In
$ \sum_{j=0}^q {q\choose j}{1\over n}\sum_{i=1}^n X_i^j(-\bar X)^{q-j} \quad \overrightarrow{a.s.} \quad \sum_{j=0}^q {q\choose j} \mathbb{E}(X^j) (-\mathbb{E}(X))^{q-j} $
using the Strong Law, why is it, that we can say that
$ \frac{1}{n}\sum_{i=1}^n(-\bar X)^{q-j} \quad\overrightarrow{a.s.} \quad (\mathbb{-E}X)^{q-j} $
The reason I am wondering is, that Slutsky's Theorem would only give me that convergence in probability is preserved under the continuous function $f(x) = x^{q-j}$ and since $\bar{X}$ are not iid it seems that using the Strong law on the complete term would not work ?