Proof. For the mean, apply linearity of expectation once per added summand:
\[ \begin{aligned} \operatorname{E}\mathopen{}\left[S_n\right]\mathclose{} &= \sum_{i=1}^n\operatorname{E}\mathopen{}\left[X_i\right]\mathclose{} && \text{(linearity of expectation, applied } n - 1 \text{ times)} \\ &= n\mu && \text{(each } X_i \text{ has mean } \mu \text{)} \end{aligned} \]
For the variance, apply the variance of a linear combination with every \(a_i = 1\). For \(i \ne j\), \(X_i\) and \(X_j\) are independent (take \(A_k = \mathbb{R}\) for every other \(k\) in the definition of independence), so \(\operatorname{Cov}\mathopen{}\left(X_i, X_j\right)\mathclose{} = 0\) for independent summands; and \(\operatorname{Cov}\mathopen{}\left(X_i, X_i\right)\mathclose{} = \operatorname{Var}\mathopen{}\left(X_i\right)\mathclose{}\) (covariance of a variable with itself):
\[ \begin{aligned} \operatorname{Var}\mathopen{}\left(S_n\right)\mathclose{} &= \sum_{i=1}^n\sum_{j=1}^n \operatorname{Cov}\mathopen{}\left(X_i, X_j\right)\mathclose{} && \text{(variance of a linear combination, all } a_i = 1 \text{)} \\ &= \sum_{i=1}^n\operatorname{Cov}\mathopen{}\left(X_i, X_i\right)\mathclose{} + \sum_{i \ne j} \operatorname{Cov}\mathopen{}\left(X_i, X_j\right)\mathclose{} && \text{(split off the terms with } i = j \text{)} \\ &= \sum_{i=1}^n\operatorname{Var}\mathopen{}\left(X_i\right)\mathclose{} + 0 && \text{(covariance with itself; independent summands)} \\ &= n\sigma^2 && \text{(each } X_i \text{ has variance } \sigma^2\text{)} \end{aligned} \]
Example 1 (The sum of five dice) A single fair die roll has the discrete uniform distribution on \(\mathopen{}\left\{1, \ldots, 6\right\}\mathclose{}\), which is flat, not bell-shaped (Figure 1). Its mean is \(\mu = 3.5\), and its variance is \(\sigma^2= \sum_{x=1}^{6} (x - 3.5)^2 / 6 = 35/12\).
dice_sum_pmf <- function(n_dice) {
totals <- rowSums(expand.grid(rep(list(1:6), n_dice)))
probs <- prop.table(table(totals))
data.frame(total = as.numeric(names(probs)), p = as.vector(probs))
}
dice_plot <-
ggplot2::ggplot(mapping = ggplot2::aes(x = total, y = p)) +
ggplot2::xlab("sum of dice (x)") +
ggplot2::ylab("Probability of outcome, Pr(X=x)") +
ggplot2::expand_limits(y = 0)
dice_plot + ggplot2::geom_col(data = dice_sum_pmf(1))
The sum of five independent rolls, \(S_5\), is already close to bell-shaped (Figure 2).
pmf_five_dice <- dice_sum_pmf(5)
dice_plot + ggplot2::geom_col(data = pmf_five_dice)
For example, the exact probability that five dice total at most 15 is \(\Pr(S_5 \le 15) = 0.3052\). The normal approximation that Theorem 1 suggests, with mean \(5 \cdot 3.5 = 17.5\) and variance \(5 \cdot 35/12 \approx 14.58\), evaluated at \(15.5\) to account for \(S_5\) taking only integer values, gives \(\Phi\mathopen{}\left((15.5 - 17.5)/\sqrt{14.58}\right)\mathclose{} \approx 0.3002\).