# Expected value of a random variable conditioned on a positively correlated event

I have a random variable $$x \in [a, b]$$ with PDF $$f(x)$$ and an event $$E$$ which satisfies the following property for any $$x'.

$$\Pr[E\mid x > x'] \geq \Pr[E]$$

My question is whether or not the following inequality holds.

$$\int_a^b uf(u)\Pr[E\mid x=u] \, du \geq \Pr[E]\int_a^b uf(u) \, du$$

The answer is yes. Indeed, let us write $$X$$ instead of $$x$$, according to standard notation, to distinguish between random variables (denoted by upper-case letters) and their values (denoted by lower-case letters). Let us write $$P$$ instead of $$\Pr$$, and then let us also write $$A$$ instead of $$E$$, to distinguish it from the expectation sign.
Then we need to show that $$$$EX\,P(A|X)\ge P(A)\,EX$$$$ given that $$$$P(A|X>y)\ge P(A)$$$$ for all $$y$$ such that $$P(X>y)\ne0$$.
Replacing $$X$$ by $$X-a$$, we may assume that $$X\ge0$$. Then $$\begin{multline*} EX\,P(A|X)=EX\,E(1_A|X)=EX\,1_A=E\Big(\int_0^X dy\Big)\,1_A \\ =E\Big(\int_0^\infty dy\,1_{X>y}\Big)\,1_A =E\Big(\int_0^\infty dy\,1_{X>y}1_A\Big) \\ =E\int_0^\infty dy\,1_{X>y,A}=\int_0^\infty dy\,E1_{X>y,A} \\ =\int_0^\infty dy\,P(X>y,A) =\int_0^\infty dy\,P(A|X>y)P(X>y) \\ \ge\int_0^\infty dy\,P(A)P(X>y)=P(A)\int_0^\infty dy\,P(X>y)=P(A)EX, \end{multline*}$$ as claimed.
(The last equality follows immediately from the special case $$EX=\int_0^\infty dy\,P(X>y)$$ of the previously proved equality $$EX\,P(A|X)=\int_0^\infty dy\,P(X>y,A)$$, with $$A$$ replaced there by an event of probability $$1$$.)
• Thanks for the answer! Can you please explain the second step in more detail? How do you get $𝐸 \int_{0}^{\infty } dy 1_{X>y,A}$ and what does that mean? – Melika May 20 '19 at 18:21
• I have added the details you requested. The meaning is this: $Y:=\int_0^\infty dy\,1_{X>y,A}$ is a random variable (r.v., which is function of the r.v. $X$), and $E\int_0^\infty dy\,1_{X>y,A}$ is the expectation of the r.v. $Y$. – Iosif Pinelis May 20 '19 at 20:41