Positive correlation of random with itself under monotone transformation

Question

Suppose $f:\mathbb R^+\to \mathbb R^+$ is a monotone increasing function and $X\ge 0$. Now clearly $\operatorname{Cov}(X,f(X))\ge 0$ but except $X$ is constant can there be equality? I really appreciate if someone could provide an example.

Answer 1

No, the equality holds iff $X$ is almost surely constant.

Here is a proof. Let $Y$ be independent from $X$ but have the same distribution. Since $f$ is increasing, we have $(Y-X)(f(Y)-f(X))\ge0$. By taking the expected value (provided that it makes sense, i.e. the random variables are sufficiently integrable), we get $$ \begin{align*} \mathbb E[(Y-X)(f(Y)-f(X))]&=\mathbb E[Yf(Y)-Xf(Y)-Yf(X)+Xf(X)]\\ &=\mathbb E[Xf(X)]+\mathbb E[Yf(Y)]-\mathbb E[Xf(Y)]-\mathbb E[Yf(X)]\\ &\ge0. \end{align*} $$

Using the fact that $\mathbb E[Xf(X)]=\mathbb E[Yf(Y)]$ (since $X$ and $Y$ have the same distribution) and $\mathbb E[Xf(Y)]=\mathbb E[X]\mathbb E[f(Y)]=\mathbb E[Yf(X)]$ (by independence), we deduce that $$ \mathbb E[(Y-X)(f(Y)-f(X))]=2\operatorname{cov}(X,f(X))\ge0, $$ with equality iff $(Y-X)(f(Y)-f(X))=0$ almost surely. The later can be rewritten as almost surely: $Y=X$ or $f(Y)=f(X)$. But since $f$ is increasing (I suppose strictly increasing) and therefore one-to-one, $f(Y)=f(X)\implies Y=X$. So if the covariance is zero then $Y=X$ almost surely, which implies that $$ \mathbb E[X^2]=\mathbb E[XY]=\mathbb E[X]\mathbb E[Y]=\mathbb E[X]^2, $$ and therefore $\operatorname{Var}(X)=0$, or equivalently, $X$ is almost surely constant.