# Probabilistic Taylor theorem for concave functions

This paper proves a probabilistic version of Taylor's theorem

$$\begin{equation*} \mathbb{E}g(X) = \sum_{k=0}^{n-1} \frac{g^{(k)}(0)}{k!} \mathbb{E}X^k + \frac{\mathbb{E}X^n}{n!} \mathbb{E} g^{(n)}(X_{(n)}), \end{equation*}$$ where $$X_{(n)}$$ is another random variable derived from $$X$$ and $$g^{(n)}$$ is the $$n$$-th derivative of $$g$$. Suppose we take $$n=4$$ and we know that $$g^{(4)} < 0$$ (e.g. due to concavity), then it seems to follow that $$\begin{equation*} \mathbb{E}g(X) = \sum_{k=0}^{3} \frac{g^{(k)}(0)}{k!} \mathbb{E}X^k + \frac{\mathbb{E}X^4}{4!} \mathbb{E} g^{(4)}(X_{(4)}) < \sum_{k=0}^{3} \frac{g^{(k)}(0)}{k!} \mathbb{E}X^k . \end{equation*}$$ Hence we can get explicit upper bound on $$\mathbb{E}g(X)$$. However, a crucial assumption in the probabilistic Taylor theorem is that $$X$$ is non-negative. This means we cannot apply the result to a centered random variable like $$X-\mu$$, where $$\mu = \mathbb{E}X$$. Suppose again that $$g^{(4)} < 0$$. Are there any results that would allow me to conclude something like the following? $$\begin{equation*} \mathbb{E}g(X-\mu) = \sum_{k=0}^{3} \frac{g^{(k)}(0)}{k!} \mathbb{E}(X-\mu)^k + \frac{\mathbb{E}(X-\mu)^4}{4!} \mathbb{E} g^{(4)}(Y) < \sum_{k=0}^{3} \frac{g^{(k)}(0)}{k!} \mathbb{E}(X-\mu)^k, \end{equation*}$$ where $$Y$$ would be another random variable linked to $$X-\mu$$.

If $$g^{(4)}\le0$$, then $$g(x)=\sum_{k=0}^3\frac{g^{(k)}(0)}{k!}\,x^k+\frac{x^4}4\, \int_0^1g^{(4)}(sx)(1-s)^3\,ds \le\sum_{k=0}^3\frac{g^{(k)}(0)}{k!}\,x^k$$ for real $$x$$.
Replacing here $$x$$ by $$X-\mu$$ and assuming that $$E|X|^3<\infty$$, we get $$Eg(X-\mu)\le\sum_{k=0}^3\frac{g^{(k)}(0)}{k!}\,E(X-\mu)^k,\tag{1}\label{1}$$ as desired.
(The strict inequality $$<$$ in \eqref{1} will not hold in general, even if $$g^{(4)}<0$$. In particular, the strict inequality in \eqref{1} will not hold if $$P(X=\mu)=1$$.)