Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

My question is about the usage of Taylor expansions when dealing with asymptotics in local polynomial fitting. The expressions that set me of are of the type: $$ g(X_i) = \sum_{j=0}^{p} \frac{g^{(j)}(x_0)}{j!}(X_i-x_0)^j + o_p((X_i-x_0)^p), $$ where $x_0$ is fixed and $X_i$ is a random design point. I have two questions concerning these type of expressions. First, the $o_p$ symbol stands (as far as I understand) for convergence in probability to 0, but there is no sequence to converge. There in fact is only 1 random variable $X_i$ ($i$ is a fixed index, in a sample of e.g. size $n$). So my question is how to make sense out of this? My second question concerns the order of the remainder term. The above fits to the real analysis intuition where we have $$ f(x) = \sum_{j=0}^{p} \frac{f^{(j)}(x_0)}{j!}(x-x_0)^j + o(|x-x_0|^p), $$ for $x\to x_0$. However, e.g. in Brockwell, Davis: Time Series: Theory and Methods (1987), chapter 6 - asymptotic theory, where ``Taylor Expansion in Probability'' is treated (p. 194/195) we have $$ g(X_n) = \sum_{j=0}^{p-1} \frac{g^{(j)}(x_0)}{j!}(X_n-x_0)^j + o_p(r_n^p), $$ with $\{X_n\}$ a sequence of rv's such that $X_n = x_0 + O_p(r_n)$, and $0<r_n\to 0$ as $n\to\infty$. Thus the sum contains only $p$ terms, instead of $p+1$ as above. Nevertheless the $o_p$ symbol here makes sense as an actual sequence is involved. So do I include $p$ or $p+1$ terms in the sum above? I appreciate any comments - if my points are not clear I can try to clarify$\ldots$

share|cite|improve this question

migrated from Aug 14 '12 at 10:44

This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.

Your Answer


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