Expectation of random variable multiplied by non-negative function of itself

Question

Suppose $\mathbb{E}[X] \neq 0$. Let $f(X) > 0$. Does it always hold that $$ \mathbb{E}[Xf(X)] \neq 0 ? $$

Answer 1

No. To construct a counterexample, let $p < 1/2$ and set $X \in \{1,-1\}$ with $\mathbb{P}[X=1] = p$ then $E[X] = p - (1-p) = 2p-1 < 0$ and now let $f(-1) = 1$ and $f(1) = a$ so $$\mathbb{E}[Xf(X)] = ap - (1-p) = (a+1)p - 1$$ so let $a$ solve $\mathbb{E}[Xf(X)] = 0$.

Clearly there is a unique $a>1$ that solves the problem...