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 $X-\mu$.