# Let $U$ and $V$ be independent random variables both having Bernoulli , Compute $E(W)$

Let $U$ and $V$ be independent random variables both having Bernoulli distribution, i.e. $U$ ~ Bernoulli($0.2$) and $V$ ~ Bernoulli($0.2$)

Let $W = U \cdot V$

(a) Compute $E(W)$

(b) Compute $E(UW)$

My attempt:

$(a)$

$$E(W) = E(UV) = E(U)E(V)$$

Because they're independent

$E(U) = E(V) = 0.2$, so $E(W) = 0.04$

$(b)$

$E(UW) = E(UUW) = E(U^2V) = E(U^2V) = E(U^2)E(V)$

$E(U^2) = 0.2^2 = 0.04$ [Not sure about this]

Thus,

$E(UW) = 0.04 \cdot 0.2 = 0.008$

• Note $U^2=U$ here so $E[U^2]=E[U]$ not $\left(E[U]\right)^2$ – Henry Jul 25 '18 at 23:53

\begin{align} E(U^2) & = 0 \cdot P(U^2 = 0) + 1 \cdot P(U^2 = 1) \\ & = 0 \cdot P(U = 0) + 1 \cdot P(U = 1) \qquad \leftarrow \text{because $U$ is either $0$ or $1$} \\ & = P(U = 1) \end{align}
$\mathsf E(UV) = \mathsf E(U)\mathsf E(V) = 0.2^2$ Yes.
$\mathsf E(U^2V) = \mathsf E(U^2)\mathsf E(V)$ Also.
However, $\require{cancel}\cancel{~~\mathsf E(U^2)=\mathsf E(U)^2~~}$ No! Rather: $\mathsf E(U^2)=\mathsf{Var}(U)+\mathsf E(U)^2$
Though, because $U$ is Bernouli, we can also use: $\mathsf E(U^2)=\mathsf P(U^2=1)=\mathsf P(U=1)=\mathsf E(U)$.