[proofplan]
We compute the mean of the sample average by finite linearity of expectation and the identical distribution of the variables. For the variance, we center each [random variable](/page/Random%20Variable) by subtracting $\mu$, expand the square of the centered sum, and separate diagonal terms from off-diagonal terms. Independence makes every off-diagonal product have expectation zero, while identical distribution makes every diagonal term equal to $\sigma^2$.
[/proofplan]
[step:Compute the expectation of the sample average]
Since each $Y_i$ has the same distribution as $Y_1$, we have $\mathbb{E}[Y_i]=\mu$ for every $i \in \{1,\dots,n\}$. The random variable $\hat{\mu}_n$ is square-integrable because it is a finite linear combination of square-integrable random variables, hence integrable. By finite linearity of expectation,
\begin{align*}
\mathbb{E}[\hat{\mu}_n]=\mathbb{E}\left[\frac{1}{n}\sum_{i=1}^{n}Y_i\right].
\end{align*}
Therefore
\begin{align*}
\mathbb{E}[\hat{\mu}_n]=\frac{1}{n}\sum_{i=1}^{n}\mathbb{E}[Y_i].
\end{align*}
Using $\mathbb{E}[Y_i]=\mu$ for each $i$, this becomes
\begin{align*}
\mathbb{E}[\hat{\mu}_n]=\frac{1}{n}\sum_{i=1}^{n}\mu=\mu.
\end{align*}
[/step]
[step:Center the variables and rewrite the variance]
For each $i \in \{1,\dots,n\}$, define the centered random variable $X_i: \Omega \to \mathbb{R}$ by
\begin{align*}
X_i(\omega)=Y_i(\omega)-\mu.
\end{align*}
Then $X_i$ is square-integrable, because $Y_i$ is square-integrable and $\mu$ is finite. Also,
\begin{align*}
\mathbb{E}[X_i]=\mathbb{E}[Y_i-\mu]=\mathbb{E}[Y_i]-\mu=0.
\end{align*}
Since $\mathbb{E}[\hat{\mu}_n]=\mu$, the definition of variance gives
\begin{align*}
\operatorname{Var}(\hat{\mu}_n)=\mathbb{E}\left[(\hat{\mu}_n-\mu)^2\right].
\end{align*}
Using the definition of $\hat{\mu}_n$ and the centered variables,
\begin{align*}
\hat{\mu}_n-\mu=\frac{1}{n}\sum_{i=1}^{n}(Y_i-\mu)=\frac{1}{n}\sum_{i=1}^{n}X_i.
\end{align*}
Hence
\begin{align*}
\operatorname{Var}(\hat{\mu}_n)=\frac{1}{n^2}\mathbb{E}\left[\left(\sum_{i=1}^{n}X_i\right)^2\right].
\end{align*}
[guided]
The variance is easiest to compute after subtracting the common mean. For each $i \in \{1,\dots,n\}$, define the centered random variable $X_i: \Omega \to \mathbb{R}$ by
\begin{align*}
X_i(\omega)=Y_i(\omega)-\mu.
\end{align*}
This centering has two useful effects. First, $X_i$ is still square-integrible, since it differs from the square-integrible random variable $Y_i$ by the constant $\mu$. Second, its expectation is zero:
\begin{align*}
\mathbb{E}[X_i]=\mathbb{E}[Y_i-\mu]=\mathbb{E}[Y_i]-\mu=0.
\end{align*}
Because the first step proved $\mathbb{E}[\hat{\mu}_n]=\mu$, the variance of $\hat{\mu}_n$ is
\begin{align*}
\operatorname{Var}(\hat{\mu}_n)=\mathbb{E}\left[(\hat{\mu}_n-\mu)^2\right].
\end{align*}
Now substitute the definition of the average:
\begin{align*}
\hat{\mu}_n-\mu=\frac{1}{n}\sum_{i=1}^{n}Y_i-\mu.
\end{align*}
Since $\mu=\frac{1}{n}\sum_{i=1}^{n}\mu$, this becomes
\begin{align*}
\hat{\mu}_n-\mu=\frac{1}{n}\sum_{i=1}^{n}(Y_i-\mu)=\frac{1}{n}\sum_{i=1}^{n}X_i.
\end{align*}
Therefore the constant factor $1/n$ is squared in the variance computation:
\begin{align*}
\operatorname{Var}(\hat{\mu}_n)=\frac{1}{n^2}\mathbb{E}\left[\left(\sum_{i=1}^{n}X_i\right)^2\right].
\end{align*}
[/guided]
[/step]
[step:Expand the square and separate diagonal from off-diagonal terms]
Because the sum is finite and each $X_i$ is square-integrible, all products $X_iX_j$ are integrable by Cauchy-Schwarz. Expanding the square pointwise gives
\begin{align*}
\left(\sum_{i=1}^{n}X_i\right)^2=\sum_{i=1}^{n}\sum_{j=1}^{n}X_iX_j.
\end{align*}
Taking expectations and using finite linearity,
\begin{align*}
\mathbb{E}\left[\left(\sum_{i=1}^{n}X_i\right)^2\right]=\sum_{i=1}^{n}\sum_{j=1}^{n}\mathbb{E}[X_iX_j].
\end{align*}
For the diagonal terms, $\mathbb{E}[X_i^2]=\operatorname{Var}(Y_i)=\sigma^2$, since $Y_i$ has the same distribution as $Y_1$. For $i \neq j$, independence of $Y_i$ and $Y_j$ implies independence of $X_i=Y_i-\mu$ and $X_j=Y_j-\mu$, so
\begin{align*}
\mathbb{E}[X_iX_j]=\mathbb{E}[X_i]\mathbb{E}[X_j]=0.
\end{align*}
Thus only the $n$ diagonal terms remain:
\begin{align*}
\mathbb{E}\left[\left(\sum_{i=1}^{n}X_i\right)^2\right]=\sum_{i=1}^{n}\sigma^2=n\sigma^2.
\end{align*}
[/step]
[step:Insert the expanded second moment into the variance formula]
Substituting the preceding identity into the variance formula gives
\begin{align*}
\operatorname{Var}(\hat{\mu}_n)=\frac{1}{n^2}n\sigma^2.
\end{align*}
Therefore
\begin{align*}
\operatorname{Var}(\hat{\mu}_n)=\frac{\sigma^2}{n}.
\end{align*}
Together with $\mathbb{E}[\hat{\mu}_n]=\mu$, this proves both asserted identities.
[/step]
