Probability of conditional distribution is uniform(0,1)

Question

Suppose we have two random variables $X$ and $Y$. Define the conditional distribution function of $X$ on $Y$ as \begin{align*} F(x) := \mathbb{P}(X \leq x | Y) \end{align*} where $x \in \mathbb{R}$. How do I show that $F(X)$ conditional on $Y$ is distributed Uniform(0,1), in other words, for any $\alpha \in (0,1)$, \begin{align*} \mathbb{P}(F(X) \leq \alpha | Y) = \alpha \end{align*}

Answer 1

We know that $F(X)$ is a uniform$(0,1)$ random variable, so to have $\mathbb{P}(F(X)\le \alpha|Y)=\alpha=\mathbb{P}(F(X)\le \alpha)$, we must have $X$ and $Y$ are independent. Conversely, of $X$ and $Y$ are independent, the conditional distribution has to be Uniform$(0,1)$ by definition.