limit of expectation of a random variable.

Question

Is $\lim_{n \to \infty}(E(\bar{z}_n)-\mu)^2$ always the same as $(\lim_{n \to \infty}E(\bar{z}_n)-\mu)^2$? Where $\mu$ is the mean and $\bar{z}_n$ is the sample mean.

Thanks advance!

Answer 1

Assuming the limit exists, such that $$\lim_{n \to \infty}E(\bar{z}_n) = L,$$ then by linearity of limits, we have $$\begin{align}\lim_{n \to \infty}\left\{(E(\bar{z}_n) - \mu) ^2\right\} &= \lim_{n \to \infty}\left\{E(\bar{z}_n)^2 - 2\mu E(\bar{z}_n) + \mu^2\right\}\\ &= \left(\lim_{n \to \infty} E(\bar{z}_n)\right)^2 - 2\mu\left(\lim_{n \to \infty} E(\bar{z}_n)\right) + \mu^2\\ &= L^2 - 2\mu L + \mu^2 \\&=(L-\mu)^2 \\ &= \left(\lim_{n \to \infty}E(\bar{z}_n) -\mu\right)^2\end{align}$$