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Let $B$ a random variable with the Bernoulli distribution and $V$ a non-negative real random variable with a density function. Thus $\mathbb{P}(B=1)=1-\mathbb{P}(B=0)=p$. I'm trying to prove that $\mathbb{E}[BV] = p\mathbb{E}[V]$.

If $B$ and $V$ are independent, then it is known that $\mathbb{E}[BV] = \mathbb{E}[B] \mathbb{E}[V] = p \mathbb{E}[V]$ as desired. But if they are dependent this may not be true in general, I guess.

I'm wondering whether my reasoning is correct:

$$ \mathbb{P}(BV\ge x) = \mathbb{P}( BV\ge x \cap ( B=0 \cup B=1 )) = \mathbb{P}( BV\ge x \cap B=0 ) + \mathbb{P}( BV\ge x \cap B=1 ) = \mathbb{P}( 0\ge x \cap B=0 ) + \mathbb{P}( V\ge x \cap B=1 ) = \mathbb{P}( 0\ge x ) (1-p) + \mathbb{P}( V\ge x ) p $$ where in the last I've used the conditional probability. Now, since $BV$ is non-negative $$ \mathbb{E}[BV] = \int_{\mathbb{R}_+} \mathbb{P}(BV\ge x)\mbox{d}x = \int_{\mathbb{R}_+} \mathbb{P}( V\ge x ) p \mbox{d}x = p \mathbb{E}[V] $$ Is this correct? What bothers me is that $B$ and $V$ are dependent but still $\mathbb{E}[BV] = \mathbb{E}[B] \mathbb{E}[V]$.

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    You claim that $P(V\geq x,B=1)=P(V\geq x)P(B=1)$ which, in general, requires independence. – 2017-02-03
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    $\mathbb{P}(B V\ge x, B=1) = \mathbb{P}(V\ge x, B=1) = \mathbb{P}(V\ge x| B=1)\mathbb{P}(B=1)$. So $\mathbb{P}(V\ge x| B=1) \neq \mathbb{P}(V\ge x)$ if they are dependent. Ok I understood my mistake. – 2017-02-03

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Take $V$ uniformly distributed over $[0,1]$ and $B = \mathbb{1} \left(V \geq \frac{1}{2} \right)$. Now $p = \mathbb{P}(B = 1) = \frac{1}{2}$.

However $$\mathbb{E}[BV] = \int_\Omega \mathbb{1} \left(V \geq \frac{1}{2} \right) V \, \mathrm{d}\mathbb{P} = \int_A V \, \mathrm{d}\mathbb{P} > p \frac{1}{2} = p \mathbb{E}[V], $$

where $A = \left\{ \omega \in \Omega \: \middle| \: V(\omega) \geq \frac{1}{2} \right\}$ and $p = \mathbb{P}(A)$ is the expectation of $B$, since $V > \frac{1}{2}$ in a set of a positive measure in $A$. Thus $\mathbb{E}[BV] \neq p \mathbb{E}[V].$

I hence think you need to control the dependence between $B$ and $V$ somehow.