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.)

  • $\begingroup$ is this question and answer related to my recent question stats.stackexchange.com/questions/560316/… ? I have tried using cumulants but was unable to replicate the results! $\endgroup$ Jan 17 at 22:11
  • $\begingroup$ @hydrologist: Yes, I wrote this question to have a baseline to answer your question. $\endgroup$
    – Ben
    Jan 18 at 2:12

1 Answer 1


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:

$$\begin{align} \mathbb{E}(H_n) &= \bar{\kappa}_1 \\[6pt] &= \sum_{i=1}^n c_i \cdot \kappa_1 \\[6pt] &= \sum_{i=1}^n c_i \cdot \mu \\[6pt] &= \mu \sum_{i=1}^n c_i \\[6pt] &= \mu S_{n,1}, \\[12pt] \mathbb{V}(H_n) &= \bar{\kappa}_2 \\[6pt] &= \sum_{i=1}^n c_i^2 \cdot \kappa_2 \\[6pt] &= \sum_{i=1}^n c_i^2 \cdot \sigma^2 \\[6pt] &= \sigma^2 \sum_{i=1}^n c_i^2 \\[6pt] &= \sigma^2 S_{n,2}, \\[12pt] \mathbb{Skew}(H_n) &= \frac{\bar{\kappa}_3}{\bar{\kappa}_2^{3/2}} \\[6pt] &= \frac{\sum_{i=1}^n c_i^3 \cdot \kappa_3}{(\sum_{i=1}^n c_i^2 \cdot \kappa_2)^{3/2}} \\[6pt] &= \frac{\sum_{i=1}^n c_i^3 \cdot \gamma \cdot \sigma^3}{(\sum_{i=1}^n c_i^2 \cdot \sigma^2)^{3/2}}, \\[6pt] &= \frac{\gamma \sum_{i=1}^n c_i^3}{(\sum_{i=1}^n c_i^2)^{3/2}} \\[6pt] &= \gamma \cdot \frac{S_{n,3}}{S_{n,2}^{3/2}}, \\[6pt] \mathbb{Kurt}(H_n) &= \frac{\bar{\kappa}_4 + 3 \bar{\kappa}_2^2}{\bar{\kappa}_2^2} \\[6pt] &= \frac{\sum_{i=1}^n c_i^4 \cdot \kappa_4 + 3 (\sum_{i=1}^n c_i^2 \cdot \kappa_2)^2}{(\sum_{i=1}^n c_i^2 \cdot \kappa_2)^2} \\[6pt] &= \frac{\sum_{i=1}^n c_i^4 \cdot (\kappa-3) \sigma^4 + 3 (\sum_{i=1}^n c_i^2 \cdot \sigma^2)^2}{(\sum_{i=1}^n c_i^2 \cdot \sigma^2)^2} \\[6pt] &= \frac{(\kappa-3) \sum_{i=1}^n c_i^4 + 3 (\sum_{i=1}^n c_i^2)^2}{(\sum_{i=1}^n c_i^2)^2}. \\[6pt] &= \frac{(\kappa-3) S_{n,4} + 3 S_{n,2}^2}{S_{n,2}^2} \\[6pt] &= 3 + (\kappa-3) \frac{S_{n,4}}{S_{n,2}^2} \\[6pt] \end{align}$$

  • $\begingroup$ This is one of the cases where the excess kurtosis has a simpler formula. $\endgroup$ Jan 18 at 9:05
  • $\begingroup$ Yes, that's correct. $\endgroup$
    – Ben
    Jan 18 at 10:40

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