Given $U,V \sim R(0,1)$. Determine covariance between $X = UV$ and $V$

Question

Given $U,V \sim R(0,1)$ independent.

Determine covariance of $X = U \cdot V$ and $V$

About notation: $U,V \sim R(0,1)$ means random variables $U,V$ are equally / uniformly distributed from interval $0$ to $1$.

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What need to calculate is $$\text{cov}(X,V) = E(XV) - E(X)E(V)$$

Now need to know what is $E(V)$ and what is $E(X) = E(U \cdot V)$

$E(V) = \frac{1}{2} \cdot(a+b)$ in this example we have interval $a=0$ and $b = 1$ so we have

$E(V) = \frac{1}{2}$

$E(U) = \frac{1}{2}$

Is this correct so far I don't know how is continued :(

Answer 1

First noting that $U,V$ are independent uniform variables on $[0,1]$ we have

\begin{align*} \mathbf{E}[U] & = \mathbf{E}[V] = \int_0^1 v dv = \frac12 \\ \mathbf{E}[V^2] & = \int_0^1 v^2 d v = \frac13 \end{align*}

Then using the independence of $U,\, V$ we have

\begin{align*} \mathbf{E}[X] & = \mathbf{E}[UV] \\ & = \mathbf{E}[U] \mathbf{E}[V] \\ & = \frac12 \times \frac12 \\ & = \frac14 \end{align*} whilst \begin{align*} \mathbf{E}[XV] & = \mathbf{E}[ U V^2 ] \\ & = \mathbf{E}[U] \mathbf{E}[V^2] \\ & = \frac12 \times \frac13 \\ & = \frac16 \end{align*} Hence \begin{align*} \text{Cov}(X,V) & = \mathbf{E}[XV] - \mathbf{E}[X]\mathbf{E}[V] \\ & = \frac16 - \left( \frac14 \times \frac12 \right) \\ & = \frac1{24} \end{align*}