What are the first four moments of a linear function of IID random variables? Consider a sequence of IID random variables $X_1,X_2,X_3,...$ from a common distribution with mean $\mu$, variance $\sigma^2$, skewness $\gamma$ and kurtosis $\kappa$ (all finite).  Given a sequence of constants $c_1,c_2,c_3,...$ we can define the linear function:
$$H_n \equiv \sum_{i=1}^n c_i X_i.$$
What are the corresponding expressions for the mean, variance, skewness and kurtosis of $H_n$?  (Note that the present question extends a related question here.)
 A: To facilitate this analysis, define the sums $S_{n,r} \equiv \sum_{i=1}^n c_i^r$.  Using these quantities the mean, variance, skewness and kurtosis of the quantity $H_n$ can be written as shown in the box below.  These formulae are valid for any case where the underlying values are IID with finite kurtosis.
$$\begin{align}
\boxed{
\quad \quad \quad \mathbb{E}(H_n) = \mu S_{n,1}
\quad \quad \quad \quad \quad \quad \quad \ \
\mathbb{V}(H_n) = \sigma^2 S_{n,2}, \\[18pt]
\quad \mathbb{Skew}(H_n) = \gamma \cdot \frac{S_{n,3}}{S_{n,2}^{3/2}}
\quad \quad \quad \quad \quad 
\quad \mathbb{Kurt}(H_n) = 3 + (\kappa-3) \frac{S_{n,4}}{S_{n,2}^2}. \quad \\}
\end{align}$$
These reults are simplest to derive via the cumulant function of the random variable of interest.  To do this, observe that the random variable $H_n$ has moment generating function:
$$\begin{align}
m_{H_n}(t)
\equiv \mathbb{E}(e^{t H_n})
= \prod_{i=1}^n \mathbb{E}(e^{t c_i X_i})
= \prod_{i=1}^n m_{X}(t c_i),
\end{align}$$
which gives the cumulant function:
$$\begin{align}
K_{H_n}(t)
= \log m_{H_n}(t)
= \sum_{i=1}^n \log m_{X}(t c_i)
= \sum_{i=1}^n K_{X}(t c_i).
\end{align}$$
Now, let $\kappa_r$ denote the $r$th cumulant of the underlying random variables $X_i$.  The cumulants of $H_n$ are related to these cumulants by:
$$\begin{align}
\bar{\kappa}_n
\equiv \frac{d^r K_{H_n}}{dt^r}(t) \Bigg|_{t=0}
= \sum_{i=1}^n c_i^r \cdot \frac{d^r K_{X}}{dt^r}(t c_i) \Bigg|_{t=0}
= \sum_{i=1}^n c_i^r \cdot \kappa_r.
\end{align}$$
Using the relationship of the cumulants to the moments of interest, we then have:
$$\begin{align}
\mathbb{E}(H_n)
&= \bar{\kappa}_1 \\[6pt]
&= \sum_{i=1}^n c_i \cdot \kappa_1 \\[6pt]
&= \sum_{i=1}^n c_i \cdot \mu \\[6pt]
&= \mu \sum_{i=1}^n c_i  \\[6pt]
&= \mu S_{n,1},  \\[12pt]
\mathbb{V}(H_n)
&= \bar{\kappa}_2 \\[6pt]
&= \sum_{i=1}^n c_i^2 \cdot \kappa_2 \\[6pt]
&= \sum_{i=1}^n c_i^2 \cdot \sigma^2 \\[6pt]
&= \sigma^2 \sum_{i=1}^n c_i^2  \\[6pt]
&= \sigma^2 S_{n,2},  \\[12pt]
\mathbb{Skew}(H_n)
&= \frac{\bar{\kappa}_3}{\bar{\kappa}_2^{3/2}} \\[6pt]
&= \frac{\sum_{i=1}^n c_i^3 \cdot \kappa_3}{(\sum_{i=1}^n c_i^2 \cdot \kappa_2)^{3/2}} \\[6pt]
&= \frac{\sum_{i=1}^n c_i^3 \cdot \gamma \cdot \sigma^3}{(\sum_{i=1}^n c_i^2 \cdot \sigma^2)^{3/2}}, \\[6pt]
&= \frac{\gamma \sum_{i=1}^n c_i^3}{(\sum_{i=1}^n c_i^2)^{3/2}} \\[6pt]
&= \gamma \cdot \frac{S_{n,3}}{S_{n,2}^{3/2}}, \\[6pt]
\mathbb{Kurt}(H_n)
&= \frac{\bar{\kappa}_4 + 3 \bar{\kappa}_2^2}{\bar{\kappa}_2^2} \\[6pt]
&= \frac{\sum_{i=1}^n c_i^4 \cdot \kappa_4 + 3 (\sum_{i=1}^n c_i^2 \cdot \kappa_2)^2}{(\sum_{i=1}^n c_i^2 \cdot \kappa_2)^2} \\[6pt]
&= \frac{\sum_{i=1}^n c_i^4 \cdot (\kappa-3) \sigma^4 + 3 (\sum_{i=1}^n c_i^2 \cdot \sigma^2)^2}{(\sum_{i=1}^n c_i^2 \cdot \sigma^2)^2} \\[6pt]
&= \frac{(\kappa-3) \sum_{i=1}^n c_i^4 + 3 (\sum_{i=1}^n c_i^2)^2}{(\sum_{i=1}^n c_i^2)^2}. \\[6pt]
&= \frac{(\kappa-3) S_{n,4} + 3 S_{n,2}^2}{S_{n,2}^2} \\[6pt]
&= 3 + (\kappa-3) \frac{S_{n,4}}{S_{n,2}^2} \\[6pt]
\end{align}$$
