This is Q6.46 of Sheldon Ross' A First Course in Probability:
The joint density of $X$ and $Y$ is$f(x, y) = cxy^3$, $0 < x < 1$, $0 < y < x^{1/4}$
Find $c$ and the conditional distribution of $X$, given $Y = y$
The first part (i.e, finding the normalising constant) was fine. The problem is finding the conditional distribution of $X$ given $Y = y$. My attempt follows below:
Given the joint, we know that the conditional density of X given $Y = y$ is
$$\frac{f(x, y)}{f_Y(y)}$$ $$=\frac{12xy^3}{\int_{0}^{1} 12xy^3 dx}$$ $$=\frac{12xy^3}{|6x^2y^3|^1_0}$$ $$=\frac{12xy^3}{6y^3}$$ $$=2x$$
My confusion is on using this to find the cumulative distribution. My understanding is that I simply need to integrate this to get the CDF - but over what exactly? I was thinking from 0 to 1, but that ignores the $0 < y < x^{1/4}$ condition which I'm not sure how to use. And regardless, I don't get anywhere close to their answer:
$$\frac{2x}{1 - y^8}, y^4$$
and I don't understand why there are two answers either.
You miscalculated the marginal of $Y$. To obtain the marginal of $Y$, we are interested in $y$ going between numerical values, but in the original path $y$ goes between functions, we must fix this in the following way \begin{align*} 0 < y < x^{1/4} \Rightarrow y^{4} < x < 1 \end{align*} and we have $0 < y< 1$. You can draw a picture of this to find everything easier.So $$ f_{Y}(y) = \int_{y^{4}}^{1} 12xy^3 dx = -6 y^3 (y^8-1), \quad 0<y<1 $$ substituting in the conditional formula $$ f_{X|Y = y}(x) = \frac{12xy^3}{-6 y^3 (y^8-1)} = \frac{2x}{1-y^8}, \quad y^4 < x < 1 $$