Variance of sample mean of n rolls of fair 6-sided die

Question

I'm trying to work on variance of a simple variable x which is defined in the work attached herewith. But the variance comes out to be a negative number. Could someone point out the mistake I'm making?

Answer 1

Because the $X_i$ are independent, the variance is also a linear operator: $$\operatorname{Var}[\bar X] = \operatorname{Var}\left[\frac{1}{n} \sum_{i=1}^n X_i \right] = \frac{1}{n^2} \operatorname{Var}\left[\sum_{i=1}^n X_i\right] \overset{\text{ind}}{=} \frac{1}{n} \sum_{i=1}^n \operatorname{Var}[X_i],$$ where the "ind" symbol above the equality means that we used independence. Because the $X_i$ are identically distributed, then each $X_i$ has the same variance, thus $$\operatorname{Var}[\bar X] = \frac{1}{n} \operatorname{Var}[X_1] = \frac{35}{12n}.$$

Your mistake in your calculation is where you split up the terms in the square of the sum, but forget that the double sum should be multiplied by $2$: $$\left(\sum_{i=1}^n X_i\right)^2 = \sum_{i=1}^n X_i \sum_{j=1}^n X_j = \sum_{i=1}^n \sum_{j=1}^n X_i X_j.$$ This admits the decomposition $$\sum_{k=1}^n X_k^2 + \sum_{i \ne j} X_i X_j,$$ but since $X_i X_j = X_j X_i$, we write this as $$\sum_{k=1}^n X_k^2 + 2 \sum_{i=1}^{n-1} \sum_{j=i+1}^n X_i X_j.$$ For instance, with $n = 3$, we have $$(x_1 + x_2 + x_3)^2 = x_1^2 + x_2^2 + x_3^2 + 2(x_1 x_2 + x_2 x_3 + x_3 x_1).$$

Answer 2

Let $S_n := X_1 + \ldots + X_n$, where $X_i$ is the outcome of the $i$th roll. Then by independence, one computes $$ \mbox{Var}(S_n/n) = (1/n^2)\mbox{Var}(S_n) = (1/n^2)\sum_{i=1}^n\mbox{Var}(X_i) = (1/n^2)\cdot n \cdot \mbox{Var}(X_1) = (1/n)\mbox{Var}(X_1). $$

Noting that $P(X_1 = k) = 1/6$ for all $k \in \{1,\ldots,6\}$, and using the law of total variance with $A_k := \{\omega \in \Omega \mid X_1(\omega) = k\}$, you can compute $\mbox{Var}(X_1)$ exactly (exercise!). Let this value be $v$.

Therefore, $\mbox{Var}(S_n/n) = v/n$.