Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


$$ \sum_{j=0}^q {q\choose j}{1\over n}\sum_{i=1}^n X_i^j(-\bar X)^{q-j} \quad \overrightarrow{a.s.} \quad \sum_{j=0}^q {q\choose j} \mathbb{E}(X^j) (-\mathbb{E}(X))^{q-j} $$

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 ?

share|cite|improve this question
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
up vote 0 down vote accepted

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 !

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.