I am trying to show for $X_n$ iid st. $E|X|^q < \infty$ that
$ \frac{1}{n} \sum^n (X_i - \bar{X})^q \to E((X-EX)^q) \quad \text{in probability} $
We note:
- $ \frac{1}{n} \sum^n X_i \to EX \quad \text{in P by WLLN}$ i.e. $ \bar{X}_n \to EX \quad \text{in P} $
- $ Y_i := (X_i - EX)^q $ are iid
i.e. we use Slutsky on $\frac{1}{n} \sum^n (X_i - \bar{X})^q $ to obtain iid random variables to then use WLLN as per below:
$ \frac{1}{n} \sum^n (X_i - \bar{X})^q \quad \overrightarrow{Slutsky} \quad \frac{1}{n} \sum^n (X_i - EX)^q \quad\overrightarrow{WLLN} \quad E(X-EX)^q $
I am wondering whether my above analysis is correct ? (if not, where do I go wrong ?)