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Consider the random variables $X,Y$ and assume that $$ E(X|Y)=0 $$ Does this imply that $$E(X|X\geq A,Y)\neq 0 ?$$

I think this holds for the truncated Normal, for example. But does it hold generically?

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    $\begingroup$ A quick answer is "of course not," as evidenced by any instance of this situation where the support of $X$ has a finite lower bound greater than $A.$ $\endgroup$
    – whuber
    Commented Jul 1, 2022 at 18:35

2 Answers 2

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There are many ways to analyze this. The methods shown in this answer are chosen for their usefulness and helpfulness in understanding expectations and conditional expectations. That is, it is hoped the journey to the result will be more interesting than the result itself.


Because "$\mid Y$" occurs everywhere, let's drop it from the notation as being superfluous. The question therefore concerns the relationship between the expectation $E[X]$ and the expectation of the truncated variable, $E[X\mid X\ge A].$

By defining $Z = X - A$ the question concerns the relation between $E[Z\mid Z \ge 0]$ and $E[Z].$

Writing $S_Z$ for the complementary distribution function of $Z,$ by definition $$S_Z(0) = \Pr(Z\ge 0) = \Pr(X\ge A)$$ and $$1-S_Z(0) = 1 - \Pr(Z \ge 0) = 1 - \Pr(X \ge A) = \Pr(X \lt A).$$

This takes us to the crux of the matter: observe that

$$E[Z \mid Z\lt 0] \le E[0 \mid Z \lt 0] = 0.$$

In other words, when $Z$ is restricted to negative values its expectation cannot exceed $0.$

Finally, notice that $S_Z(0) \le 1$ because $S_Z(0)$ is a probability.

All these observations are trivial. But we may combine them and exploit the linearity properties of expectation (first and last lines below) and the law of iterated expectations (second line) to deduce

$$\begin{aligned} E[X]-A &= E[X-A] = E[Z] \\&= E[Z\mid Z \lt 0]\Pr(Z \lt 0) +E[Z\mid Z\ge 0]\Pr(Z \ge 0)\\ &= \color{red}{E[Z\mid Z \lt 0](1-S_Z(0)) + E[Z\mid Z\ge 0] S_Z(0)}\\ &\le 0(1-S_Z(0)) + E[Z\mid Z\ge 0]S_Z(0)\\ &= E[Z\mid Z\ge 0]S_Z(0)\\ &\le E[Z\mid Z\ge 0]\\ &= E[X-A\mid X \ge A]\\ &= E[X\mid X \ge A] - A. \end{aligned}$$

(The expression in red on the third line is particularly memorable and useful: it says the expectation of any random variable $Z$ is a weighted average of the conditional expectations that it is positive and non-positive.)

Upon adding $A$ to both sides the inequality is preserved, becoming

$$E[X] \le E[X\mid X \ge A].$$

By definition, the support of $X$ is the smallest closed set of real numbers $\mathcal{D}_X$ for which $\Pr(X\in \mathcal{D}_X)=1.$ When the support has a lower bound and $A$ is less than or equal to this lower bound, $\Pr(X \ge A)=1$ and so (since $S_Z(0)=1$) the previous inequality becomes an equality. But otherwise--again by the very definition of support--when $A$ exceeds the greatest lower bound of the support, $\Pr(X \ge A)\lt 1$ and the previous inequality is strict.

This analysis can be summarized in these terms as

The conditional expectation of $X$ given $X\ge A$ strictly exceeds the expectation of $X$ when $A$ exceeds all lower bounds of the support of $X.$ Otherwise, the conditional expectation and expectation are equal.

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Using the law of total probability and $E_{p(x|x<a, y)} \leq a \leq E_{p(x|x\geq a, y)}$, we have \begin{align*} E_{p(x|y)}x &= p(x\geq a) E_{p(x|x\geq a, y)}x + p(x<a) E_{p(x|x<a, y)}x \\&\leq p(x\geq a) E_{p(x|x\geq a, y)}x + p(x<a) E_{p(x|x\geq a, y)}x \\&= (p(x\geq a)+p(x<a)) E_{p(x|x\geq a, y)}x \\&= E_{p(x|x\geq a, y)}x. \end{align*}

The assumption $E_{p(x|y)}x = 0$ is not needed for this result, but it implies $E_{p(x|x\geq a, y)}x \geq 0$.

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