# basic application of Strong Law of Large Numbers

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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There is no dependence on i in $\bar{X}$ so pull it out of the sum. The $\frac{1}{n}$ and the sum cancel (in the second case, or associate with the $X_i^j$ terms in the first case) so you are just looking at the limit as n goes to infinity of $-\bar{X}^{q-j}$. Discarding a set of measure zero, you can assume that $\bar{X}$ converges to $\mathbb{E}$X everywhere at which point that just becomes the statement that if $f$ is continuous and $x_n \to x$ then $f(x_n)\to f(x)$ –  Chris Janjigian Feb 4 '12 at 18:21
@PeeJay It's not clear to me whether $\bar{X}$ in the sum refers to $\bar{X}_n$ or $\bar{X}_i$. –  Ben Derrett Feb 4 '12 at 18:47
@Chris so in the last line of your comment, is that not using Slutsky s Theorem ? ( because if it is, then I could only conclude that convergence in probability follows right ? ) –  Beltrame Feb 4 '12 at 18:58
@BenDerrett the $\bar{X}$ is supposed to refer to the sample mean, i.e. it s referring to n. –  Beltrame Feb 4 '12 at 19:00
@PeeJay That is correct, you do not need Slutsky's theorem for that part. Chris's comment above explains it very well. –  Byron Schmuland Feb 4 '12 at 20:49