Suppose I have a continuous random variable $Y$ and a random Bernoulli variable $T$ such that $P(T=1\mid Y)$ is monotonically increasing in $Y$. How can I show that $E[Y\mid T=1]>E[Y\mid T=0]$?
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:
$=\frac{1}{p(1-p)}(E[YZ] - E[Z]E[Y])$
$=\frac{1}{p(1-p)}Cov(Y,Z)$
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:
$=E[Y \mid T = 1]>E[Y \mid T = 0]$