X ∼ Uniform[0, 1]
Conditional on X, Y∼ Uniform[0, X]
I've then found the joint density of function X and Y to be 1/X.
As the joint density only contains X and not Y, can we say that X and Y are independent and therefore, the marginal density of Y is the same as the conditional density of Y given X, which is 1/X ?
joint density of $X$ and $Y$ depends on $Y = y$:
$$f_{X,Y}(x,y) = f_X(x) f_{Y|X}(y|x) = 1 \cdot \frac{1}{x} \cdot \left[ y \le x \right]$$
Thus, marginal distribution $f_Y(y) = \int \limits_y^1 \frac{1}{x}dx = -\ln(y)$
you can notice that $f_Y(y) \ne f_{Y|X}(y|x)$ thus variables are not independent.
You can also understand that $Y$ and $X$ depend, simply because knowing $X=x$ already gives you knowledge that $Y$ can't be higher than $x$