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?

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


1 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.

  • $\begingroup$ Thank you @HJ Liang. Your answer is very nice. But I think P(G=1)EY(0∗Y|G=1) can not be written as 0 without any information about Y. There will be a problem if Y does not have finite marginal expectations i.e. if Y can be infinite with some probability. $\endgroup$ Apr 1, 2021 at 17:45
  • $\begingroup$ Hello @MdAshiqurRahman, You don't have to consider Y since 0*Y = 0 a.s. $\endgroup$
    – HJ Liang
    Apr 2, 2021 at 0:36
  • $\begingroup$ isn't 0*infinity is indeterminate? Here is a related post:math.stackexchange.com/questions/45327/…. $\endgroup$ Apr 2, 2021 at 6:44
  • $\begingroup$ This is a good question! It seems that we only care about almost surely finite random variables in probability theory. I'm not sure in which book I have read this statement, and it may be an advanced probability textbook written by Kai Lai Chung or someone else. $\endgroup$
    – HJ Liang
    Apr 2, 2021 at 7:40

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