The two-dimensional random vector $(X, Y)$ has a distribution with joint density: $$g(x,y) = 8xy \ \text{ for: } 0 \leq x \leq y \leq 1$$
Determine the distribution of the variable $\frac{X}{Y}$. Is this variable independent of $Y$?
I don't know how to do that. I've calculated PDFs:
- $f_X(x) = 4x(1 - x^2)$ for $x \in [0,1]$ and $0$ else,
- $f_Y(y) = 4y^3$ for $x \in [0,1]$ and $0$ else.
Also I calculated $Cov(X, Y) = \frac{56}{25}$, $Var(X) = \frac{11}{225}$, $Var(Y) = \frac{2}{75}$.
I know that $X$ and $Y$ are not independent since: $f_{X,Y} (x,y) \neq f_X(x) \cdot f_Y(y)$
However, I don't know how to answer the question.
Edit:
For: $f_X(x)$ I did: $\int_x^1 8xy \ dx$
For: $f_Y(y)$ I did: $\int_0^y 8xy \ dx$
For: $Cov(X,Y)$ I did: $E(XY) - E(X)E(Y)$, therefore:
- for: $E(XY)$ I did: $\int_0^1 \int_0^y xy \cdot 8xy \ dx dy $
- for: $E(X)$ I did: $\int_0^1 x \cdot 4x(1-x^2) \ dx$
- for: $E(Y)$ I did: $\int_0^1 y \cdot 4y^3 \ dy$
For $Var(X)$ I did: $\int_0^1 \left(x - \frac{8}{15} \right)^2 \cdot 4x(1 - x^2) \ dx$
For $Var(Y)$ I did: $\int_0^1 \left(y - \frac{4}{5} \right)^2 \cdot 4y^3 \ dy$
Yes $U = X/Y$ and $V = Y$ are in fact independent. To see this do a multivariate pdf transformation from $(X, Y)$ to $(U, V)$.
Let $T : \mathbb R^2 \to \mathbb R^2$ be $T(x, y) = \begin{cases} (x/y, y) &y \neq 0 \\ 0 &y = 0 \end{cases}$. $(X, Y)$'s support $\{(x, y) \in \mathbb R^2 : 0 \leq x \leq y \leq 1\}$ is mapped almost everywhere to $\{(u, v) \in \mathbb R^2 : 0 \leq u \leq 1, 0 \leq v \leq 1\} = [0, 1] \times [0, 1]$ which must be $(U, V)$'s support. The inverse of $T$ almost everywhere is $T^{-1} : \mathbb R^2 \to \mathbb R^2$, $T^{-1}(u, v) = (uv, v)$ with Jacobian $JT^{-1}(u, v) = \begin{pmatrix} v &u \\ 0 &1 \end{pmatrix}$. So the pdf of $(U, V)$ is $$ f_{U, V}(u, v) = f_{X, Y}(T^{-1}(u, v))\big|JT^{-1}(u, v)\big| = 8uv^3 \text{ for } (u, v) \in [0, 1] \times [0, 1] $$ and $0$ elsewhere.
You can easily note that $U, V$ are independent as their joint support is rectangular and their pdf is separable. And finally, deriving the pdf of $U = X/Y$ is straightforward from the above.