If my random variables $X_1,...,X_n$ are i.i.d. $N(\mu,\sigma^2)$, why isn't $\bar{X}\sim N(\mu,0)$?

Question

If my random variables $X_1,...,X_n$ are i.i.d. $N(\mu,\sigma^2)$, why isn't $\bar{X}\sim N(\mu,0)$?

In other words, if, as I understand it, $X_1,...,X_n$ all have the same mean, $\mu$, how can there be any variance at all in $\bar{X}$?

Answer 1

You draw a sample consisting of $n$ observations, you can compute a sample mean.

You draw another sample consisting of another $n$ observations, you can compute another sample mean.

We do not expect the first sample mean to be equal to the second sample mean, in fact, it is unlikely that either of them would be equal to $\mu$.

Each sample mean is random and not deterministic. It depends on the sample drawn.

Answer 2

Let's look at a simple example in which $n=3.$ And instead of a normal distribution we'll say $$ X_i = \begin{cases} 1 \\ {} & \text{each with probability }1/2. \\ 0 \end{cases} $$ Then $\mu= \operatorname E(X_i) = 1/2$ and $\sigma= \sqrt{\operatorname{var}(X_i)} = 1/2.$

As in the question, we let $\overline X =\dfrac{X_1+X_2+X_3} 3.$

What is $\operatorname{var}(\,\overline X\,)$?

We have $$ (X_1,X_2,X_3) = \begin{cases} (0,0,0) & \text{so that } \overline X = 0 \\[6pt] (0,0,1) & \text{so that } \overline X = 1/3 \\ (0,1,0) & \text{so that } \overline X = 1/3 & \\ (1,0,0) & \text{so that } \overline X = 1/3 \\[6pt] (0,1,1) & \text{so that } \overline X = 2/3 \\ (1,0,1) & \text{so that } \overline X = 2/3 \\ (1,1,0) & \text{so that } \overline X = 2/3 \\[6pt] (1,1,1) & \text{so that } \overline X = 1 \end{cases} \text{ each with probability }1/8. $$ What then is the variance of $\overline X$?