Consider four random variables $W, X,Q,Y$, where $Q$ and $Y$ are binary. Assume $$ \begin{aligned} & (1) \quad E(Q(X+W)|Y=1)\geq 0\\ & (2) \quad E(W|Y=1)=0\\ & (3) \quad \Pr(Q=1|Y=1,W)=1 \end{aligned} $$ Note that (2) and (3) imply $$ (4) \quad E(QW|Y=1)=0 $$ Hence, by combining (1) and (4), we conclude $$ (5) \quad E(QX|Y=1)\geq 0. $$

Question: I want to obtain (5) under a condition weaker than (3), ideally, something like $$ (3')\quad \Pr(Q=1|Y=1,W)> \Pr(Q=0|Y=1,W). $$ To this end, I am willing to also modify (2), where I suspect we might need something like $$ (2')\quad E(W|Y=1)\leq 0. $$ I cannot modify (1). If this strategy does not seem meaningful, could you advise on an alternative successful strategy? Also, I cannot assume mean independence between $W,Q$ or impose $E(W|Q=1)=0$.

  • 1
    $\begingroup$ What is the support of $W$? Is $W$ non-negative? Non-positive? Arbitrary? Can its various conditional supports be affected/restricted, or this should not/cannot happen? $\endgroup$ Commented Oct 14, 2023 at 13:27
  • $\begingroup$ $W$ is absolutely continuous, with support equal to the real line. $\endgroup$
    – Star
    Commented Oct 14, 2023 at 15:14
  • $\begingroup$ Its support conditions can be costrained. $\endgroup$
    – Star
    Commented Oct 14, 2023 at 16:06

2 Answers 2


Note that all the $Y=1$ conditions can be dropped, as everything is conditioned on $Y=1$, and $Y$ appears nowhere except in the conditioning part of the statements.

Replacement condition:

(2'). $E(W|Q=1)=0$

This will do the job as a weaker replacement for (2) and (3) together, as a) $E(QW|Q=0) =0$ by construction and b) $Q$ binary together with (2') imply (4), which, as you observe, combined with (1) implies (5).

Note that (2') is a necessary condition for (4) to hold, as

$$E(QW) = 0\cdot p(Q=0) + E(W|Q=1)\cdot p(Q=1)$$

so for $E(QW) = 0$, it must be that $E(W|Q=1) = 0$ or $p(Q=1)=0$, the latter of which simplifies the problem considerably!

This assumption is much weaker than the original (2) and (3); it makes no assumption about $p(Q=1)$, as does (3), and it does not assume the existence of moments for $W$, as does (2), except when $Q=1$.

  • $\begingroup$ Thanks, but I cannot assume mean independence between $W,Q$ $\endgroup$
    – Star
    Commented Oct 14, 2023 at 8:44
  • $\begingroup$ Apologies for not having specified this!!! $\endgroup$
    – Star
    Commented Oct 14, 2023 at 8:45
  • 1
    $\begingroup$ 1) I am not assuming mean independence between $W$ and $Q$. 2) Note that whatever conditions you impose, they have to end up with $E(W|Q=1) = 0$, as $E(WQ) = 0\cdot p(Q=0) + E(W|Q=1)\cdot p(Q=1)$, so, for $E(WQ) = 0$, it must be that $E(W|Q=1) = 0$ or $p(Q=0) = 1$. $\endgroup$
    – jbowman
    Commented Oct 14, 2023 at 15:12
  • $\begingroup$ Ok, understood. $\endgroup$
    – Star
    Commented Oct 14, 2023 at 16:05

We must have that $W$ and $Q$ are not mean-independent, so they are certainly statistically dependent.

But conditional independence does not imply, nor is implied by, independence.

Assume then conditional independence of $Q$ from $W$ given $Y=1$: $$A1: \Pr(Q \mid Y=1, W) = \Pr(Q \mid Y=1)$$

From the linearity of expected value we have $$(1):\; E(Q(X+W)|Y=1)= E(QX|Y=1) + E(QW|Y=1) \geq 0$$ $$\implies E(QX|Y=1) \geq -E(QW|Y=1).$$

So a sufficient condition for what we want is $$E(QW|Y=1) \leq 0.$$

Under $A1$,

$$E(QW|Y=1) = E(Q|Y=1) \cdot E(W|Y=1) = E(Q|Y=1) \cdot 0 = 0.$$

Whether the $A1$ assumption is considered a "weaker" assumption than OP's eq. $(3)$, depends on the particulars of the case.


In a later comment the OP informed that the conditional supports can be different than the unconditional ones. In such a case:

We need only to assume $(1)$ and

$$A2: W \mid \{Y=1\}\, \in (-\infty, 0]. $$

From $(1)$ we still get that a sufficient condition for what we want is $E(QW|Y=1) \leq 0$. Since $Q \in \{0,1\}$ and since, under $\{Y=1\}$, $W$ is non-positive, it follows that $$QW \mid \{Y=1\} \leq 0 \implies E(QW|Y=1) \leq 0$$

$$\implies E(QX|Y=1) \geq 0.$$


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