Find expectation and covariance of W= V^2/R? V and R have uniform distributions

Question

So I have $W$ = $\frac{V^2}{R}$

With $V$ ~ $U(10,20)$ and $R$ ~ $U(1,2)$

Find the expectation $E(W)$ and covariance of $W$.

EDIT: Since covariance makes no sense here, I'm pretty sure deviation was asked instead.

How do I solve this?

Answer 1

If you can assume that V and R are independent from each other, then $E[W] = E[V^2]*E[\frac{1}{R}]$

To find $E[V^2]$ you can use $var(V)=E[V^2] - (E[V])^2$ and solve for $E[V^2]$. Since you know $V$ is uniform, you can find $var(V)$ and $(E[V])^2$ easily.

To find $E[\frac{1}{R}]$ use the general form of the expected value rule: $E[g(x)] = \int_{-\infty}^{\infty} g(x)*f_X(x) dx$ where $g(x) = \frac{1}{R}$ and $f_X(x)$ is calculated from the given uniform distribution for $R$. The bounds on the integral get replaced by the upper and lower bounds for $R$.

To find covariance, you need to specify 2 random variables, not one. So $cov(W)$ is incomplete.