Suppose we have a continuous random variable $Y$ and a random Bernoulli variable $T$ such that $P(T=1|Y)$ is monotonically increasing in $Y$.

How can we show that $E[Y|T=1]>E[Y|T=0]$?

To me, it makes intuitive sense, but I can't prove it mathematically.


I've managed to answer the question myself. I'm including the solution here for other people that might have the same question.

I want to show that the quantity below is greater than $0$.

$=E[Y \mid T = 1]-E[Y \mid T = 0]$

$=\int^{\infty}_{-\infty}{yP(Y=y \mid T=1)dy} -\int^{\infty}_{-\infty}{yP(Y=y \mid T=0)dy}$

Let $p = P(T=1)$ and $f$ be the PDF of $Y$. Then:

$=\int^{\infty}_{-\infty}{y\left(\frac{P(T=1 \mid Y=y)f(y)}{p}-\frac{(1-P(T=1 \mid Y=y))f(y)}{1-p}\right)dy}$

$=\int^{\infty}_{-\infty}{f(y)y\left(\frac{P(T=1 \mid Y=y) - p}{p(1-p)}\right)dy}$

$=\frac{1}{p(1-p)}(\int^{\infty}_{-\infty}{f(y)y P(T=1 \mid Y=y)dy} - p\int^{\infty}_{-\infty}{f(y)ydy})$

Let $Z=P(T=1 \mid Y)$. Then:

$=\frac{1}{p(1-p)}(E[YZ] - E[Z]E[Y])$


We know that $Z$ is increasing in $Y$, and the covariance of a random variable and an increasing function of that random variable is always positive (see reference). Therefore:

$=E[Y \mid T = 1]>E[Y \mid T = 0]$


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