Two dimensional random variables and conditional pmf if $f(x)=24x^2, 0<x<\frac{1}{2} \text{ and} f(y|x=X)=\frac{y}{2x^2}, \text{ if} 0< y<2x$

Question

My solution:

$$f_Y=\int^{0.5}_{0.5y}12y dx=\left. 12xy \right|^{0.5}_{0.5y}=6y-6y^2$$

$$f(x|Y=y)=\frac{12y}{6y(1-y)}=\frac{2}{1-y}$$

In general, I think I don't understand what I have to do at all. Can anyone assist?

Answer 1

As you have written $f_{X,Y}(x,y)=f_X(x)\cdot f_{Y|X}(x,y)$. The notation is not always uniformly. In a simplified way $f_X$, $f(x)$ and $f_X(x)$ have the same meaning. They represents the pdf of the random variable $X$.

For calculation we can input the terms:

$$f_{X,Y}(x,y)=24x^2\cdot \frac{y}{2x^2}=12y$$

From this result we can re-calculate $f_X(x)$ by integrating. For this purpose we need the bounds for y. We know that $y<2x$ and we also know that $0<y$

$$f_X(x)=\int_{0}^{2x} 12y \, dy=...$$

To calculate $f_Y(y)$ you integrate $f_{X,Y}(x,y)$ as well. We know that $y<2x$ and therefore $\frac12y<x<0.5$

$$f_Y(y)=\int_{\frac12y}^{0.5} 12y \, dx=...$$