This [paper](https://reader.elsevier.com/reader/sd/pii/0167715294901104?token=A61379C3631ADCAC256F64615D513FEE297A7F515D6873C962AA17DFBF8B8992D4BBB4DE58A9018CA4D09009E587C0E9&originRegion=us-east-1&originCreation=20220914163053) 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 $X-\mu$.



Probabilistic Taylor theorem for concave functions
If $g^{(4)}\le0$, [then](https://en.wikipedia.org/wiki/Taylor%27s_theorem#Explicit_formulas_for_the_remainder)
$$g(x)=\sum\_{k=0}^3\frac{g^{(k)}(0)}{k!}\,x^k+\frac{x^4}4\,
\int\_0^1g^{(4)}(sx)(1-s)^3\,ds
\le\sum\_{k=0}^3\frac{g^{(k)}(0)}{k!}\,x^k$$
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 $P(X=\mu)=1$.)


