Let $(X, Y)$ be a point chosen at random on the unit square $[0, 1] × [0, 1]$. Find $Var[XY]$.
My Attempt (I think I have the right answer. I just want some verification. Thank you)
First, we want to find the c.d.f. so we can find the p.d.f.
The c.d.f. is $$ F_{X,Y}(x,y)= \begin{cases} 0 & \text{if } x\leq 0, y\leq 0\\ xy & \text{if } 0<x\leq 1, 0<y\leq 1\\ 1 & \text{if } x>1,y>1 \end{cases} $$ Then the joint p.m.f. is $$f_{X,Y}(x,y)=1\quad \text{for } 0<x\leq 1,0<y\leq 1$$ Integrating with respects to $y$ gives the p.d.f.for $X$, and with respects to $x$ for $Y$. So $$f_X(x)= \begin{cases} 1 &\text{if }0<x\leq 1\\ 0 &\text{otherwise} \end{cases}$$ $$f_Y(y)= \begin{cases} 1 &\text{if }0<y\leq 1\\ 0 &\text{otherwise} \end{cases}$$ It is clear that the marginal mean is $\mathbb{E}[X]=\mathbb{E}[Y]=\frac{1}{2}$. Since the joint p.d.f. is the product of the marginals, we know that they are independent, so the expectation of $XY$ is the product of their marginal expectations: $$\mathbb{E}[XY]=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$$ Now we need to find $\mathbb{E}[(XY)^2]$. $$\mathbb{E}[(XY)^2]=\int^1_0\int^1_0(xy)^2dxdy=\frac{1}{9}$$ Using the definition of variance, we have $$Var(XY)=\mathbb{E}[(XY)^2]-\mathbb{E}[XY]^2=\frac{1}{9}-\frac{1}{16}=\frac{7}{144}$$
Verified. You have the right answer. All your work checks out, and in short, since the variables are independent and identically uniformly distributed over $(0..1)$ then:...
$$\begin{align}\mathsf {Var}(XY) &=\mathsf E(X^2)\mathsf E(Y^2)-\mathsf E(X)^2\mathsf E(Y)^2 \\ &= \left(\int_0^1 s^2\mathrm ds\right)^2-\left(\int_0^1 s\mathrm d s\right)^4\\ &= \left(\dfrac 13\right)^2-\left(\dfrac{1}{2}\right)^4\\&=\dfrac{2^4-3^2}{3^22^4}\\&=\dfrac{7}{144}\end{align}$$