Probabilistic Taylor theorem for concave functions

Question

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 $X-\mu$.

Answer 1

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.

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(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 $P(X=\mu)=1$.)