional probability when X is uniform on [-1, 1]

Question

Let X be a real random variable that is uniformly distributed between -1 and +1. Intuitively, it is obvious that, for every y in R with 0<y<=1, we have

P(X=x|X^2=y) = 0 for x^2≠y

and

P(X=x|X^2=y) = 1/2 for x^2=y

Shouldn’t it be P(X=x|X^2=y) = 1 for x^2=y ?

Answer 1

The basic idea of the conditional probability $P(.|B)$ is to set the probability measure to zero outside B and to rescale it inside B so that $P(B|B) = 1$

So we have $P(B|B)=1$ and $P(B^c|B)=0$ ("$B^c$ "=not B).

$A: X$

$B: X^2=y$ , $B^c: X^2 \neq y$

$A|B=${$-\sqrt y ,\sqrt y$ }

$X=x$ is the event that the random variable X takes a value x

Since each value in $A|B$ has the same probability and $P(B|B)=1$ we have:

$P(X=x|X^2=y)=1/2$