<s>[INST] Find $g(x|y=\frac{1}{2})$, the conditional pdf of $X$ given $Y = \frac{1}{2}$ (Need confirmation) Let X and Y be continuous random variables having the joint pdf $$f(x,y) = 8xy , 0\leq{y}\leq{x}\leq{1}$$ I found that the marginal pdf of Y is $f_2(y) = 4y - 4y^3$. Does $g(x|y=\frac{1}{2}) = \frac{f(x,\frac{1}{2})}{f_2(\frac{1}{2})} = \frac{8x}{3}$ or $g(x|y=\frac{1}{2}) = \frac{f(x,y)}{f_2(\frac{1}{2})} = \frac{16xy}{3}$. A little confused here with the definition. Sorry for the tiny fractions, I don't know how to enlarge them.
 [/INST] $$f_Y(y) = \int_{y}^{1} 8xy \: \text{dx} = 4y - 4y^3$$
$$ f_Y(y) =
\begin{cases}
4y - 4y^3,  & 0 \leq y \leq 1 \\
0, & \text{otherwise}
\end{cases}
$$
$$g(x \mid y=\frac{1}{2}) = \frac{f_{X,Y}(x,y=\frac{1}{2})}{f_Y(y=\frac{1}{2})} = \frac{8x}{3}$$
</s>