# Inequalities involving expectations

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

• 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? Commented Oct 14, 2023 at 13:27
• $W$ is absolutely continuous, with support equal to the real line.
– Star
Commented Oct 14, 2023 at 15:14
• Its support conditions can be costrained.
– Star
Commented Oct 14, 2023 at 16:06

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 $$Q=1$$.

• Thanks, but I cannot assume mean independence between $W,Q$
– Star
Commented Oct 14, 2023 at 8:44
• Apologies for not having specified this!!!
– Star
Commented Oct 14, 2023 at 8:45
• 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 $p(Q=0) = 1$. Commented Oct 14, 2023 at 15:12
• Ok, understood.
– 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.

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 $$QW \mid \{Y=1\} \leq 0 \implies E(QW|Y=1) \leq 0$$
$$\implies E(QX|Y=1) \geq 0.$$