Does uniform conditional distribution imply independence? Take two random variable $X,Y$ and suppose $X$ is distributed uniformly on $[0,1]$ conditional on $Y$. Does this imply that $X$ is independent of $Y$? Could you make an example?
 A: Independence would mean that knowing the value of $Y$ gives no information on the value of $X$.
So here $X$ will be independent of $Y$ only if $X$ has a uniform marginal distribution on $[0,1]$, and the conditional distribution $X|Y$ is uniform on $[0,1]$ independent of the value of $Y$.
An example be a uniform (joint) distribution over the unit square.

Here are some examples using Tetris blocks:
For the "S" block

we have $p[X|Y=\text{middle}]=p[X]=\text{uniform}$, but $X$ is certainly not independent of $Y$.
While for the "O" block

we have $p[X|Y]=p[X]=\text{uniform}$, so $X$ is independent of $Y$.
A: If the conditional pdf of $X$ given $Y$ is the same density function for all values of $Y$ (in the support of $f_Y(y)$), that is, $f_{X\mid Y}(x\mid y)$ equals $g(x)$ where the value of $g$ does not depend on $y$ at all, then
$$f_X(x) = \int f_{X,Y}(x,y) \mathrm dy  = \int f_{X\mid Y}(x\mid y)\cdot f_Y(y)\mathrm dy = g(x)\int f_Y(y)\mathrm dy = g(x),$$ that is, the unconditional pdf of $X$ is the same as the common conditional pdf
of $X$ given $Y$. In particular,
uniformity does not have anything to do with it at all: what we need is that it is always the same density function regardless of the value of $y$.
Suppose that $f_{X,Y}(x,y)$ has value $1$ on (the interior of) the unit square. Then $f_{X\mid Y}(x\mid y) \sim U(0,1)$ and $X$ and $Y$ are both
independent  $U(0,1)$ random variables.
Suppose that $f_{X,Y}(x,y)$ has value $2y$ on (the interior of) the unit square. Then $f_{X\mid Y}(x\mid y) \sim U(0,1)$, and $X \sim U(0,1)$ also. Note that $X$ and $Y$ are independent random variables but $Y$ is not a $U(0,1)$ random variable.
Suppose that $f_{X,Y}(x,y)$ has value $2x$ on (the interior of) the unit square. Then $f_{X\mid Y}(x\mid y) = 2x\mathbf 1_{\{x\colon x \in (0,1)\}}$ and the unconditional density of $X$ is the same density.  $X$ and $Y$  are independent random variables but $X$ is not a $U(0,1)$ random variable. Uniform distribution of $X$ is not needed: what is needed is that $f_{X\mid Y}(x\mid y)$ is the same for all choices of
$y$.
