How to calculate Expectation of variance

Question

Assume we have an estimator $\bar{\theta}$ for a parameter $\theta$. The bias of the estimator is the difference between the true value of the estimator, and its expected value:

$$\operatorname{Bias}(\bar{\theta}) = \operatorname{E}(\bar{\theta} - \theta)$$

If $\operatorname{Bias}(\bar{\theta})= 0$, then $\bar{\theta}$ is called unbiased. The variance of the estimator $\bar{\theta}$ is the expected square deviation from its expected value:

$$\operatorname{var}(\bar{\theta}) = \operatorname{E}[(\bar{\theta} − E[\bar{\theta} ])^2]$$

The mean squared error of the estimator $\bar{\theta}$ is

$$\operatorname{Error}(\bar{\theta}) = \operatorname{E}(\bar{\theta} − \theta)^2 = \operatorname{Bias}(\bar{\theta})^2+\operatorname{var}(\bar{\theta})$$

Let $X_1, \ldots , X_N$ be a sample of i.i.d random variables. Assume that $X_i$ has mean $\mu$ and variance $\sigma^2$ . Calculate the bias, variance and mean squared error of the following mean estimators:

(a) $\bar{\mu} =\frac 1 N \sum_{i=1}^N X_i $(i.e. the sample mean)

Bias I calculated as Below

\begin{align} & \operatorname{E}(\bar{\mu})-\mu \\[10pt] = {} & \operatorname{E} \left( \frac 1 N \sum_{i=1}^N X_i \right) - \mu \\[10pt] = {} & \frac 1 N \sum_{i=1}^N \operatorname{E}(X_i) - \mu \\[10pt] = {} & \frac N N \mu - \mu \\[10pt] \operatorname{Bias}(\bar{\mu}) = {} & 0 \end{align}

But which identity i need to use for calculating the variance??

Answer 1

\begin{align} & \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) \\[8pt] = {} & \frac 1 {N^2} \sum_{i=1}^N \operatorname{var}(X_i) = \cdots \end{align}