2
$\begingroup$

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?

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

$\endgroup$
2
+50
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$ – Md Ashiqur Rahman Apr 1 at 17:45
  • $\begingroup$ Hello @MdAshiqurRahman, You don't have to consider Y since 0*Y = 0 a.s. $\endgroup$ – HJ Liang Apr 2 at 0:36
  • $\begingroup$ isn't 0*infinity is indeterminate? Here is a related post:math.stackexchange.com/questions/45327/…. $\endgroup$ – Md Ashiqur Rahman Apr 2 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 at 7:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.