From conditional to unconditional expectation

Question

Consider a random variable $Y$ and a random variable $G$. $G$ can only take value $1$ or $0$.

Is it true that $$ E(Y|G=0)\geq 0 \Leftrightarrow E((1-G)Y) \geq 0 \quad ?$$

My thought is yes and below I report the proof (I imagine that $Y$ is discrete for simplicity). Is it correct? What is really that I'm leveraging on for this result?

$$ E(Y|G=0)=\sum_{y} y Pr(Y=y|G=0)\geq 0 \Leftrightarrow \sum_{y} y Pr(Y=y|G=0)Pr(G=0)\geq 0 \Leftrightarrow \sum_{y} y Pr(Y=y,G=0)\geq 0 \Leftrightarrow E((1-G)Y)\geq 0 $$

Answer 1

You can prove this proposition by law of iterated expectations. \begin{align} \mathbb{E} (1-G)Y &= \mathbb{E}_G \mathbb{E}_Y( (1-G)Y | G)\\ &= \mathbb{P}(G=1) \mathbb{E}_Y(0*Y|G=1) + \mathbb{P}(G=0)\mathbb{E}_Y(1*Y| G=0)\\ &= \mathbb{P}(G=0) \mathbb{E}_Y(Y|G=0). \end{align} It is clear that your proposition is true. The key to proving this proposition is to use the $1-G$ as an indicator function.