Notation question for variance I'm a little confused what the difference between these equations are:
$$\begin{array}{rcl} \operatorname{Var}(\bar{Y}) & = & \frac{\sigma^2} n = \frac 1 n \sum (x_i - \mu_X)^2 \\ s_x^2 & = & \frac 1 {n-1} \sum (x_i - \mu_X)^2 \end{array}$$
I was also under the impression that $s_x^2$ was the same as $\operatorname{Var}(\bar{Y})$, but that seems wrong since the equations are different - one has a degrees of freedom adjustment, and one deosn't. Any help would be appreciated!
 A: You've got lots of little details wrong.
\begin{align}
\mu & = \operatorname{E}(X_i) \text{ for } i = 1,\ldots,n \\[10pt]
\bar X & = \frac{X_1+\cdots+X_n} n \\[10pt]
\sigma^2 & = \operatorname{var}(X_i) = \operatorname{E}\Big( (X_i-\mu)^2\Big) \text{ for } i = 1,\ldots,n \\[10pt]
S^2 & = \frac 1 {n-1}\left( (X_1-\bar X)^2 + \cdots +(X_n-\bar X)^2 \right) \\[10pt]
T^2 & = \frac 1 n \left( (X_1-\mu)^2 + \cdots + (X_n-\mu)^2 \right) & \longleftarrow & \text{This one is not standard} \\
& & & \text{notation but I am adopting} \\
& & & \text{it for use below.} \\[10pt]
& \text{From the above, it follows that} \\[10pt]
& \operatorname{E}(S^2) = \sigma^2 \\[10pt]
& \operatorname{E}(T^2) = \sigma^2
\end{align}
It is the use of $\bar X$ rather than $\mu$ in the definition of $S^2$ that makes it necessary to divide by $n-1$ rather than by $n$, in order to make the expected value $\sigma^2$, whereas in the definition of $T^2$ I divided by $n$.
However, the alleged need to make the expected value of an estimator of $\sigma^2$ equal to $\sigma^2$ is at best exaggerated. The mean squared estimation error $\operatorname{E}\Big((T^2-\sigma^2)^2\Big)$ of the biased estimator is actually (slightly) smaller that the mean squared estimation error $\operatorname{E}\Big( (S^2-\sigma^2)^2\Big)$ of the unbiased estimator $S^2$.

Also, you should be careful about distinguishing between $x$ and $X$.
A: $\operatorname{Var}(\bar Y)$ is the population variance and the $s^2_x$ is the sample variance. we use $n-1$ for sample variance, but $n$ for population variance ($\sigma^2$), because the sample variance is the unbiased estimator of the population variance, meaning that $\operatorname{E}(s^2_x) = \sigma^2$ where $\sigma^2$ is the population variance. 
