I assume to have two random variables $X,Y$ with pdfs $p_x,p_Y$, respectively. Unfortunately, I don't have much background knowledge on probability theory. I am wondering whether assuming $p(x | y)$ is identical for all $y \in Y$ automatically implies that $X$ and $Y$ are (stochastically) independent?
My feeling is that this is not automatically fulfilled. It is clear to me that the implication for the other direction (independency implies the same conditional density) is satisfied. Thanks a lot!
This just duplicates my comment as an answer: If the conditional PDF $f_{X|Y}(x|y)$ only depends on $x$, then there is a function $g$ such that $$ f_{X|Y}(x|y)=g(x) \quad\forall x \in \mathbb{R}, \forall y \in S_Y$$ where $S_Y$ is the support of the PDF of $Y$. Then for all $x\in\mathbb{R}$ we have \begin{align} f_X(x) &= \int_{y\in S_Y}f_{X|Y}(x|y)f_Y(y)dy \\ &=\int_{y\in S_Y} g(x)f_Y(y)dy\\ &=g(x)\int_{y\in S_Y}f_Y(y)dy\\ &=g(x) \end{align} and so $$f_{X|Y}(x|y)=f_X(x) \quad \forall x\in \mathbb{R}, y \in S_Y$$ which implies independence.
Note that random variables $X, Y$ are defined to be independent if $$P[X\leq x, Y\leq y] = P[X\leq x]P[Y\leq y] \quad x,y\in\mathbb{R}$$ although if a joint PDF exists and factors into a product of marginal PDFs, so $f_{X,Y}(x,y)=f_X(x)f_Y(y)$ for all $x,y\in\mathbb{R}$, this is a sufficient condition for independence.