Does $\operatorname{Cov}(X,Y) = 0$ mean $\operatorname{Cov}(X,\log Y) = 0$?

Question

Suppose $X,Y$ are positive random variables with $\operatorname{Cov}(X,Y)=0$. Define $Z= \log Y$. Does it necessarily follow that $\operatorname{Cov}(X,Z) = 0$? I know it's true for linear transformations of $Y$, but now I have something more complicated. Conceptually I feel like it should be true, however there could be a counterexample I haven't thought of.

Answer 1

$\newcommand{\Cov}{\operatorname{Cov}} \newcommand{\E}{\mathbb{E}}$

The answer to you question is NO, in general.

Let $g(Y)$ be a function of $Y$. Then, $$\Cov[X,g(Y)] = \E[(X-\mu_X)(Y-\mu_{g(Y)})] %= \E[Xg(Y)-\mu_{g(Y)}X-\mu_Xg(Y)+\mu_X\mu_{g(Y)}] = \E[Xg(Y)]-\mu_X\mu_{g(Y)}.$$

Having $\Cov[X,g(Y)]=0$ only means $$\E[Xg(Y)]=\mu_X\mu_{g(Y)},$$ which does not hold in general (and your specific $g(Y)=\operatorname{log}{Y}$).

Remark: If $X$ and $g(Y)$ are independent, then $\Cov[X,g(Y)]=0$. This is definitely correct when $X$ and $Y$ are independent.

Answer 2

A counterexample requires that $X$ and $Y$ not be independent. Let $(X,Y)$ be uniform on the three points $(1,a)$, $(3,b)$, and $(4,c)$. Then $E(X)=8/3$, $E(Y)=(a+b+c)/3$, and $E(XY)=(a+3b+4c)/3$. To have $\mathrm{Cov}(X,Y)=0$ means $E(XY)=E(X)E(Y)$, which means $$3(a+3b+4c)=8a+8b+8c\quad\Longleftrightarrow\quad b+4c=5a\;.$$ You can find positive $(a,b,c)$ to satisfy the above, but then (in general) $(\log a, \log b, \log c)$ won't satisfy the same equation, which would be necessary if $\mathrm{Cov}(X,\log Y)=0$, since $(X,\log Y)$ is uniform on the three points $(1,\log a)$, $(3,\log b)$, $(4, \log c)$.

Example: $a=6$, $b=10$, $c=5$ wil do. But $\log b + 4\log c-5\log a=\log(bc^4/a^5)$ is not zero.

Answer 3

Let's see if we can make an example really simple. Let $$ Y = \begin{cases} 1 \\ 2 \\ 3 \end{cases} $$ each with probability $1/3$, and let $X=(Y-2)^2+1$. Then $\operatorname{cov}(X,Y)=0$ and $\operatorname{cov}(X,\log Y)\ne0$. The difference happens because the simple arithmetic progression $1,2,3$ is replaced by $\log1=0, \log2, \log3$, which is not an arithmetic progression: the amount you add to the first term to get the second is not the same as the amount you add to the second to get the third.