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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 ?

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    @ByronSchmuland : ok, that makes it clear. Tks !2012-02-04

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Ok, so the bottom line is that using the strong law we do not need Slutsky or any other result like it, because convergence a.s. is preserved under continuous transformations.

This is actually pretty easy to show just using the sequential definition of continuity and applying it to the random variable. Convergence a.e. of the composition follows immediately an no further Results from measure theoretic probability are required.

Hopefully this is an adequate summary, if not then comments are most welcome !