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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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1  
That shouldn't work. Once you apply Slutsky the n has been taken to infinity so it should not be in the second expression. –  guy Jan 30 '12 at 21:11
    
thanks, I guess you are right ... :( ! –  Beltrame Jan 30 '12 at 23:40
    
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. :) –  cardinal Jan 31 '12 at 19:45

1 Answer 1

up vote 2 down vote accepted

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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(+1) Clean and elegant. I believe we should be able to get a similar result for nonintegral $q > 1$ and $|X_i - \bar X|^q$, instead, but it would require a different argument. –  cardinal Jan 31 '12 at 4:31
    
@cardinal Thanks for the kind words. I was also thinking how to handle averages of the type ${1\over n}\sum_{i=1}^n f(X_i,\bar X)$, without success. –  Byron Schmuland Jan 31 '12 at 14:58
    
Apologies for not having thought this through carefully, but...one can write $f(X_i,\bar X) = f(X_i,\bar X) - f(X_i,\mu) + f(X_i,\mu)$ Then, the last term converges by the SLLN as long as $f$ is "nice" and so we only need to control a term like $|n^{-1} \sum_i (f(X_i,\bar X)-f(X_i,\mu))|$. For the case of $f(x,y)=|x-y|^q$, I think you can split on $\{|\bar X| \leq \epsilon\}$ and its complement. The expectation on the complement should go to zero by dominated convergence and so only the expectation of the first term is left. I suspect that can be controlled by some standard inequalities. –  cardinal Jan 31 '12 at 15:32
    
I forgot to mention that in the specific example I mentioned, we can assume wlog that the mean is zero. In your more general formulation, that's not true, of course. –  cardinal Jan 31 '12 at 15:43
2  
@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. –  Byron Schmuland Feb 4 '12 at 20:22

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