A point $(X, Y)$ is randomly selected from the interior of a square with vertices $(0, 0),(3, 0),(3, 3),(0, 3)$. $Z$ in an area of a rectangle with vertices $(0, 0), (X, 0), (X, Y), (0, Y)$. Determine the distribution of the variable $Z$, expected value of $Z$ and variance of $Z$.
As I understand, we know that there are two random variables. One is $X$, which follows a uniform distribution on the interval $[0,3]$ and the other one is $Y$. It follows a uniform distribution on the same interval, that is $[0,3]$. The two random variables are independent. There is also a rectangle constructed for which the lengths of two adjacent sides are $X$ and $Y$.
We have a following picture – we choose a random point $(X,Y)$ somewhere in a red square:
Therefore we know that $Z = XY$, so:
PDFs of a $X$ and $Y$ with uniform distribution is equal to $\frac{1}{3-0} = \frac{1}{3}$ for $x, y \in [0,3]$ $0$ otherwise.
Therefore PDF of Z is equal to $\frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9}$ for $z \in [0,3]$ and $0$ otherwise. PDF is unique for a distribution therefore we have our distribution.
$$E(Z) = E(XY)$$
$$E(Z) = E(X) + E(Y) + Cov(X,Y)$$
We know that variables X and Y are independent, therefore $Cov(X,Y) = 0$ and we have:
$$E(Z) = E(X) + E(Y)$$
$E(X) = E(Y) = \frac{0 + 3}{2} = \frac{3}{2}$ and we get that:
$$E(Z) = \frac{3}{2} + \frac{3}{2} = 3$$
I don't know how to calculate variance of $Z$.
Best Answer
I would say
$E[XY] = \int_0^3 \int_0^3 (\frac 13 x)(\frac 13 y) \ dy \ dx = \frac 19 \int_0^3 x\ dx \int_0^3 y\ dy = (\frac {1}{9})(\frac 92)(\frac 92) = \frac {9}{4}$