Covariance Issue with a joint distribution

Question

Here is the problem:

Let $X$ and $Y$ be continuous random variables with joint density function:

$$f(x,y) = \frac{8}{3}xy\text{ for }0 \le x \le 1\text{ and }x \le y \le 2x $$

I'm trying to find the covariance of $X$ and $Y$. I found the best way to do this online, my question is actually about the way that I was trying to work out the problem and how to get around the wall i found.

I know that $\operatorname{Cov}(x,y) = E(XY) - \mu_x\mu_y$.

For this problem I solved $E(XY)$, and I created $F_x(X)$ and found the average through that, Is there a way to solve for the $E(Y)$ from $F_y(Y)$ since it equals $\frac {28}{9}x^3$?

Answer 1

Directly, using the joint density function,

$$E(Y) = \int_{0}^{1}\left(\int_{x}^{2x}\frac{8}{3}xy^2dy\right)dx=\frac{56}{45}$$

The marginal density function is

$$f_Y(y) = \int_{y/2}^{y} \frac{8}{3} xy \,dx=y^3 \,\,\,(0\leq y\leq1)\\f_Y(y)= \int_{y/2}^{1} \frac{8}{3} xy \,dx=\frac{4}{3}\left(y-\frac{y^3}{4}\right)\,\,\,(1<y\leq2)$$

Alternatively,

$$E(Y) = \int_{0}^{2}yf_Y(y)\,dy=\int_{0}^{1}y^4\,dy+\int_{1}^{2}\frac{4}{3}\left(y^2-\frac{y^4}{4}\right)\,dy=\frac{56}{45}$$

Answer 2

Hint: $f_Y(y)=\int_{-\infty}^{\infty} f(x,y) \ dx$ and $E(Y)=\int_{-\infty}^{\infty} y \cdot f_Y(y) \ dy$

$f_Y(y)$ means here the marginal pdf of Y.

Edit: I´m little bit late. :)

Answer 3

$f(x,y) = \begin{cases}\frac 8 3 xy & 0\leq x\leq 1 , x\leq y\leq 2x \\ 0 & \text{elsewhere} \end{cases} = \begin{cases}\frac 8 3 xy & 0\leq y\leq 2 , \frac y 2 \leq x\leq y \\ 0 & \text{elsewhere}\end{cases}$

The expectation of any function $g$ of X, Y is: $\mathbb{E}[g(X,Y)] = \int_0^1 \int_x^{2x} g(x,y) f(x,y) \;\mathrm{d}y \; \mathrm{d} x$

So: ...

$\begin{align}\mathbb{E}(X) & = \frac 8 3 \int_0^1 x^2 \int_x^{2x} y \;\mathrm{d}y\; \mathrm{d} x \\ & = \frac 4 3 \int_0^1 x^2 y^2|_{y=x}^{y=2x}\;\mathrm{d} x \\ & = 4 \int_0^1 x^4\;\mathrm{d} x \\ & = \frac 4 5\end{align}$

$\begin{align}\mathbb{E}(Y) & = \frac 83 \int_0^1 x \int_x^{2x} y^2 \;\mathrm{d}y\; \mathrm{d} x \\ & = \frac 8 9\int_0^1 x y^3|_{y=x}^{y=2x}\;\mathrm{d} x \\ & = \frac {56} 9\int_0^1 x^4\;\mathrm{d} x \\ & = \frac {56} {45}\end{align}$

$\begin{align}\mathbb{E}(XY) & = \frac 8 3 \int_0^1 x^2 \int_x^{2x} y^2 \;\mathrm{d}y \mathrm{d} x \\ & = \frac 8 9 \int_0^1 x^2 y^3|_{y=x}^{y=2x}\;\mathrm{d} x \\ & = \frac {56} 9\int_0^1 x^5\;\mathrm{d} x \\ & = \frac {28} {27}\end{align}$