Proof of Expectation and Variance of Sample Mean

E (\overset{x}{ˉ}) Var (\overset{x}{ˉ}) = E (X) = \frac{Var ( X )}{N}

For the first equation (Expectation of the sample mean):

E (\overset{x}{ˉ}) = E (\frac{1}{N} j = 1 \sum N x_{j}) = \frac{1}{N} j = 1 \sum N E (x_{j}) = \frac{1}{N} j = 1 \sum N E (X) = \frac{1}{N} \cdot N E (X) = E (X) (Definition of sample mean) (Linearity of expectation) (Identically distributed: E (x_{j}) = E (X)) (Summing N identical terms) (Simplification)

For the second equation (Variance of the sample mean):

Var (\overset{x}{ˉ}) = Var (\frac{1}{N} j = 1 \sum N x_{j}) = \frac{1}{N ^{2}} j = 1 \sum N Var (x_{j}) = \frac{1}{N ^{2}} j = 1 \sum N Var (X) = \frac{1}{N ^{2}} \cdot N Var (X) = \frac{Var ( X )}{N} (Definition of sample mean) (Independence + Var (a X) = a^{2} Var (X)) (Identically distributed: Var (x_{j}) = Var (X)) (Summing N identical terms) (Simplification)

Key Notes:

Expectation Proof:
- The second equality uses the linearity of expectation (no independence required).
- The third equality relies on the identical distribution of $x_{j}$ (i.e., $E (x_{j}) = E (X)$ ).
Variance Proof:
- The second equality requires independence to decompose $Var (\sum x_{j})$ into $\sum Var (x_{j})$ .
- The third equality uses identical distribution (i.e., $Var (x_{j}) = Var (X)$ ).

Optional: Derivation of Variance of Sample Mean

To prove the following equation:

Var (\frac{1}{N} j = 1 \sum N x_{j}) = \frac{1}{N ^{2}} j = 1 \sum N Var (x_{j})

We note above that we use $Independence$ and $Var (a X) = a^{2} Var (X)$ , which are explained below:

Let $x_{1}, x_{2}, \dots, x_{N}$ be independent random variables.
The variance of their sum is: $Var (j = 1 \sum N x_{j}) = j = 1 \sum N Var (x_{j}) + i \neq = j \sum Cov (x_{i}, x_{j}) .$
Since $x_{1}, x_{2}, \dots, x_{N}$ are independent, $Cov (x_{i}, x_{j}) = 0$ for all $i \neq = j$ .
Therefore: $Var (j = 1 \sum N x_{j}) = j = 1 \sum N Var (x_{j}) .$

Let $a$ be a constant and $X$ be a random variable.
The variance of $a X$ is: $Var (a X) = E [(a X - E [a X])^{2}] .$
Since $E [a X] = a E [X]$ , we have: $Var (a X) = E [(a X - a E [X])^{2}] = E [a^{2} (X - E [X])^{2}] = a^{2} E [(X - E [X])^{2}] = a^{2} Var (X) .$

In our case, the sample mean $\overset{x}{ˉ}$ is defined as: $\overset{x}{ˉ} = \frac{1}{N} j = 1 \sum N x_{j} .$
Applying the scaling property ( $a = \frac{1}{N}$ ) and the variance of a sum property (due to independence), we get: $Var (\overset{x}{ˉ}) = Var (\frac{1}{N} j = 1 \sum N x_{j}) = (\frac{1}{N})^{2} Var (j = 1 \sum N x_{j}) .$
Substituting the variance of the sum: $Var (\overset{x}{ˉ}) = \frac{1}{N ^{2}} j = 1 \sum N Var (x_{j}) .$