The question:
Let $ (x, y) \in \mathbb{R}^2$ and $0 < y \leq 1, 0 \leq x \leq y $. Assume that the distribution of the vector $(X,Y)$ is continuous with density $$ f(x,y)=\frac{1}{y} $$
is a probability density and then calculate the average vector and covariance matrix for the vector $(X, Y )$.
The problem I solved the question but I feel that I have problems in:
1) is it enough to prove that function is a density by showing that the double integral = 1.
$2)$the defining the limits of both integrals in each case among the three required cases
The solution
$*$ Now, integrating the function gives $1$ as
$\int_0^1 \int_0^y 1/y~ dx dy = \int_0^1 1 dy = 1$.
$*$ Following, to compute the mean vector, I used the formula: $E[<X,Y>] = < E(x) E(y)> ~$ where $E(X) = \int_0^1 \int_0^y \frac{x}{y}~ dx dy~ = \frac{1}{2}$ and $~ E(Y) = \int_0^1 \int_0^y \frac{y}{y}~ dx dy = \frac{1}{2}$
$*$ Finally, to compute the covariance matrix, it is necessary to find $\sigma^2(x), \sigma^2(y), cov(x,y)$.
$\sigma^2(x) = E(X^2)-E^2(X) = $ which are both calculated as above in the mean vector. For $cov(x,y) = E(XY) - E(X) E(Y)$ where $E(XY) = \int_0^1 \int_0^y \frac{xy}{y}~ dx dy = \frac{1}{6}$
Sorry for taking long Also, when computing the marginals they seem strange! $f_Y(y)= \int_0^y \frac{1}{y}~ dx = 1 $ and $f_X(x)= \int_0^1 \frac{1}{y}~ dy = \log(1)- \log(0)$!