4
$\begingroup$

Equation (11.1) of these notes on concentration inequalities contains the equality $$\sum_{i=1}^n\Bbb E[Z-Z_i]^2_{+/-}=\sum_{i=1}^n\Bbb E\left[Z-\Bbb E^{\{i\}}[Z]\right]^2$$ but the $+/-$ subscript notation under the expectation is not explained.

Here $Z$ is a function of random variables $X_i,i\le n$, and $^{\{i\}}$ means swapping $X_i$ for an independent copy $X_i'$ but keeping all $X_j,j\ne i$ the same.

The equality implies that somehow, the notations $Z_i$ (which means a replacement of $Z$ by $X_i'$ at its $i$th value) and $+/-$ together mean a conditional expectation on $Z$. The proofs further down the page do not use this notation and instead prove the inequality for the other expression.

A subscript $+$ by itself can mean a positive domain only, and likewise for $-$, but I have never seen them used together before. I was suggested a possible "signed measure" interpretation; however it is usually denoted with a superscript and doesn't seem applicable in this context. Any clarification will be appreciated.

$\endgroup$

1 Answer 1

Reset to default
4
+50
$\begingroup$

From (10.3) and (10.5) in the previous notes we have $$\frac12\Bbb E(Z - Z_i)^2 =\Bbb E(Z - Z_i)_-^2 =\Bbb E(Z - Z_i)_+^2 =\Bbb E\bigl(Z -\Bbb E^{\{i\}}Z_i\bigr)^2.$$ Comparing that with (11.1) suggests that $\Bbb E(Z - Z_i)_{+/-}^2$ means $\Bbb E(Z - Z_i)_+^2$ or $\Bbb E(Z - Z_i)_-^2$.

$\endgroup$
1
  • $\begingroup$ That makes sense. Good thinking to check the previous chapter. $\endgroup$ Commented Nov 16, 2021 at 7:30

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.