Q:

Expansion of cumulant transform

Verify the following expansion for a cumulant generating function of a random variable $X$.
  \begin{align}
\kappa(t) & = \mu t + \frac{1}{2}\sigma^2t^2+\frac{1}{6}\rho_3\sigma^3t^3 + \frac{1}{24}\rho_4\sigma^4t^4 + \cdots \\
& = \mu t + \frac{1}{2}\sigma^2t^2+\frac{1}{6}\kappa_3 t^3 + \frac{1}{24}\kappa_4 t^4 + \cdots, \tag{$*$}
\end{align}
  where $\mu=\mathbb EX$ and $\rho_r=\frac{\kappa_r}{\sigma^r}$ is the standardized cumulant.

By applying Taylor expansion twice, I got:
\begin{align}
\kappa(t)  & := \log\mathbb Ee^{tX} = \log\mathbb E\left( 1+tX+\frac{t^2X^2}{2} + \frac{t^3X^3}{3!} + \frac{t^4X^4}{4!} + \cdots\right) \\
& = \log \left( 1+t\mathbb EX + \frac{t^2\mathbb EX^2}{2} + \frac{t^3\mathbb EX^3}{3!} + \frac{t^4\mathbb EX^4}{4!} + \cdots \right) \\
& = t\mathbb EX + \frac{t^2\mathbb EX^2}{2} + \frac{t^3\mathbb EX^3}{3!} + \frac{t^4\mathbb EX^4}{4!} + \cdots \\
& =\mu t+\frac{t^2\mathbb EX^2}{2} + \frac{t^3\mathbb EX^3}{6} + \frac{t^4\mathbb EX^4}{24} + \cdots \tag{$**$}
\end{align}
It is clear that $(*)$ and $(**)$ are not the same since cumulant and moment are not equal. Where did I get wrong, please? Thank you!

A:

The Taylor expansion of $\log(1+x)$ is $x-x^2/2+x^3/3-\cdots$. Therefore up to second order we get
$$
\newcommand{\EE}{\mathbb{E}}
\newcommand{\VV}{\mathbb{V}}
\log\left(1+t\EE[X]+\frac{t^2}{2}\EE[X^2]+O(t^3)\right) \\ =
\left(t\EE[X]+\frac{t^2}{2}\EE[X^2]+O(t^3)\right)-\frac{1}{2}\left(t\EE[X]+O(t^2)\right)^2 \\ =
t\EE[x] + \frac{t^2}{2}\EE[X^2]-\frac{t^2}{2}\EE[X]^2 + O(t^3) \\ =
t\EE[x] + \frac{t^2}{2}\VV[X] + O(t^3).
$$

