Let $y_i$ be iid for all i. Suppose $E(y_t) = \mu$ and $var(y_t) = \sigma^2<\infty$ . Consider the estimator $\hat{y}=\sum_{t=1}^{n} \frac{y_t}{n}$.
I want to directly show $\hat{y}$ converges to $\mu$ in quadratic mean (i.e. $\hat{y} \rightarrow_{q.m.} \mu$)
By definition, I need to prove $E[(\hat{y} \ - \ \mu)^2] = 0$ as $n \rightarrow 0$.
I expanded the thing and got $E[\hat{y}^2] - 2 \mu^2 + \mu$ and then I'm confused about how to deal with $E[\hat{y}^2]$.
Thank you for the help!
Since $E(\hat{y})=\mu$, we have $$E[(\hat{y} \ - \ \mu)^2]= Var(\hat{y})=\frac{\sigma^2}n$$ (this is where independence is needed), which goes to $0$ as $n$ grows.