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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.

Carlos Fernandez
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  • Just a few remarks:

    I do believe that some additional assumptions are necessary in order to secure the existence of $\frac{\partial }{\partial \mu} Cov(Y,T)$.

    Also, if $f(y)$ is increasing and strictly convex, then we have $f(y) \rightarrow \infty$ as $y\rightarrow \infty$, which is problematic if $f(y)$ is supposed to be a probability.

     – Leander Tilsted Kristensen May 27 '20 at 08:59
  • Thanks for your help @LeanderTilstedKristensen. What additional assumptions do you have in mind? That may help me find a counterexample.

    Regarding your other comment about $g$, I have removed $T$ from the original problem formulation, so that $g$ is no longer a probability.

     – Carlos Fernandez May 27 '20 at 11:21
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    First of all in order to write $Cov(Y,g(Y))$ as a function of $\mu$, then one would have to assume, that $Y$ belongs to a family of probability distributions, which is uniquely determined by $\mu$. For instance the poisson distribution is uniquely determined by the mean, but the normal distribution is not (there is also a variance parameter). Secondly, usually such derivatives "tricks" such as $$\frac{\partial}{\partial \mu} \mathbb{E}[h(Y)] = \frac{\partial}{\partial \mu} \int_{-\infty}^\infty h(y) f(y | \mu) : dy = \int_{-\infty}^\infty h(y) \frac{\partial}{\partial \mu}f(y|\mu) : dy$$ – Leander Tilsted Kristensen May 27 '20 at 13:07
  • I have a preliminary answer that I believe doesn't need to assume any specific probability distribution (though I'm not 100% sure the answer is correct). – Carlos Fernandez May 27 '20 at 14:28

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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.

Carlos Fernandez
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    The decomposition $Y=\mu + X$ can of course always be made, simply choose $X=Y- \mu$, but the problem is that $X$ may depend on $\mu$. Consider the following: If $Y\sim N(\mu , 1)$, then $Y-\mu \sim N(0,1)$ and your method would work, but if we instead look at $Y\sim N(\mu , \mu^2)$ then $Y-\mu \sim N(0,\mu^2)$, and thus $Y-\mu$ is not independent of $\mu$. – Leander Tilsted Kristensen May 27 '20 at 15:05
  • When that is said, then i do think your argument holds, but only when $Y-\mu$ does not depend on $\mu$. Such a parameter is called a location parameter by the way https://en.wikipedia.org/wiki/Location_parameter . – Leander Tilsted Kristensen May 27 '20 at 15:15
  • In general i advice against writing derivatives with respect to random variables. All derivatives should simply be with respect to $\mu$. – Leander Tilsted Kristensen May 27 '20 at 15:19
  • I see your point. Thanks for the reference as well. That said, even if there is dependence, the argument will hold as long as $\frac{\partial Y}{\partial \mu}$ is positive, and I'm having a hard time picturing a case where this would not be the case. – Carlos Fernandez May 27 '20 at 15:21
  • No i do not believe it generalizes. Also, usually $\frac{\partial Y}{\partial \mu}$ doesn't have any meaning. The influence of $\mu$ on $Y$ is only on the distribution of $Y$ usually not on $Y$ itself. Also expression such as $\frac{\partial }{\partial Y}Cov(Y,g(Y))$ make no sense. $Cov(Y,g(Y))$ is not a function of $Y$ (at least not in the usual sense). – Leander Tilsted Kristensen May 27 '20 at 15:45
  • Thanks a lot for taking the time to answer me. I guess I'll think about how to reformulate the argument above by assuming that no other parameters depend on the location parameter. – Carlos Fernandez May 27 '20 at 15:58
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    @LeanderTilstedKristensen thanks a lot of your help! The discussion helped me to find a counterexample and describe a specific situation in which the statement is true. – Carlos Fernandez May 27 '20 at 21:25