Is covariance preserved under monotone functions?

Question

Suppose that you have to random variables $X$ and $Y$ such that the covariance $C[X,Y] > 0$. Let $f$ be an increasing function. Can you then conclude that $C[X,f(Y)] > 0$ as well? Is the converse true if $f$ is decreasing?

Answer 1

Let's use the formula:

$$\mbox{cov}(X,Y) = E[XY] - E[X]E[Y].$$

What can happen when replacing $X$ with $f(X)$? We can make $E[f(X)]$ very big without effecting $E[f(X)Y]$ very much. Let $(X,Y)$ have joint distribution:

$$p((X,Y) = (1,0)) = 0.05$$ $$p((X,Y) = (0,0.1)) = 0.05$$ $$p((X,Y) = (0.5,0.5)) = 0.9$$

Then $E[XY] = 0.9 \times (0.5 \times 0.5) = 0.225$, $E[X] = 0.05 + 0.5\times 0.9 =0.5$, $E[Y] = 0.1 \times 0.05+0.9\times 0.5 = 0.445$. Then the formula above gives covariance $0.0025 > 0$.

Now, let's have $f$ s.t. $f(0) = 0$, $f(0.5) = 0.5$ and $f(1) = 100$.

$$p((f(X),Y) = (100,0)) = 0.05$$ $$p((f(X),Y) = (0,0.1)) = 0.05$$ $$p((f(X),Y) = (0.5,0.5)) = 0.9$$

Note that $E[f(X)Y]$ and $E[Y]$ haven't changed. But now $E[f(X)] = 100\times0.05 + 0.5\times0.9 = 5.45$. So the formula gives covariance $-2.20025$.