Let $X$ be any random variable with any distribution. Given that we know $\mathbb{E}[X]$, $\mathbb{E}[\max(0,X)]$, and $\mathbb{E}[\min(0,X)]$, can you write a formula for $\mathbb{E}[f(X)]$ where $f$ is any twice differentiable function?
I am guessing the solution might have something to do with the arbitrage free pricing of $f(X)$ given price of $X$, price of call option on $X$, and price of put option on $X$, but I am not an expert in pricing.
Any ideas?
I don't think one can answer your question. Suppose $X=e^{\mu+\sigma Z}$ is log-normal, i.e. positive. Thus, $\mathbb{E}[\max\{0,X\}]=\mathbb{E}[X] $ and $\mathbb{E}[\min\{0,X\}]=0$. From just knowing $\mathbb{E}[X]$, you cannot conclude what $\mathbb{E}[f(X)]$ is for an arbitrary function $f$. An implication of your statement is that the mean fully characterises the distribution of positive random variables, which is not true in general.
It would work for distributions that only have one parameter (are characterised by their mean) like the Poisson distribution or exponential distribution. Depending on your function, you might get some information about $\mathbb{E}[f(X)]$ from Jensen's inequality or a first-order Taylor expansion.
Famously, Carr and Madan derived a static replication formula, see @Gordon's answer here:
\begin{align*} f(x) &= f(a) + f'(a) (x-a) + \int_a^{\infty}(x-u)^+f''(u)\text du + \int_{0}^a(u - x)^+f''(u)\text du. \end{align*} This allows you to calculate expectations of the form $\mathbb{E}[f(X)]$ - but you do need a continuum of option prices (at all positive strikes). This finding echoes the result from Breeden-Litzenberger (1978): A complete list of option prices fully characterises the distribution of the underlying asset at maturity. However, you need to know much more than just the mean of the positive and negative part of $X$.
