Suppose I have a continuous random variable $Y$ with $\mu=E[Y]$ and $g(Y)$ is strictly convex and increasing in $Y$. Does it follow that $\frac{\partial}{\partial\mu}Cov(Y,g(Y))>0$?
To me, it makes intuitive sense, but I can't prove it mathematically.
Here's my reasoning. Since $g(Y)$ has a positive slope (i.e., it is increasing), $Cov(Y, g(Y))$ is positive, and a larger slope for $g$ implies a larger covariance. Moreover, since $g(Y)$ is strictly convex, then $\frac{\partial}{\partial\mu}Cov(Y,g(Y))$ should also be positive because the slope of $g$ is larger when $Y$ is larger (and $Y$ should be larger when $\mu$ is larger).
Any ideas on how to proceed? Alternatively, a counterexample where the above is not true would also be greatly appreciated.
I've an answer to this question (thanks to the discussion with Leander).
In general, it does not follow that $\frac{\partial}{\partial \mu}Cov(Y, g(Y))$ if $g$ is increasing and convex.
Here is a counterexample. Let $Y=\frac{Z}{\mu}+ \mu$, and $Z$ follows standard normal distribution. Moreover, let $g(y)=y^2$ which is convex and increasing (when $\mu$ is relatively large). Then:
$Cov(Y, g(Y))=Cov(\frac{Z}{\mu}+ \mu, \frac{Z^2}{\mu^2} + 2Z + \mu^2)$ $=\frac{1}{\mu^3} Cov(Z, Z^2) + \frac{2}{\mu} Cov(Z,Z)$ $= \frac{2}{\mu}$
Then:
$\frac{\partial}{\partial \mu}Cov(Y, g(Y))=-2\mu^{-2}$
Which is negative.
That said, the statement is true when $Y$ can be decomposed into a location parameter and a random variable that is independent of the location parameter.
Let $Y=\mu + X$, where $X$ is a random variable that does not depend on $\mu$ and for which $E[X]=0$. Then:
$Cov(Y, g(Y))=Cov(X, g(X+\mu))$
And we have that:
$\frac{\partial}{\partial \mu}Cov(X, g(X+\mu))$
$=\frac{\partial}{\partial \mu}E[Xg(X+\mu)]-E[X]\frac{\partial}{\partial \mu}E[g(X+\mu)]$
$=E[Xg'(X+\mu)]-E[X]E[g'(X+\mu)]$
$=Cov(X,g'(X+\mu))$
$=Cov(Y,g'(Y))$
Which is positive because: $g$ is strictly convex, so $g'$ is increasing in $Y$, and the covariance of a random variable and an increasing function is positive.