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

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.

• I edited your title so that it summarises your question, instead of just being the first line of it. Feel free to revert if you think it's worse this way. – Ben Millwood Jun 10 '15 at 22:35
• I think your edit is better. Thank you :) – Brenton Jun 10 '15 at 22:36
• If $X$ and $Y$ are independent random variables, then $\iint (X-\bar X)(Y-\bar Y)\,\rho(x,y)dxdy=0$ and $\iint (X-\bar X)(\log Y-\bar \log Y)\,\rho(x,y)dxdy=\int (X-\bar X)\rho_X(x)dx\int (\log Y-\bar \log Y)\rho_Y(x)dy=0$ also. If $X$ and $Y$ are not independent, then one cannot conclude generally that $\text{Cov}(X,Y)=0\implies \text{Cov}(X,\log Y)=0$. – Mark Viola Jun 10 '15 at 22:50
• My answer isn't any good. Will try again. – Clarinetist Jun 10 '15 at 23:21

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.

• Why does $\log Y$ have uniform distribution? – Sina Jun 11 '15 at 0:04
• @Sina $\log Y$ is uniform on the three points $\log a$, $\log b$, $\log c$, since $Y$ is uniform on the three points $a$, $b$, $c$. – grand_chat Jun 11 '15 at 0:06
• (Never mind, my previous comment was wrong.) – Clarinetist Jun 11 '15 at 0:35
• @Clarinetist True. But if the pair $(X,Y)$ has a discrete uniform distribution, then the answer is yes, in general (barring duplicate values for $XY$). In the above example $XY$ takes three values: $a$, $3b$, $4c$ each with prob $1/3$. – grand_chat Jun 11 '15 at 0:38
• Great counterexample. +1 – Clarinetist Jun 11 '15 at 0:40

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

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.

• My question would be, then, do you have an example of uncorrelated but dependent positive random variables? – Clarinetist Jun 10 '15 at 23:35
• I do not actually, because I realized in my work, I made a small mistake: $Y$ is positive but $X$ does not have to be (in fact, $X$ is a normal r.v.). I believe I might be able to show that $Cov(X, logY) = 0$, but was hoping that would tell me something about $Cov(X,Y)$ – Brenton Jun 10 '15 at 23:37
• You need a counter example to validate your conclusion. – Mark Viola Jun 11 '15 at 0:02
• I couldn't come up with a counterexample. This book might be helpful: Counterexamples in probability and statistics Author: Joseph P Romano; Andrew F Siegel – Bob Jun 11 '15 at 0:28

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.