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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 ?)

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    Byron provides a nice, simple answer to your question for the only reasonable choices of $q$ that do not require inserting absolute values. In addition to accepting his answer, I'd encourage you to upvote it as well. Cheers. :)2012-01-31

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If $q$ is a non-negative integer, then your strategy works. Using a combination of the weak law of large numbers and Slutsky's Theorem we get $\begin{eqnarray*} {1\over n}\sum_{i=1}^n(X_i-\bar X)^q &=&{1\over n}\sum_{i=1}^n \sum_{j=0}^q {q\choose j} X_i^j(-\bar X)^{q-j}\\ &=&\sum_{j=0}^q {q\choose j}{1\over n}\sum_{i=1}^n X_i^j(-\bar X)^{q-j}\\ &\to&\sum_{j=0}^q {q\choose j} \mathbb{E}(X^j) (-\mathbb{E}(X))^{q-j}\\ &=&\mathbb{E}\left[(X-\mathbb{E}(X))^q\right] \\ \end{eqnarray*}$ The convergence is in probability, or even almost surely (by the strong law of large numbers).

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    @PeeJay Since $\bar X\to \mathbb{E}(X)$, then $f(\bar X)\to f(\mathbb{E}(X))$, almost surely, for any continuous function $f$. It has nothing to do with Slutsky, or even probability for that matter.2012-02-04